Play with
Graphs
With Sessionwise Theory & Exercises
Play with
Graphs
With Sessionwise Theory & Exercises
Amit M. Agarwal
ARIHANT PRAKASHAN (Series), MEERUT
ARIHANT PRAKASHAN (Series), MEERUT
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PREFACE
It is a matter of great pleasure and pride for me to introduce to you this
book “Play with Graphs”. As a teacher, guiding the Engineering aspirants
for over a decade now, I have always been in the lookout for right
approach to understand various mathematical problems. I had always felt
the need of a book that can develop and sharpen the ideas of the
students within a very short span of time.
The book in your hands, aims to help you solve various mathematical
problems by the use of graphs. Ways to draw different types of graphs are
very easy and can be understood by even an average student. I feel glad
to mention that the use of graphs in solving various mathematical
problems has been well illustrated in this book.
I would like to take this opportunity to thank
M/s Arihant Prakashan for assigning this work to me.
It is their inspiration that has encouraged me to bring this book in this
present form.
I would also like to thank Arihant DTP Unit for the nice laser typesetting.
Valuable suggestions from the students and teachers are always
welcome, and these will find due places in the ensuing editions.
Amit M. Agarwal
CONTENTS
1.
INTRODUCTION TO GRAPHS
1-50
1.1 Algebraic functions
1. Polynomial function
2. Rational function
3. Irrational function
4. Piecewise functions
1.2 Transcendental functions
1. Trigonometric function
2. Exponential function
3. Logarithmic function
4. Geometrical curves
1.3 Trigonometric inequalities
1.4 Solving equations graphically
2.
CURVATURE AND TRANSFORMATIONS
51-137
2.1 Curvature
2.2 Concavity, convexity and points of inflexion
2.3 Plotting of algebraic curves using concavity
2.4 Graphical transformations
2.5 Sketching h(x)= maximum {f(x), g(x)} and h(x)= minimum {f (x), g(x)}
2.6 When f(x), g(x) — f(x) + g(x) = h(x)
2.7 When f(x), g(x) — f(x). g(x) = h(x)
3.
ASYMPTOTES, SINGULAR POINTS
AND CURVE TRACING
138-165
3.1 Asymptotes
3.2 Singular points
3.3 Remember for tracing cartesian equation
HINTS & SOLUTIONS
166-182
INTRODUCTION OF
GRAPHS
1
➥ In this section, we shall revise some basic curves which are given as.
Polynomial
Rational
Modulus
Algebraic
Irrational
Introduction of Graphs
er
t
ap
h
C
Signum
Piecewise
Greatest integer function
Fractional part function
FUNCTIONS
Least integer function
Trigonometric
Exponential
Logarithmic/Inverse of exponential
Transcendental
Geometrical curves
Inverse trigonometric curves
1.1
ALGEBRAIC FUNCTIONS
1. Polynomial Function
A function of the form:
f ( x ) = a 0 + a1 x + a 2 x 2 + … + a n x n ;
where n ∈ N and
a 0, a1 , a 2,…, a n ∈ R.
Then, f is called a polynomial function. “f ( x) is also called polynomial in x”.
1
Play with Graphs
Some of basic polynomial functions are
(i) Identity function/Graph of f (x) = x
A function f defined by f ( x) = x for all x ∈ R, is called the
identity function.
Here, y = x clearly represents a straight line passing through
the origin and inclined at an angle of 45° with x-axis shown as:
The domain and range of identity functions are both
equal to R.
y
y=x
45°
Fig. 1.1
(ii) Graph of f (x) = x 2
A function given by f ( x) = x 2 is called the square function.
The domain of square function is R and its range is R + ∪ { 0}
or [0, ∞)
Clearly y = x 2 , is a parabola. Since y = x 2 is an even
function, so its graph is symmetrical about y-axis, shown as:
y
y=x2
x
O
(iii) Graph of f (x) = x 3
A function given by f ( x) = x 3 is called the cube function.
The domain and range of cube are both equal to R.
Since, y = x 3 is an odd function, so its graph is symmetrical
about opposite quadrant, i.e., “origin”, shown as:
Fig. 1.2
y
y = x3
O
(iv) Graph of f (x) = x 2 n ; n ∈ N
If n ∈ N, then function f given by f ( x) = x 2n is an even function.
So, its graph is always symmetrical about y-axis.
Also, x 2 > x 4 > x 6 > x 8 > … for all x ∈ ( −1, 1)
and x 2 < x 4 < x 6 < x 8 < … for all x ∈ ( −∞, − 1) ∪ (1, ∞ )
Graphs of y = x 2 , y = x 4 , y = x 6 ,…, etc. are shown as:
y=x4 y=x6
y
y = x6
O
1
y = –x
–1
x
O
x
Fig. 1.3
y = x2
y=x
x
Fig. 1.4
(v) Graph of f (x) = x 2 n −1; n ∈ N
If n ∈ N, then the function f given by f ( x) = x 2n −1 is an odd function. So, its graph is
symmetrical about origin or opposite quadrants.
2
x ∈ (1, ∞ )
x < x3 < x5 < …
x ∈ ( 0, 1)
x > x3 > x5 > …
x ∈ ( −1, 0)
x< x < x <…
x ∈ ( −∞, − 1)
x > x3 > x5 > …
3
y=x 5 y=x 3
y=x
y
x
O
–1
5
Graphs of f ( x) = x, f ( x) = x 3 , f ( x) = x 5 ,… are shown as in
Fig. 1.5
1
Fig. 1.5
2. Rational Expression
A function obtained by dividing a polynomial by another polynomial is called a rational function.
P ( x)
⇒
f ( x) =
Q ( x)
Domain ∈ R − { x | Q( x) = 0}
domain ∈R except those points for which denominator = 0.
i.e.,
Graphs of some Simple Rational Functions
1
(i) Graph of f (x) =
x
1
is called the reciprocal
A function defined by f ( x) =
x
function or rectangular hyperbola, with coordinate axis as
1
asymptotes. The domain and range of f ( x) = is R – { 0}.
x
Since, f ( x) is odd function, so its graph is symmetrical
about opposite quadrants. Also, we observe
lim f ( x) = + ∞
x → 0+
lim
and
x → 0–
f ( x) = − ∞
.
Introduction of Graphs
Here, comparison of values of x, x 3 , x 5 ,…
for
y
(1,1)
1
O 1
–1
(–1,–1) –1
x
Fig. 1.6
as x → ± ∞ ⇒ f ( x) → 0
1
Thus, f ( x) = could be shown as in Fig. 1.6.
x
1
(ii) Graph of f (x) = 2
x
1
Here, f ( x) = 2 is an even function, so its graph is symmetrical about y-axis.
x
y
Domain of f ( x) is R − { 0} and range is (0, ∞).
and
Also, as
and
y→∞
as
y→0
as
Thus, f ( x) =
lim
f ( x)
lim
f ( x).
x → 0+
x→ ±∞
or
lim
x → 0−
f ( x) .
1
O
x
Fig. 1.7
could be shown as in Fig. 1.7.
2
x
3
(iii) Graph of f (x) =
Play with Graphs
Here, f ( x) =
1
x
1
x 2n – 1
2n − 1
; n∈N
1
y=
x y =1
y
x3
is an odd function, so its graph is
symmetrical in opposite quadrants.
Also, y → ∞
when
y→−∞
when
lim
f ( x) and
lim
f ( x).
x → 0+
x → 0−
1 (1,1) A
1
1
; f ( x) = 5 , …, etc. will
x3
x
1
be similar to the graph of f ( x) = which has asymptotes
x
as coordinate axes, shown as in Fig. 1.8
1
(iv) Graph of f (x) = 2 n ; n ∈ N
x
1
We observe that the function f ( x) = 2n is an even
x
function, so its graph is symmetrical about y-axis.
Thus, the graph for f ( x) =
Also,
y → ∞ as
y → 0 as
and
lim
x → 0+
lim
x → −∞
f ( x)
or
f ( x) or
lim
x → 0−
lim
x→ + ∞
O
–1
(–1,–1)
–1
B
x
1
Fig. 1.8
1
y y = x2
y = 14
x
f ( x)
(–1,–1)
B 1
A
(1,1)
f ( x).
–1
O
x
1
The values of y decrease as the values of x increase.
1
1
Thus, the graph of f ( x) = 2 ; f ( x) = 4 , … , etc. will be
Fig. 1.9
x
x
1
similar as the graph of f ( x) = 2 , which has asymptotes as coordinate axis. Shown as in Fig. 1.9.
x
3. Irrational Function
The algebraic function containing terms having non-integral rational powers of x are called
irrational functions.
Graphs of Some Simple Irrational Functions
(i) Graph of f (x) = x1/ 2
Here; f ( x) = x is the portion of the parabola y 2 = x, which
lies above x-axis.
Domain of f ( x) ∈ R + ∪ { 0} or [ 0, ∞ )
and
y=x
(1, 1)
y= x
1
range of f ( x) ∈ R + ∪ { 0} or [ 0, ∞ ).
Thus, the graph of f ( x) = x1/2 is shown as;
Note
If f (x) = x n and g (x) = x1/n , then f (x) and g (x) are inverse of
each other.
∴ f (x) = x n and g (x) = x1/n is the mirror image about y = x.
4
y
O
1
Fig. 1.10
x
As discussed above, if g( x) = x 3 . Then f ( x) = x1/3
is image of g( x) about y = x.
1
O
–1
range of f ( x) ∈ R.
and
y=x
y =x1/3
domain f ( x) ∈ R.
where
y=x3
y
x
1
–1
Thus, the graph of f ( x) = x1/ 3 is shown in Fig.
1.11;
Fig. 1.11
(iii) Graph of
f (x) = x1/ 2 n ; n ∈ N
Here, f ( x) = x1/ 2n is defined for all x ∈ [ 0, ∞ ) and the values
taken by f ( x) are positive.
So, domain and range of f ( x) are [ 0, ∞ ).
Here, the graph of f ( x) = x1/ 2n is the mirror image of the
graph of f ( x) = x 2n about the line y = x, when x ∈ [ 0, ∞ ).
Thus, f ( x) = x1/ 2 , f ( x) = x1/ 4 , … are shown as;
(iv) Graph of f (x) = x1/ 2 n −1 , when n ∈ N
Here, f ( x) = x1/ 2n −1 is defined for all x ∈ R. So,
domain of f ( x) ∈ R, and range of f ( x) ∈ R. Also the
graph of f ( x) = x1/ 2n −1 is the mirror image of the graph
of f ( x) = x 2n −1 about the line y = x when x ∈ R.
Thus, f ( x) = x1/ 3 , f ( x) = x1/ 5 , …, are shown
as;
Note
We have discussed some of the simple curves
for Polynomial, Rational and Irrational
functions. Graphs of the some more difficult
rational functions will be discussed in
chapter 3. Such as;
y=
x2 + x + 1
1
x
, y= 2
,… .
y= 2
x+1
x −x+1
x −1
y=x4 y=x2
y=x
y
y = x1/2
y = x1/4
1
O
x
1
Introduction of Graphs
(ii) Graph of f (x) = x1/ 3
Fig. 1.12
y = x5 y=x3
y
y=x
y =x1/3
y = x1/5
1
O
–1
1
x
–1
Fig. 1.13
4. Piecewise Functions
As discussed piecewise functions are:
(a) Absolute value function (or modulus function), (b) Signum function.
(c) Greatest integer function.
(d) Fractional part function.
(e) Least integer function.
(a) Absolute value function (or modulus function)
 x, x ≥ 0
y = |x|= 
− x , x < 0
5
Play with Graphs
y
y = –x
“It is the numerical value of x”.
“It is symmetric about y-axis” where domain ∈R
and
range ∈[0, ∞).
Properties of modulus functions
135°
45°
or
x
O
(i) |x | ≤ a ⇒ − a ≤ x ≤ a ; ( a ≥ 0)
(ii) |x | ≥ a ⇒ x ≤ − a
(iii) |x ± y| ≤ |x| + |y|
y=x
x ≥ a ; ( a ≥ 0)
Fig. 1.14
(iv) |x ± y| ≥ |x| − |y | .
(b) Signum function; y = Sgn(x)
It is defined by;
x
|x|
;
or

y = Sgn( x) =  x
|x|
 0 ;
y-axis
 +1, if
x≠0 
=  −1, if
x = 0.  0, if
1
x> 0
x< 0
x=0
–1
Domain of f ( x) ∈ R.
Range of f ( x) ∈ { −1, 0, 1}.
Here,
and
x-axis
O
Fig. 1.15
(c) Greatest integer function
[x] indicates the integral part of x which is nearest and smaller
integer to x. It is also known as floor of x.
Thus, [ 2.3] = 2, [ 0.23] = 0, [ 2] = 2, [ −8.0725] = − 9, … .
In general;
n ≤ x < n + 1 ( n ∈ Integer) ⇒ [ x] = n.
Here, f ( x) = [ x] could be expressed graphically as;
[x] = n
n
n+1
x
Fig. 1.16
y-axis
3
x
[x]
0≤ x< 1
0
1≤ x< 2
1
2≤ x< 3
2
Thus, f ( x) = [ x] could be shown as;
Properties of greatest integer function
(i) [ x] = x holds, if x is integer.
(ii) [ x + I] = [ x] + I, if I is integer.
(iii) [ x + y ] ≥ [ x] + [ y ].
(iv) If [ φ ( x)] ≥ I, then φ ( x) ≥ I.
(v) If [ φ ( x)] ≤ I, then φ ( x) < I + 1.
(vi) [ − x] = − [ x], if x ∈ integer.
(vii) [ − x] = − [ x] − 1, if x ∉integer.
“It is also known as stepwise function/floor of x.”
6
2
1
–3
–2
–1
O
1
2
–1
–2
–3
Fig. 1.17
3
4 x-axis
x = [ x] + { x} = I + f ; where I = [ x] and f = { x}
y = { x} = x − [ x] , where 0 ≤ { x} < 1; shown as:
∴
y
O
1
3
x–
x+2
2
−2 ≤ x < − 1
–1
x–
x+1
–2
1
−1 ≤ x < 0
–3
x–
x−2
x
2≤ x< 3
1
1
x−1
x+
1≤ x< 2
2
x
x+
0≤ x< 1
3
{x}
x+
x
2
x
3
Fig. 1.18
Properties of fractional part of x
(i) { x} = x ; if
0≤ x< 1
(ii) { x} = 0 ; if x ∈ integer.
(iii) { − x} = 1 − { x} ; if x ∈ integer.
(e) Least integer function
Introduction of Graphs
(d) Fractional part of function
Here, {.} denotes the fractional part of x. Thus, in y = { x}.
y = ( x) = x ,
( x) or x indicates the integral part of x which is nearest and greatest integer to x.
It is known as ceiling of x.
(x) = x = n +1
[x ] = n
Thus, 2.3023 = 3, ( 0.23) = 1, ( −8.0725) = − 8, ( −0.6) = 0
In general,
n < x ≤ n + 1 ( n ∈ integer))
or
i. e.,
( x) = n + 1
x
shown as;
Here, f ( x) = ( x) = x, can be expressed graphically as:
n
n+1
x
Fig. 1.19
y-axis
x
x = ( x)
3
−1 < x ≤ 0
0
2
0< x≤ 1
1
1
1< x≤ 2
2
−2 < x ≤ − 1
−1
−3 < x ≤ − 2
−2
O 1
–4 –3 –2 –1
–1
x-axis
2
3
4
–2
–3
Properties of least integer function
Fig. 1.20
(i) ( x) = x = x, if x is integer.
(ii) ( x + I) = x + I = ( x) + I ; if I ∈ integer.
(iii) Greatest integer converts x = I + f to [ x] = I while x converts to ( I + 1).
Note
We shall discuss the curves:
y = {sin x} , y = {x3 } , y = {sin−1 (sin x)} y = [sin x] , etc. in chapter 2. (Curvature and
Transformations).
7
1.2 TRANSCENDENTAL FUNCTIONS
Play with Graphs
1. Trigonometric Function
(a) Sine function
Here, f ( x) = sin x can be discussed in two ways i.e., Graph diagram and Circle diagram where
Domain of sine function is “R” and range is [–1, 1].
Graph diagram
y
(On x-axis and y-axis)
3
π
π, 1
– ,1
increases
f ( x) = sin x,
2
1
2
y =1
strictly from –1 to 1 as x increases
B
D
π
π
h1
h1
from − to , decreases strictly
2
2
x
2π
π
OA π
π
3π
π
π
3π
2π C
from 1 to –1 as x increases from
2
2
2
2
2
3π
y = –1
to
and so on. We have graph
π , –1 –1
3π, –1
2
2
2
as;
Fig. 1.21
Here, the height is same after
every interval of 2π. (i.e., In above figure, AB = CD after every interval of 2π).
∴ sin x is called periodic function with period 2π.
Circle diagram
π ,1
5π , 1 …
π/2
=
2
2
(On trigonometric plane or using
y=1
γ
quadrants). Let a circle of radius ‘1’,
i.e., unit circle.
β
a
Then,
sin α = ,
b
1
1
1
a
b
α
sin β = ,
O
0=y
1
…, 5π, 3π, π y = 0
3π
x = 0, 2π, 4π ,…
c
2
c
d
sin γ = − ,
1
d
2π–δ
sin δ = − , … , shown as.
1
y = –1
∴ sin x generates a circle of
3π/2, 7π/2, …
radius ‘1’.
(b) Cosine function
Here, f ( x) = cos x
The domain of cosine function is R and the range is [–1, 1].
Graph diagram (on x-axis and y-axis)
As discussed, cos x decreases strictly
(–2π, 1)
from 1 to –1 as x increases from 0 to π,
increases strictly from –1 to 1 as x
increases from π to 2 π and so on. Also,
cos x is periodic with period 2 π.
–π
–2π –3π/2
Fig. 1.22
y
–π/2
O
π/2
(–π, –1)
8
(2π, 1)
(0, 1)
π
(π, –1)
Fig. 1.23
3π/2 2π
x
y=1
Let a circle of radius ‘1’, i.e., a unit circle.
cos α =
Then,
cos γ =
∴
a
b
, cos β = − ,
1
1
1
b
δ
c
d
, cos δ = −
1
1
y=0
a
α
O
γ
1
d
β1
y=0
1
x
c
cos x generates a circle of radius ‘1’.
y = –1
(c) Tangent function
f ( x) = tan x
The domain of the function y = tan x is;
 π
3π
5π 
R − ± , ±
,±
,…
2
2

 2
π


i. e., R − ( 2n + 1) 
2

and Range ∈ R or ( −∞, ∞ ).
The function y = tan x increases strictly
from − ∞ to + ∞ as x increases from
π
π π
3π 3π
5π
to
−
to , to
,
, … and so on.
2
2 2
2
2
2
The graph is shown as :
Note
Fig. 1.24
y
1
–π
x = –3π/2
–
π
4
O π
x
π
Introduction of Graphs
y
Circle diagram
4
–1
x = –π/2
x = π/2
x = 3π/2
Fig. 1.25
Here, the curve tends to meet at x = ±
π
3π
5π
±
,±
, … but never meets or tends to
2
2
2
infinity.
∴
x=±
π
3π
5π
… are asymptotes to y = tan x.
, ±
, ±
2
2
2
(d) Cosecant function
f ( x) = cosec x
y
y = cosec x
y = cosec x
y = cosec x
π ,1
2
1
–2π –3π/2 –π
–
y = cosec x
–π/2
O
π ,–1
2
–1
π/2
y = cosec x
Fig. 1.26
π
3π/2
y = sin x
x
2π
3π ,–1
2
y = cosec x
9
Here, domain of y = cosec x is,
Play with Graphs
R − { 0, ± π, ± 2 π, ± 3 π, …}
and range ∈ R − ( − 1, 1).
R − { nπ| n ∈ z}
i. e.,
as shown in Fig. 1.26.
The function y = cosec x is periodic with period 2 π.
(e) Secant function
f ( x) = sec x


π
domain ∈ R − ( 2n + 1)
n ∈ z
2


Range ∈ R − ( −1 , 1)
Here,
Shown as:
y
y = sec x
(–2π, 1)
y = sec x
(2π, 1 )
(0, 1 )
1
–2π – 3π
2
–π
(–π, –1 )
–
π
2
O
π
2
–1
π
3π
2
x
2π
y = cos x
(π, –1 )
Fig. 1.27
The function y = sec x is periodic with period 2 π .
Note
(i) The curve y = cosec x tends to meet at x = 0 , ± π , ± 2 π , … at infinity.
∴
x = 0, ± π , ± 2π , …
x = nπ , n ∈ integer are asymptote to y = cosec x.
π 3π
(ii) The curve y = sec x tends to meet at x = ± ±
, … at infinity.
2
2
π
3π
5π
π
, ±
, … or x = (2 n + 1) , n ∈ integer are asymptote to y = cosec x.
∴x = ± , ±
2
2
2
2
or
Here, we have used the notation of asymptotes of a curve in the context of special curves,
but we would have a detailed discussion in chapter 3.
(f) Cotangent function
f ( x) = cot x
Here,
domain ∈R − { nπ| n ∈ z} Range ∈ R.
which is periodic with period π, and has x = nπ, n ∈ z as asymptotes. As shown in Fig. 1.28;
10
y
3π
2
–π
–
π
2
O
π
2
3π
2
π
2π
x
asymptotes
–
Introduction of Graphs
–2π
y = cot x
Fig. 1.28
2. Exponential Function
Here, f ( x) = a x, a > 0, a ≠ 1, and x ∈ R, where domain ∈R,
Range ∈ ( 0, ∞ ).
Case I. a > 1
Here, f ( x) = y = a x increase with the increase in x, i.e., f ( x) is increasing function on R.
y
y = a x, a > 1
(0,1)
O
x
Fig. 1.29
4x
3x
2x
y-axis
For example;
y = 2 x , y = 3 x , y = 4 x ,… have;
2x < 3x < 4x < … for x > 1
and
(0,1)
2 > 3 > 4 > … for 0 < x < 1.
x
x
x
x-axis
O
and they can be shown as;
Fig. 1.30
Case II. 0 < a < 1
Here, f ( x) = a x decrease with the increase in x, i.e., f ( x) is
decreasing function on R.
“In general, exponential function increases
decreases as ( a > 1) or ( 0 < a < 1) respectively”.
3. Logarithmic Function
or
y=ax
y-axis
0<a<1
(0,1)
O
x-axis
Fig. 1.31
(Inverse of Exponential)
The function f ( x) = log a x; ( x , a > 0) and a ≠ 1 is a logarithmic function.
Thus, the domain of logarithmic function is all real positive numbers and their range is the set R
of all real numbers.
We have seen that y = a x is strictly increasing when a > 1 and strictly decreasing when 0 < a < 1.
11
Play with Graphs
If 0 < a < 1
y-axis
Thus, the function is invertible. The
inverse of this function is denoted by log a x,
we write
y = a x ⇒ x = log a y ;
where
x ∈ R and y ∈( 0, ∞ )
writing y = log a x in place of x = log a y ,
we have the graph of y = log a x .
Thus, logarithmic function is also
known as inverse of exponential function.
Properties of logarithmic function
If a > 1
y-axis
(1,0)
x-axis
O
(1,0)
1. log e ( ab) = log e a + log e b
{ a, b > 0}
 a
2. log e   = log e a − log e b
 b
{ a, b > 0}
3. log e a m = m log e a
4. log a a = 1
{a > 0
{a > 0
5. log m a =
{ a, b > 0, b ≠ 1 and m ∈ R}
1
log b a
m
1
6. log b a =
log a b
log m a
7. log b a =
log m b
loga m
8. a
=m
9. a logc b = b logc a
and m ∈ R}
and a ≠ 1}
{ a, b > 0
and a, b ≠ 1}
{ a, b > 0 ≠ {1} and m > 0}
 x > y , if
⇒
 x < y , if
which could be graphically shown as;
If m > 1 (Graph of log m a)
10. If log m x > log m y
logm x
m>1
{ a, m > 0 and a ≠ 1}
{ a, b, c > 0 and c ≠ 1}
{ m, x, y , > 0 and m ≠ 1}
0< m < 1
logm a
Again if 0 < m < 1. (Graph of log m a)
logm x
logm y
O
x-axis
Fig. 1.32
b
logm y
1
y
x
O
Fig. 1.33
11. log m a = b ⇒
a=m
b
and
m > 1.
⇒ log m x > log m y ; when x < y and 0 < m < 1.
{ a, m > 0 ; m ≠ 1 ; b ∈ R}
 a > m ; if m > 1
12. log m a > b ⇒ 
b
 a < m ; if 0 < m < 1
b
 a < m b ; if m > 1
.
b
 a > m ; if 0 < m < 1
13. log m a < b ⇒ 
x y1
Fig. 1.34
⇒ log m x > log m y when x > y
12
O
y
(a) Straight line
ax + by + c = 0 (represents general equation of straight line). We
know,
c
when x = 0
y =–
b
c
and
when y = 0
x=−
a
joining above points we get required straight line.
0, – c
b
c
– ,0
a
x
O
Fig. 1.35
(b) Circle
We know,
(i)
x 2 + y 2 = a 2 is circle with centre ( 0, 0) (ii)
and radius r.
( x − a) 2 + ( y − b) 2 = r 2 , circle
centre (a, b) and radius r.
y
with
y
C (0, 0)
(r, 0)
(a, b)
r
C
x
Introduction of Graphs
4. Geometrical Curves
x
Fig. 1.37
Fig. 1.36
(iii)
x 2 + y 2 + 2gx + 2fy + c = 0 ;
centre ( −g, − f ); radius g 2 + f 2 − c .
y
(iv)
( x − x1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) = 0;
End points of diameter are ( x1 , y 1) and
( x 2 , y 2).
A
(x1, y1)
r
C
(–g, –f )
O
B
(x2, y2)
x
O
Fig. 1.39
Fig. 1.38
(c) Parabola
(i) y 2 = 4 ax
Vertex
Focus
Axis
Directrix
(ii) y 2 = – 4 ax
:
:
:
:
(0, 0)
(a, 0)
x-axis or y = 0
x=−a
Vertex
Focus
Axis
Directrix
: (0, 0)
: (– a, 0)
: x-axis or y = 0
: x=a
13
y
y
V
(a, 0) Focus
Vertex
Directrix
Play with Graphs
y 2= – 4ax
F
x
y 2 = 4ax
x=a
x = –a
Fig. 1.41
Fig. 1.40
(iii) x 2 = 4 ay
Vextex
Focus
Axis
Directrix
x
V (0, 0)
(–a, 0)
(iv) x 2 = − 4 ay
( 0, 0)
( 0, a)
y-axis or x = 0
y =−a
:
:
:
:
( 0, 0)
( 0, − a)
y-axis or x = 0
y =a
:
:
:
:
y
x 2 = 4ay
y
F
Vertex
Focus
Axis
Directrix
y=a
(0, a)
V (0, 0)
x
x
V (0, 0)
Directrix
(0, –a)
F
y = –a
x 2 = –4ay
Fig. 1.42
Fig. 1.43
(v) (y − k) 2 = 4a (x − h)
Vertex
Focus
Axis
Directrix
:
:
:
:
( h, k )
( h + a, k )
x=h
x=h−a
y=k
V (h, k) F (h + a, k)
x = h–a x = h
directrix
Fig. 1.44
(d) Ellipse
x2 y2
(i) 2 + 2 = 1 (a 2 > b 2 )
a
b
Centre
: (0, 0)
Focus
: ( ±ae, 0)
Vertex
: ( ±a, 0)
Eccentricity : e = 1 −
Directrix
14
: x=±
a
e
a 2> b 2
b
b2
–a
O
a
–b
a2
x = –a/e
x = a/e
Fig. 1.45
x2 y2
+ 2 =1
a2
b
(a 2 < b 2 )
y
(iii)
(x − h) 2
a2
+
(y − k) 2
b2
Introduction of Graphs
(ii)
= 1 (a 2 > b 2 )
y
b2> a2
y = b/e
(0, b)
(h, k +b)
y=k
O
(–a, 0)
x
(a, 0)
(a+h, k)
A′
(a–h, k)
(h,k)
A
(h, k–b)
x
O
(0,–b)
x=h
y = –b/e
directrix
Fig. 1.47
Fig. 1.46
(e) Hyperbola
x2 y2
(i) 2 − 2 = 1
a
b
Centre
: (0, 0)
Focus
: ( ±ae, 0)
Vertices
: ( ±a, 0)
Eccentricity : e = 1 +
Directrix
: x= ±
y
b
y =– x
a
ym
as
b
(–ae,0) (–a, 0)
2
x
(a, 0) (ae, 0)
O
a2
as
ym
p
a
e
x = –a/e
In above figure asymptotes are y = ±
(ii) −
b
y= x
a
te
pto
x2 y2
+ 2 =1
a2
b
x = a/e
tot
e
Fig. 1.48
b
x.
a
(iii)
(x − h) 2
a2
−
(y − k) 2
b2
=1
y
y
te
to
p
(0, b)
ym
as
(h, k)
O
x
x
ot
pt
ym
as
(0, –b)
y=k
e
x=h
Fig. 1.49
Fig. 1.50
15
(v) xy = c 2
As asymptotes are perpendicular. Therefore,
called rectangular hyperbola.
Here, the asymptotes are x-axis and y-axis.
y
asymptote
Play with Graphs
(iv) x 2 − y 2 = a 2 (Rectangular hyperbola)
y=x
te
to
mp
y
as
O
O
asymptote
x
(–c, –c)
as
ym
pto
te
y = –x
Fig. 1.51
Note
(c, c)
Fig. 1.52
In above curves we have used the name asymptotes for its complete definition see
chapter 3.
Inverse Trigonometric Curves
As we know trigonometric functions are many one in their domain, hence, they are not
invertible.
But their inverse can be obtained by restricting the domain so as to make invertible.
Note
Every inverse trigonometric is been converted to a function by shortening the domain.
Let f ( x) = sin x
We know, sin x is not invertible for x ∈ R.
In order to get the inverse we have to define domain as:
 π π
x ∈ − , 
 2 2
 π π
∴ If f : − ,  → [ −1, 1] defined by f ( x) = sin x is invertible and inverse can be represented
 2 2
For example:
by:
y = sin −1 x.
π
 π
−1
 − ≤ sin x ≤ 
 2
2
Similarly,
16
y = cos x becomes invertible
when
f : [ 0, π] → [ −1, 1]
y = tan x ; becomes invertible
when
 π π
f :  − ,  → ( − ∞, ∞ )
 2 2
y = cot x ; becomes invertible
when
f : ( 0, π ) → ( − ∞, ∞ )
y = sec x; becomes invertible
when
 π
f : [ 0, π] –   → R − ( −1, 1)
 2
y = cosec x; becomes invertible
when
 π π
f : – ,  – { 0} → R – (–1, 1)
 2 2
π/2
 π π
y ∈ − , 
 2 2
and
f (x) = sin x
1
x ∈ [ −1, 1]
(–π, 0)
y=x
y = sin–1x
O (1,0) (π/2,0)
(–π/2,0) (–1,0)
(π, 0)
x
–1
As the graph of f −1 ( x) is mirror image of
f ( x) about y = x.
–π/2
Fig. 1.53
(ii) Graph of y = cos −1 x ;
y = cos–1x
y
π
Here,
π/2
domain ∈ [ −1, 1]
1
π/2
Range ∈ [ 0, π]
–1
O
π
1
x
f (x) = cos x
(iii) Graph of y = tan −1 x ;
Here,
y=x
Fig. 1.54
 π π
Range ∈  − ,  .
 2 2
domain ∈R,
y
Introduction of Graphs
y
(i) Graph of y = sin −1 x ;
where,
y
y = tan x
y = π/2
y = tan–1x
x
O
x
O
y = –π/2
x = π/2
x = –π/2
Fig. 1.55
As we have discussed earlier, “graph of inverse function is image of f ( x) about y = x” or “by
interchanging the coordinate axes”.
(iv) Graph of y = cot −1 x ;
We know that the function f : ( 0, π ) → R, given by f(θ) = cot θ is invertible.
∴ Thus, domain of cot –1 x ∈ R and Range ∈ ( 0, π ).
y
y = cot x
y
(0, π)
O
(π/2,0)
(π,0)
π/2
x
O
y=π
y = cot –1x
x
x=π
Fig. 1.56
17
Play with Graphs
(v) Graph for y = sec −1 x;
 π
The function f : [ 0, π] −   → ( −∞, − 1] ∪ [1, ∞ ) given by f(θ) = sec θ is invertible.
 2
 π
−1
∴ y = sec x, has domain ∈ R − ( −1, 1) and range ∈[ 0, π] −   : shown as
 2
y
y
y=π
1
–1
y = sec x
(0, 1)
π/2
O
y = π/2
x-axis
π
(π,–1)
–1
–1
1
O
x
x = π/2 x = π
y = sec–1 x
y = sec x
Fig. 1.57
(vi) Graph for y = cosec −1x;
As we know,
 π π
f : − ,  − { 0} → R − ( −1, 1) is invertible given by f(θ) = cos θ.
 2 2
y = cosec −1 x ; domain ∈ R − ( −1, 1)
∴
 π π
Range ∈ − ,  − { 0}.
 2 2
y
y
y = π/2
1
x
–1
1
O
x
–1
y = – π/2
x = – π/2
x = π/2
x = –1
y = cosec x
x=1
y = cosec –1x
Fig. 1.58
Note
If no branch of an inverse trigonometric function is mentioned, then it means the principal
value branch of that function.
In case no branch of an inverse trigonometric function is mentioned, it will mean the
principal value branch of that function. (i.e.,)
18
Domain
Range
Principal value branch
π
π
≤ y ≤ , where y = sin −1 x
2
2
1.
sin −1 x
[ −1, 1]
 π π
− 2 , 2
−
2.
cos −1 x
[ −1, – 1]
3.
tan −1 x
R
[ 0, π]
 π π
− , 
 2 2
0 ≤ y ≤ π, where y = cos −1 x
π
π
− < y < , where y = tan −1 x
2
2
4.
cosec –1 x
( −∞, − 1] ∪ [1, ∞ ) − π , π  – {0}
 2 2
−
5.
sec –1 x
( −∞, − 1] ∪ [1, ∞ ) [ 0, π] −  π 
 
 2
 π
0 ≤ y ≤ π ; y ≠   , where y = sec −1 x
 2
6.
cot −1 x
R
π
π
≤ y ≤ ; y ≠ 0, where y = cosec −1 x
2
2
0 < y < π ; where y = cot −1 x.
( 0, π )
1.3 TRIGONOMETRIC INEQUALITIES
To solve trigonometric inequalities including trigonometric functions, it is good to practice
periodicity and monotonicity of functions.
Thus, first solve the inequality for the periodicity and then get the set of all solutions by adding
numbers of the form 2nπ ; n ∈ z, to each of the solutions obtained on that interval.
EXAMPLE
1
Introduction of Graphs
Function
1
Solve the inequality; sin x > − .
2
SOLUTION As the function sin x has least positive period 2 π. {That is why it is sufficient to solve
inequality of the form sin x > a, sin x ≥ a, sin x < a, sin x ≤ a first on the interval of length 2 π,
and then get the solution set by adding numbers of the form 2 πn, n ∈ z, to each of the solutions
 π 3π 
obtained on that interval}. Thus, let us solve this inequality on the interval − ,
, where
 2 2 
1
graph of y = sin x and y = − are taken two curves on x-y plane.
2
y
1
π
2
π
6
sin x >
1
2
π
2
O
π
x
3π
2
7π
6
–1/2
y=–
–1
y = sin x
2π
1
2
Fig. 1.59
y = sin x
and
y =−
1
2
19
1
π
7π
when − < x <
.
2
6
6
Thus, on generalising above solution;
π
7π
; n ∈ z.
2nπ − < x < 2nπ +
6
6
which implies that those and only those values of x each of which satisfies these two inequalities
for a certain n ∈z can serve as solutions to the original inequality.
Play with Graphs
From above figure, sin x > −
EXAMPLE
2
Solve the inequality:cos x ≤ −
As discussed in previous
example, cos x is periodic with period 2 π .
So, to check the solution in [ 0 , 2 π].
1
It is clear from figure, cos x ≤ − when;
2
2π
4π
.
≤ x≤
3
3
On generalising above solution;
4π
2π
≤ x ≤ 2nπ +
; n ∈z
2nπ +
3
3
1
∴ Solution of cos x ≤ −
2
1
.
2
y
SOLUTION
1
1/2
π
2
π
2
O
2π
3
π 4π
3
EXAMPLE
–1/2
y = –1/2
–1
1
–
2
cos x
Fig. 1.60
3
Solve the inequality:tan x < 2 .
We know tan x is periodic with period π .
 π π
So, to check the solution on the interval  − ,  .
 2 2
y
SOLUTION
is
clear
from
figure, tan x < 2
π
− < x < arc tan 2
2
π
< x < tan −1 2 or
2
⇒ General solution
π
2nπ − < x < 2nπ + tan −1 2
2
−
⇒
π


n ∈  2nπ − , 2nπ + arc tan 2


2
when;
1,
1
–π/2
–π/4
O
EXAMPLE
(t an–1 2, 2)
π/4
π/2
SOLUTION
Solve the inequality:
x
–1
tan x < 2
x = π/2
x = –π/2
1
π
 3x
sin 
+ <
.
 2

12
2
1
3x
π
π
 3x
Here, sin 
+ <
; put
+
=t
 2
12
2
12
2
4
y=2
π
4
x = tan–12 = arc tan 2
Fig. 1.61
20
x
2π
4 π
2π

, 2nπ +
; n ∈ z.
x ∈ 2nπ +
3 
3

⇒
It
3π
2
1
2
with period 2π, thus to check on
 π 3π 
 π 5π 
or − ,
 2 , 2 
2 
 2
From figure,
1
3π
9π
.
sin t <
, when
< t<
4
4
2
∴ generalsolution
9π
3π
< t < 2nπ +
; n ∈z
2nπ +
4
4
3x
π
Substituting t =
+
2
12
2nπ +
EXAMPLE
SOLUTION
⇒
∴
5
y= 1
2
π
2
O
3π
4
3π
2
π
–1
2π 9π
4
sin t <
t = 3π/4
5π
=t
2
t
1
2
t = 9π/4
Fig. 1.62
π
9π
3π 3x
<
+
< 2nπ +
4
4
2
12
Solve the inequality :
3π
π
≤ n ≤ nπ + ; n ∈ z.
nπ –
8
8
EXAMPLE
cos 2x − sin 2x ≥ 0.
Here, cos 2x − sin 2x can be reduced to,
π
π
1

 1


2
2  cos cos 2x − sin sin 2x
⇒
cos 2x −
sin 2x
4
4


2

 2
π

2 cos  + 2x
4

π
π

or
cos  + 2x ≥ 0 ; put 2x + = t
cos 2x − sin 2x ≥ 0
4

4
∴ cos ( t ) ≥ 0, solving graphically,
π
π
Clearly;
− ≤ t≤
2
2
π
π
or 2nπ − < t < 2nπ +
2
2
π
where
t = 2n +
4
π
π
π
∴ 2nπ − ≤ 2n + ≤ 2nπ +
2
4
2
6
y
1 cos t ≥0
–π
t=
π
2
O
–1
π
=t
2
π=t
3π
=t
2
y =0
2π= t
y = cos t
Fig. 1.63
If A + B + C = π , then prove that; cos A + cos B + cos C <
3
; where A , B, C
2
distinct.
SOLUTION
∴
y = sin t
1
1
2
4
4
13
4
π + πn < x <
π + nπ ; n ∈ z.
9
3
9
3
⇒
are
y
, now sin t is periodic
Introduction of Graphs
∴ sin t <
Here, we have the three trigonometric functions as cos A , cos B and cos C.
let f ( x) = cos x ;
which can be plotted as;
21
y
1
Play with Graphs
cos x
–π
–π/2
O
π/2
x
π
–1
Fig. 1.64
Now, let us suppose any three points x = A , x = B, x = C on f ( x) = cos x. So that A + B + C = π or
on the interval of length π.
y
A + B + C, cos A + B + C
where G, be centroid of ∆ given by
3
3
 A + B + C cos A + cos B + cos C 
1
,
P


(
B
,cos
B
)


3
3
Thus, from figure points Q, G, P are
collinear,
where; ordinate of GQ < ordinate of PQ.
G
(A,cos A)
(C,cos C)
–π/2
–π
cos A + cos B + cos C
 A + B + C
< cos 



3
3
x =A
Note
π/2
π
x
x =A+B+C
3
 π
⇒ cos A + cos B + cos C < 3 cos  
 3
⇒ cos A + cos B + cos C <
Q x= B x= C
O
f (x) = cos x
Fig. 1.65
3
2
Here, a particular case arises when A = B = C (i.e., when A, B, C are non-distinct)
cos A = cos B = cos C and A + B + C = π
⇒
A+ A+ A= π
or
1
cos A = cos B = cos C =
2
3
cos A + cos B + cos C =
2
∴
∴
EXAMPLE
Solve the inequality:
sin x cos x +
(only when A = B = C)
1
tan x ≥ 1.
2
SOLUTION
Here; left hand side is defined for all x, except x = nπ +
∴
2 sin x cos x + tan x ≥ 2
⇒
⇒
22
7
A = π /3.
2 tan x
1 + tan 2 x
2y
1 + y2
+ tan x ≥ 2
+y ≥2
π
, where n ∈ z.
2
[Let, tan x = y ]
(1 + y 2 )
∴
2y + y (1 + y 2 ) − 2 (1 + y 2 ) ≥ 0
⇒
y 3 − 2y 2 + 3y − 2 ≥ 0
⇒
y 2 ( y − 1) − y ( y − 1) + 2 ( y − 1) ≥ 0
or
( y − 1) ( y 2 − y + 2) ≥ 0
⇒
{Q 1 + y 2 ≥ 0}
≥0
y − 1≥ 0
1

{Q y 2 − y + 2 =  y − 

2
2
+
7
> 0, for all y}
4
∴ tan x ≥ 1, shown as:
from given figure;
π
π
≤ x<
4
2
π
π
nπ + ≤ x < nπ +
;
4
2
or
∴
π
π

x ∈ nπ + , n π +  ; n ∈ z
4
2

tan x ≥ 1
y
y=1
O
n ∈z
x = –π/2
x
π/4
x = π/2
Fig. 1.66
Introduction of Graphs
2y + y (1 + y 2 ) − 2 (1 + y 2 )
⇒
If A + B + C = π, then prove that
A
B
C
tan 2
+ tan 2 + tan 2 > 1.
2
2
2
A
B
C
and tan 2
are three same function. So consider
SOLUTION Here, tan 2 , tan 2
2
2
2
x
y
f ( x) = tan 2 , whose period is 2 π.
2
x
∴ plotting tan 2 for x ∈ ( − π, π ).
T
R
2
G
In given curve let us consider any three points
1
A , B, C such that
A + B + C = π.
M
S
x
x=C
x= B
O
x=A
Now, centroid of ∆RST;
B
C

2 A
+ tan 2
+ tan 2 
 A + B + C tan
2
2
2
x= –π
x= π
G
,
3
3


Fig. 1.67


A + B + C
 A + B + C 
also,
where; GN > MN.
M
, tan 2 

 2( 3)  

3
A
B
C
+ tan 2 + tan 2
tan 2
A
B
C
2
2
2 > tan 2  A + B + C 
⇒
⇒ tan 2 + tan 2 + tan 2 > 1.




2
2
2
3
6
EXAMPLE
8
23
Play with Graphs
1.4 SOLVING EQUATIONS GRAPHICALLY
Here, we sketch both left hand and right hand side of equality and the numbers of intersections
are required solutions.
EXAMPLE
SOLUTION
9
Find the number of solutions of; sin x =
x
.
10
x
10
–1 ≤ sin x ≤ 1
Here, let f ( x) = sin x and g( x) =
also we know;
x
≤1
⇒
10
Thus, to sketch both curves when x ∈ [ −10, 10]
∴
−1 ≤
− 10 ≤ x ≤ 10
y
1
2π
(2π, 10 )
π
(π, 10)
O
–3π
–10
g(x) =
–2π
–π
π
(0,0) π/2
3π
(3π, 10 ) (10,1)
x
g (x ) =
10
2π
–1
x
10
10 3π/2
3π
f (x) = sin x
Fig. 1.68
From above figure f ( x) = sin x and g( x) =
x
intersect at 7 points. So, numbers of solutions
10
are 7.
EXAMPLE
SOLUTION
10
Find the least positive value of x, satisfying tan x = x + 1 lies in the interval.
Let; f ( x) = tan x and g( x) = x + 1; which could be shown as:
g(x) =x +1
y
Least +ve(x)
1
3π
2
π
2
–1
O π
4
π
2
3π
2
x
Fig. 1.69
From the above figure tan x = x + 1 has infinitly many solutions but the least positive value of
 π π
x ∈ ,  .
 4 2
24
SOLUTION
11
Find the number of solutions of the equation,
sin x = x 2 + x + 1.
1

Let; f ( x) = sin x and g( x) = x 2 + x + 1 =  x + 

2
2
+
3
4
which could be shown as;
y
g (x) = x 2+ x + 1
1
f (x) = sin x
1 3
– ,
2 4
x
π
O
–1/2
Fig. 1.70
which do not intersect at any point, therefore no solution.
EXAMPLE
SOLUTION
12
Find the number of solutions of: e x = x 4 .
Introduction of Graphs
EXAMPLE
Let; f ( x) = e x and g( x) = x 4 , which could be shown as;
x4
y
ex
x
Fig. 1.71
From the figure, it is clear they intersect at two points, therefore two solutions.
EXAMPLE
SOLUTION
13
Find the number of solutions of; log 10 x = x.
Let; f ( x) = log 10 x and g( x) = x; which could be shown as;
y
f (x) = log10x
1
O
–1
1
10
x
g (x) = x
Fig. 1.72
From above figure, it is clear they intersect at one points, therefore 1 solution.
25
Play with Graphs
EXAMPLE
Sketch the graph for y = sin −1 ( sin x).
1
As, y = sin −1 (sin x) is periodic with period 2 π.
∴ to draw this graph we should draw the graph for one interval of length 2 π and repeat for entire
values of x.
π
π

− ≤ x≤
x ;
−1
2
2
As we know;
sin (sin x) = 
3π 
π
π 
π
( π − x) ; − ≤ π − x <
 i. e., ≤ x ≤


2
2
2
2

SOLUTION
π
π

− ≤ x≤
x ,
2
2
or
sin (sin x) = 
3π
π
π − x ,
,
≤ x≤

2
2
which is defined for the interval of length 2 π, plotted as;
−1
y
2π
x–
y=
x
2π
x+
y = π/2
x or y = 0
x
3π
2
–
3π
π
y=
x
π
2
x
O
π–
–
–π
π
2
Repeated Curve
y=
y=
y=
Main Curve
π/2
y=
Repeated Curve
y = –π/2
–π/2
Fig. 1.73
Thus, the graph for y = sin −1 (sin x), is a straight line up and a straight line down with slopes 1
 π π
and –1 respectively lying between − ,  .
 2 2
Note : Students are adviced to learn the definition of sin –1 (sin x) as,

 x + 2π

– π – x


y = sin –1 (sin x) =  x

π–x

 x – 2π

26
;
;
;
;
;
5π
3π
≤ x≤ –
2
2
π
3π
–
≤ x≤ –
2
2
π
π
– ≤ x≤
2
2
π
3π
≤ x≤
2
2
3π
5π
≤ x≤
... and so on
2
2
–
SOLUTION
2
Sketch the graph for y = cos −1 ( cos x).
As, y = cos −1 (cos x) is periodic with period 2 π.
∴ to draw this graph we should draw the graph for one interval of length 2 π and repeat for entire
values of x of length 2 π.
As we know;
0≤ x≤ π
 x;
cos −1 (cos x) = 
;
−
≤ 2 π − x ≤ π,
2
0
π
x

0≤ x≤ π
x ;
cos −1 (cos x) = 
 2 π − x ; π ≤ x ≤ 2 π.
or
Thus, it has been defined for 0 < x < 2 π that has length 2 π. So, its graph could be plotted as;
y
2π
2π
x
x–
x
π
O
–
2π
x
x
x+
–x
–
–
x+
y=π
2π
4π
π
–2
4π
(π, π)
3π
4π
x or y = 0
Introduction of Graphs
EXAMPLE
Fig. 1.74
Thus, the curve y = cos –1 (cos x).
EXAMPLE
3
Sketch the graph for y = tan −1 ( tan x).
As y = tan −1 (tan x) is periodic with period π.
∴ to draw this graph we should draw the graph for one interval of length π and repeat for entire
values of x.
π
π

As we know;
tan −1 (tan x) =  x ; − < x < 
2
2

π
π
Thus, it has been defined for − < x < that has length π. So, its graph could be plotted as;
2
2
SOLUTION
y
y = π/2
x
–3π/2
–π –π/2
O π/2
π
3π/2
2π 5π/2
x
y = –π/2
Fig. 1.75
π
Thus, the curve for y = tan −1 (tan x), where y is not defined for x ∈ ( 2n + 1) .
2
27
EXAMPLE
4
Sketch the graph for y = cosec −1 ( cosec x) .
As y = cosec −1 ( cosec x) is periodic with period 2 π.
∴ to draw this graph we should draw the graph for one interval of length 2 π and repeat for entire
values of x.
As we know;
π
π

or 0 < x ≤
x ; − ≤ x< 0

2
2
cosec −1 ( cosec x) = 
π
π
 π − x ; − ≤ π – x < 0 or 0 < π − x ≤

2
2

 π
 π 
x ∈ − , 0 ∪  0, 
 x ;
 2
 2 
−1
or
cosec ( cosec x) = 
 π − x ; x ∈  π , π ∪  π, 3π 
 2  

2 
 π 3π 
Thus, it has been defined for − ,
− { 0, π} that has length 2 π. So, its graph could be
2 
 2
plotted as
Play with Graphs
SOLUTION
y
y = π/2
–2π 3π –π
–π/2
x
π–x
O
π
2
y = cosec–1(cosec x)
3π
2
2π
x
y = –π/2
Fig. 1.76
EXAMPLE
5
Sketch the graph for y = sec −1 ( sec x).
As y = sec −1 ( sec x) is periodic with period 2π .
∴to draw this graph we should draw the graph for one interval of length 2 π and repeat for entire
values of x.
As we know;
SOLUTION

 π
π 
x ∈ 0,  ∪  , π
 x ;

2 
2

sec ( sec x) = 
π 
π


 2 π − x ; 2 π − x ∈ 0,  ∪  , π
 2
 2 

−1
or
π

x;
0≤ x<

2
sec −1 ( sec x) = 
3π
 2π − x ; π ≤ x <

2
π
< x≤ π
2
.
3π
or
< x ≤ 2π
2
or
 π 3π 
Thus, it has been defined for[ 0, 2 π] −  ,
 that has length 2 π. So, its graph could be plotted
2 2 
as;
28
(π, π/2)
(–π,π/2)
y = –x
–5π/2 –2π –3π/2 –π
–π/2
O
π/2
y = π/2
(2π–x)
x
π
3π/2 2π
5π/2
x
Fig. 1.77
Thus, the curve for y = sec −1 ( sec x).
EXAMPLE
SOLUTION
6
Sketch the graph for y = cot −1 ( cot x).
As, y = cot −1 (cot x) is periodic with period π.
∴ to draw this graph we should draw the graph for one interval of length π and repeat for entire
values of x.
As we know
cot –1 (cot x) = { x ; 0 < x < π}
Introduction of Graphs
y
which is defined for length π, i.e., x ∈ ( 0, π ) and x ∉{ nπ, n ∈ z}.
So, its graph could be plotted as;
y
y=
x
π
+2
y=
–2π –3π/2 –π
x
x
y=
+π
–π/2
π/2
y
π, π
2 2
O π/2
π
–
=x
y=π
π
3π/2
y= π
2
2π
x
Fig. 1.78
y = cot −1(cot x ).
Thus, the curve for
EXAMPLE
7
Sketch the graph for:
(i) sin ( sin −1 x)
(iv) cosec ( cosec −1 x)
SOLUTION
(ii) cos ( cos −1 x)
(v) sec ( sec −1 x)
(iii) tan ( tan −1 x)
(vi) cot ( cot −1 x)
As we know, all the above mentioned six curves are non-periodic, but have
restricted domain and range.
So, we shall first define each curve for its domain and range and then sketch these curves.
(i) Sketch for y = sin (sin −1 x )
We know; domain, x ∈ [ −1, 1]
and
(i.e., −1 ≤ x ≤ 1)
range y = x ⇒ y ∈ [ −1, 1]
29
Hence, we should sketch y = sin (sin −1 x) only when x ∈ [ −1, 1] and y = x. So, its graph could
be plotted as shown in figure.
Play with Graphs
y
1
y=
–1
y=1
–1 x )
n
(si
sin
x
1
O
y = –1
–1
x=1
x = –1
Fig. 1.79
Thus,
the graph for y = sin (sin −1 x).
(ii) Sketch for the curve y = cos (cos −1 x ).
We know,
domain, x ∈ [ −1, 1]
(i.e., −1 ≤ x ≤ 1)
range y = x ⇒ y ∈ [ −1, 1]
and
Hence, we should sketch y = cos (cos −1 x) = x only when x ∈ [ −1, 1]. So, its graph could be
plotted as shown in Fig. 1.80.
y
)
y=
–1
y=1
–1 x
1
s
o
s(c
co
1
O
x
y = –1
–1
x=1
x = –1
Fig. 1.80
Thus, the graph for y = cos(cos −1 x).
(iii) Sketch for the curve y = tan (tan −1 x )
We know,
y
domain, x ∈ R (i.e., −∞ < x < ∞)
Range
–1 x
)
n
and
(ta
n
1
y = x ⇒ y ∈ R.
y
=
ta
Hence, we should sketch
y = tan (tan −1 x) = x, ∀ x ∈ R.
So, its graph could be plotted as shown;
Thus, the graph for y = tan (tan −1 x).
30
(0, 0)
–1
–1
Fig. 1.81
1
x
y
–1 x
s
o
(c
y=
co
y=1
1
and range y = x ⇒ y ∈ R − ( −1, 1).
Hence, we should sketch
–1
y = cosec ( cosec −1 x) = x only when
x ∈ ( −∞, − 1] ∪ [1, ∞ ).
=
1 x)
x
x
1
O
y = –1
–1
–
c
So, its graph could be plotted as
shown in Fig. 1.82;
(c
ec
y=
e
os
s
co
x = –1
y = cosec ( cosec −1 x).
x=1
Fig 1.82
(v) Sketch for y = sec ( sec −1 x)
y
We know, domain ∈ R − ( −1, 1)
–1 x
(i.e., −∞ < x ≤ − 1 or 1 ≤ x < ∞)
y=
(1, 1)
and range y = x ⇒ y ∈ R − ( −1, 1).
Hence, we should sketch
)
–1 x
So, its graph could be plotted as shown
in Fig. 1.83.
Thus, the graph for
y = sec ( sec −1 x) = x
(vi) Sketch for y = cot (cot −1x )
c(
se
)=
c
se
y=1
x
=x
y = –1
(–1, –1)
ec
(s
y=
c
se
x = –1
x=1
Fig. 1.83
)
–1 x
y
We know; Domain ∈ R (i. e., – ∞ < x < ∞)
y=
Range y = x ⇒ y = R.
co
ot
t(c
or y = x
1
Hence, we should sketch
y = cot (cot –1 x) = x, ∀ x ∈ R.
Shown as in Fig. 1.84.
Thus, the graph for y = cot (cot −1 x ).
Note
x
O
y = sec ( sec −1 x) = x,
only when x ∈ ( − ∞, – 1] ∪ [1, ∞ )
and
x
c
se
(i.e., – ∞ < x ≤ − 1 or 1 ≤ x < ∞)
Thus, the graph for
)=
ec
Introduction of Graphs
(iv) Sketch for y = cosec (cosec −1x )
We know;
domain ∈ R − ( −1, 1)
O
–1
1
x
–1
Fig. 1.84
From previous discussions, we learn that if:
(i) The function is periodic then find period and trace the curve.
(ii) If non-periodic, then define for their domain and find range to trace the curve.
Now, before going ahead you must revise previous curves of inverse trigonometry as;
y = sin−1 x, y = cos−1 x, y = tan−1 x, y = cot −1 x, y = cosec−1x, y = sec−1x
with their domain and range.
31
Play with Graphs
EXAMPLE
SOLUTION
Sketch the graph for:
 2x 
1 − x 2 
(ii) cos −1 
(i) sin −1 


2
1 + x 
1 + x 2 
 3x − x 3 
(v) sin −1 ( 3x − 4x 3 )
(iv) tan −1 

 1 − 3x 2 
8
 2x 
(iii) tan −1 

1 − x 2 
(vi) cos −1 ( 4x 3 − 3x).
As we know, all the above mentioned six curves are non-periodic, but have
Restricted domain and Range.
So, we shall first define each curve for its domain and range and then sketch these curves.
 2x 
(i) Sketch for y = sin −1 
.
1 + x 2 
2x
Here, for domain
1 + x2
≤1
⇒
2 |x| ≤ 1 + x 2
⇒
⇒
⇒
|x|2 − 2 |x| + 1 ≥ 0
(|x|2 − 1) 2 ≥ 0
x ∈ R.
 2x 
y = sin −1 

1 + x 2 
For range:
 π π
y ∈ − , 
 2 2
⇒
or
or
{Q x 2 = |x|2 }

−1
 as ; y = sin θ

 π π 
⇒ y ∈ − , 
 2 2
x = tan θ
π

; 2θ >
π − 2θ
2

π
π
−1
{See Ex. 1}
y = sin (sin 2 θ) =  2 θ
; − ≤ 2θ ≤
2
2

− π − 2θ ; 2θ < − π

2
π

−1
−1
 π − 2 tan x ; tan x > 4

π
π
; − ≤ tan −1 x ≤
y =  2 tan −1 x
{Q tan θ = x ⇒ θ = tan −1 x}
4
4

 − π − 2 tan −1 x ; tan −1 x < − π

4
−1
 π − 2 tan x
; x>1

…(i)
y =  2 tan −1 x
; −1 ≤ x ≤ 1
 − π − 2 tan −1 x ; x < − 1

Defining the curve:
⇒
{Q 1 + x 2 > 0 for all x}
Let,
 2x 
 π π
Thus, y = sin −1 
 is defined for x ∈ R, where y ∈ − ,  , so the graph for Eq. (i)
2
 2 2
1 + x 
could be shown as in Fig. 1.85.
32
(1, π/2)
y = π/2
–1
x
π/2
y=
2
ta
n
y=π
–1
y = –π–2tan–1x
(–1,
x
1
O
π
)
2
–2ta –1
n x
y = –π/2
–π/2
x = –1
x=1
Fig. 1.85
 2x 
Thus, the graph for y = sin −1 
.
1 + x 2 
As in later section (i.e., chapter 2) we shall discuss that functions having sharp edges and
gaps are not differentiable at that point.
 2x 
 , we know it has sharp edge at x = − 1 and x = 1.
So, in previous curve y = sin−1 
2
1 + x 
So, not differentiable.
Note
Introduction of Graphs
y
1 – x 2 
(ii) Sketch for y = cos –1 

1 + x 2 
1 – x2
Here, for domain
1 + x2
⇒
≤1
|1 – x 2| ≤ 1 + x 2
{Q 1 + x 2 > 0, ∀ x ∈ R}
which is true for all x; as 1 + x 2 > 1 – x 2
∴
x ∈R
For range:
1 – x 
y = cos –1 

1 + x 2 
Define the curve :
Let,
2
∴
y ∈ ( 0, π )
x = tan θ
 1 – tan 2 θ
y = cos –1 
 = cos –1 (cos 2 θ)
2
 1 + tan θ
 2θ;
=
 – 2θ;
⇒
⇒
2θ ≥ 0
2θ < 0
 2 tan –1 x;
y =
 – 2 tan –1 x;
.
{See Example 2}
tan –1 x ≥ 0
tan
–1
x< 0
{Q tan θ = x ⇒ θ = tan –1 x}
33
 1 – x 2   2 tan –1 x;
So, the graph of y = cos –1 
 =
 1 + x 2   – 2 tan –1 x;
x≥ 0
is shown as:
x < 0,
Play with Graphs
y
y=
–2
tan –1
x, x
<0
(–1,π/2)
–1
–1 x, x > 0
y=π
y = 2 tan
(1,π/2)
O
x
1
Fig. 1.86
 1 – x 2   2 tan –1 x,
Thus, the graph for y = cos –1 
 =
 1 + x 2   – 2 tan –1 x,
From above figure it is clear
x≥ 0
x< 0
.
1 – x 2 
y = cos –1 
 is not differentiable at x = 0.
1 + x 2 
 2x 
(iii) Sketch for y = tan –1 

1 – x 2 
2x
Here, for domain
∈ R except; 1 – x 2 = 0
1 – x2
x ≠ ±1
i.e.,
or
x ∈ R – {1, – 1}
For range
 2x 
y = tan –1 

1 – x 2 
⇒
 π π
y ∈ – , 
 2 2

 π π
–1
 as y = tan θ ⇒ y ∈  – ,  
2 2 

Defining the curve
Let
⇒
34
x = tan θ
π

2θ < –
 π + 2θ;
2

π
π

–1  2 tan θ 
–1
{See Example 3}
y = tan 
– < 2θ <
 = tan (tan 2 θ) =  2θ;
2
2
 1 – tan 2 θ

π
 – π + 2θ;
2θ >

2
π

–1
tan –1 x < –
 π + 2 tan x;
4

π
π

–1
–1
{as tan θ = x ⇒ θ = tan –1 x}
=  2 tan x;
– < tan x <
4
4

π
 – π + 2 tan –1 x;
–1
tan x >

4
So, the graph of;
 π + 2 tan –1 x;
x< – 1



2x
–1
–1
y = tan 
– 1 < x < 1 is shown as;
 =  2 tan x;
1 – x 2  
–1
x>1
 – π + 2 tan x;
y
y=
2 ta
2t
n
π+
y = π/2
an –
1
x
–1 x
–1
O
x
1
π+2
y=–
–1 x
tan
y = –π/2
Introduction of Graphs
 π + 2 tan –1 x;
x< −1

–1
– 1< x< 1
=  2 tan x;

–1
x>1
 – π + 2 tan x;
Fig. 1.87
 π + 2 tan –1 x, x < – 1
 2x  
Thus, the graph for y = tan –1 
 =  2 tan –1 x; – 1 < x < 1
1 – x 2  
–1
x>1
 – π + 2 tan x,
which is neither continuous nor differentiable at x = {– 1, 1}.
 3x – x 3 
(v) Sketch for the curve y = tan –1 
.
 1 – 3x 2 
 3x – x 3 
y = tan –1 

 1 – 3x 2 
Here, for domain
⇒ x ∈ R except
1 – 3x 2 = 0
x≠±
1
3
 1 
x ∈ R – ±

3

∴
 3x – x 3 
y = tan –1 

 1 – 3x 2 
For range
 π π
y ∈ – , 
 2 2
⇒
Defining the curve:
⇒
Let;

 π π
–1
 as y = tan θ ⇒ y ∈  – ,  
2 2 

x = tan θ
35
Play with Graphs
⇒
π 
π

–1
–1
 π + 3θ; 3θ < – 2  π + 3 tan x; tan x < – 6


π
π
π
π


y = tan –1 (tan 3θ) =  3θ; – < 3θ <
=  3 tan –1 x; – < tan –1 x <
2
2
6
6


 – π + 3θ; 3θ > π
 – π + 3 tan –1 x; tan –1 x > π


6
2
1

–1
x< –
 π + 3 tan x;
3

1
1

–1
=  3 tan x; –
< x<
3
3


1
–1
x>
 – π + 3 tan x;
3

So, the graph of;
1

–1
x< –
 π + 3 tan x;
3

3


3x – x
1
1

y = tan –1 
< x<
 =  3 tan –1 x; –
2
 1 – 3x  
3
3

1
–1
x>
 – π + 3 tan x;
3

y
1
an –
3t
3t
y=
y=
y=
π+
–1 x
x
–1 x
an
√3
1
√3
O
1
√3
+
–π
y = π/2
n
3 ta
x
√3
y = –π/2
x = –1/√3
x = 1/√3
Fig. 1.88
1

–1
 π + 3 tan x, x < –
3

3


x
x
3
–
1
1

Thus, the curve for y = tan –1 
< x<
 =  3 tan –1 x, –
 1 – 3x 2  
3
3

1
–1
x>
 – π + 3 tan x;
3



which is neither continuous nor differentiable for x =  ±
(v) Sketch the curve y = sin –1 ( 3x – 4x 3 )
Defining the curve: Let x = sin θ,
36
1 
.
3
∴
1

–1
≤ x≤1
 π – 3 sin x;
2

1
1

y = sin –1 ( 3x – 4x 3 ) =  3 sin –1 x;
– ≤ x≤
2
2

 – π – 3 sin –1 x; – 1 ≤ x ≤ – 1

2
For domain
y = sin –1 ( 3x – 4x 3 )
⇒
x ∈[– 1, 1]
For range
y = sin –1 ( 3x – 4x 3 )
⇒
 π π
y ∈ – , 
 2 2

–1
 π – 3 sin x;


So, the graph of; y = sin –1 ( 3x – 4x 3 ) =  3 sin –1 x;

 – π – 3 sin –1 x;

is shown as:
1
≤ x≤1
2
1
1
– ≤ x≤
2
2
1
– 1≤ x≤ –
2
Introduction of Graphs
π
π
π
3π


≤ 3θ ≤
π – 3 sin –1 x;
≤ sin –1 x ≤
 π – 3 θ;

2
2
6
2


π
π
π
π


–1
–1
–1
⇒ y = sin (sin 3 θ) =  3 θ;
– ≤ 3θ ≤
=  3 sin x;
– ≤ sin x ≤
2
2
6
6


 – π – 3 θ; – 3π ≤ 3 θ ≤ – π  – π – 3 sin –1 x; – π ≤ sin –1 x ≤ – π

2
2 
2
6
–1
3s
in
y=
1
2
3
2
–1 x
n
O
–1 x
sin
3
2
3si
π–
π–3
y=–
–1
y = π/2
y=
x
y
1
x
y = –π/2
x = –1
x = –1/2
x = 1/2
x=1
Fig. 1.89
Thus, the curve for
which is not differentiable at
y = sin –1 ( 3x – 4x 3)
 1
x = ± .
 2
(vi) Sketch the curve y = cos –1 ( 4x 3 – 3x )
Here, domain ∈[– 1, 1]
range ∈[ 0, π]
Now, defining the curve
Let
x = cos θ
37
Play with Graphs
2π
π

–1
–1
 2 π – 3 cos x; 3 ≤ cos x ≤ 3
π ≤ 3θ ≤ 2 π 
 2 π – 3 θ;
π


⇒ y = cos –1 (cos 3 θ) =  3 θ;
0 ≤ 3 θ ≤ π =  3 cos –1 x;
0 ≤ cos –1 x ≤
3
 –2 π + 3 θ; – π ≤ 3 θ ≤ 0


 – 2 π + 3 cos –1 x; – π ≤ cos –1 x ≤ 0

3
1
1

–1
– ≤ x≤
 2 π – 3 cos x;
2
2

1

=  3 cos –1 x;
≤ x≤1
2

 –2 π + 3 cos –1 x; – 1 ≤ x ≤ – 1

2
π
π
1
{Q If 0 ≤ θ ≤
⇒ cos ≤ cos θ ≤ cos 0 or
≤ cosθ ≤ 1. Here, the interval changed
3
3
2
since, cos x is decreasing in [ 0, π]}
1
1

–1
– ≤ x≤
 2 π – 3 cos x;
2
2

1

–1
3
–1
So, the graph of; y = cos ( 4x – 3x) =  3 cos x;
≤ x≤1
2

1
 –2 π + 3 cos –1 x;
–1≤ x≤ – ,

2
is shown as;
y
(1/2, π)
y=π
–1 x
y = π/2
–1 x
os
+ 3c
os
3c
–2π
s
co
–3
π
2
y = (0, π/2)
y=
y=
–1 x
3
2
x = –1
(– 1 ,0)
2
x = –1/2
3
2
x = 1/2
x
x=1
Fig. 1.90
Thus, the curve for y = cos –1 ( 4x 3 – 3x),
 1
which is not differentiable at
x = ± .
 2
EXAMPLE
9
Sketch the graph for:
(i) sin x. cosec x
SOLUTION
(ii) cos x ⋅ sec x
As we know for the above curves each is equal to 1, but for different domain as;
(i) y = sin x ⋅ cosec x = 1; ∀ x ∈ R – { nπ; n ∈ z}
38
(iii) tan x ⋅ cot x
π


x ∈ R – ( 2n + 1) ; n ∈ z
2


(iii) y = tan x ⋅ cot x = 1;
π


x ∈ R –  nπ, nπ + ; n ∈ z .
2


Thus, they could be plotted as:
y
y
1
–π
1
y=1
O
π
2π
x
π
2
3π
2
y = sin x.cosec x
O
y=1
π
2
3π
2
x
y = cos x.sec x
y
1
3π –π
2
y=1
π O
2
π
2
π
3π
2
Introduction of Graphs
(ii) y = cos x ⋅ sec x = 1;
x
y = tan x.cot x
Fig. 1.91
Note
From above example it becomes clear that y = sin x ⋅ cosec x = 1, y = cos x ⋅ sec x = 1,
y = tan x ⋅ cot x = 1 but they are not equal, as their domains are different.
∴ Equal functions : Those functions which have same domain and range are equal
functions.
EXAMPLE
Sketch the graph for:
1 + x 2 
|sin x|
|cos x|
(i)
(ii)
(iii) sin –1 

sin x
cos x
 2x 
1 1

(iv) log 1/ 4  x –  + log 4 (16x 2 – 8x + 1)

4 2
(v) 1 + 3 ( log|sin x| + log |cosec x|)
(vi) 1 + 3 ( log sin x + log cosec x)
10
As we know, to plot above curves we must check periodicity domain and range;
| sin x |
(i) y =
sin x
SOLUTION
Here,
1; sin x > 0
y =
 – 1; sin x < 0
39
n ∈z
2nπ < x < ( 2n + 1) π;
1;
y =
 –1; ( 2n + 1)π < x < ( 2n + 2) π; n ∈ z
Play with Graphs
y
So, from above,
1
domain ∈ R – { n π; n ∈ z}
2nπ < x < ( 2n + 1) π
 1;
Range ∈ 
 – 1; ( 2n + 1) π < x < ( 2n + 2) π
–3π
–2π
–π
O
π
2π
3π
4π
x
–1
∴ it could be plotted as shown in
Fig. 1.92.
y = Isin xI
sin x
Fig. 1.92
(ii) Sketch for y =
Here,
| cos x |
cos x
 1; cos x > 0
y =
 –1; cos x < 0

 1;
y =
 –1;

⇒
So, it could be plotted as:
π
π
< x < 2nπ +
2
2
π
3π
2nπ + < x < 2nπ +
2
2
2nπ –
y
1
–5π/2
–3π/2
–π/2 O
π/2
3π/2
5π/2
x
–1
y=
I cos x I
cos x
Fig. 1.93
1 + x 2 
(iii) Sketch for y = sin –1 

 2x 
Here
when;
⇒
⇒
⇒
⇒
⇒
40
1 + x 2 
y = sin –1 
 is defined;
 2x 
1 + x2
≤1
2x
1 + x 2 ≤ 2 |x|
x 2 – 2 |x| + 1 ≤ 0
|x|2 – 2 |x| + 1 ≤ 0
(|x| – 1) 2 ≤ 0
(|x| – 1) 2 = 0
{as;
sin –1 x is defined when |x| ≤ 1}
{as; 1 + x 2 > 0}
{as; x 2 = |x|2 }
{as; (|x| – 1) 2 < 0 is not possible}
x=±1
Domain ∈ { ± 1}
1 + x 2 
For range y = sin –1 
 , where
 2x 
∴
∴
y = sin –1 (1)
and
π
2
y = sin –1 (–1)
⇒
y =±
∴
 π
Range ∈  ± 
 2
Hence, the graph for
1 + x 2 
y = sin –1 

 2x 
x = + 1, – 1
y
A (1, π/2)
is only two
B
(–1,–π/2)
points. Shown as:
Thus, the sketch for
1 + x 2 
y = sin –1 

 2x 
x
O
Graph for sin–1 1 + x
2x
is only two
2
Fig. 1.94
points A and B.
1 1

(iv) Sketch for y = log 1/ 4 x –  + log 4 (16x 2 – 8x + 1)

4 2
Here,
1 1

y = log 1/ 4  x –  + log 4 ( 4x – 1) 2

4 2
⇒
1
1
1 1


y = log 1/ 4  x –  + log 4 16 + log 4  x – 


2
4
4 2
or
2
1
1 1


y = – log 4  x –  + log 4 4 2 + log 4  x – 


2
4
4 2
⇒
1
1 2


y = – log 4  x –  + log 4  x –  + log 4 4


4
4 2
Introduction of Graphs
⇒
∴
2

 as;

log b n a m =
m

log b a
n

1

⇒ y = 1, whenever;  x –  > 0 {as; log a x exists only when a, x > 0 and a ≠ 1}

4
Thus,
1 1

y = log 1/ 4  x –  + log 4 ( 4x – 1) 2

4 2
⇒
1 
Domain ∈  , ∞
4 
Range ∈{1}
Thus, the graph is shown as:
Thus, the graph for
1 1

y = log 1/ 4  x –  + log 4 ( 4x – 1) 2 .

4 2
y
1
O
1
4
x
Fig. 1.95
(v) Sketch for y = 1 + 3 (log | sin x | + log | cosec x | )
Here
y = 1 + 3 [log(|sin x|⋅|cosec x|)]
41
|sin x| ≠ 0 and
whenever
y = 1 + 3 (log 1);
i.e.,
Play with Graphs
|cosec x| ≠ 0
whenever
⇒
y =1
∴
Domain ∈ R – { nπ; n ∈ z}
x ∉ nπ; n ∈ z.
{as;
log 1 = 0}
Range ∈{1}
∴ it could be plotted as:
y
1
–4π
–3π
–2π
π
O
–π
2π
3π
4π
x
Fig. 1.96
Thus, the curve for y = 1 + 3 (log|sin x| + log|cosec x|).
(vi) Sketch for y = 1 + 3 (log sin x + log cosec x )
Here
⇒
or
∴
y = 1 + 3 (log sin x ⋅ cosec x)
y = 1 + 3 log 1;
y = 1;
whenever
sin x > 0 and cosec x > 0
x ∈ ( 2nπ, ( 2n + 1) π )
x ∈ ( 2nπ, ( 2n + 1)π )
whenever
y = 1 + 3 (log sin x + log cosec x) = {1; 2nπ < x < ( 2n + 1) π is shown as;
y
1
–4π
–3π
–2π
–π
O
π
2π
3π
Fig. 1.97
Thus, the curve for y = 1 + 3 (log sin x + log cosec x)
E X A M P L E 11 Sketch the curve for cos y = cos x.
cos y = cos x
⇒
y = 2nπ ± x; n ∈ z
SOLUTION
Here,
∴
cos y = cos x, represents two straight lines;
 x + 2nπ;
y =
 – x + 2nπ;
42
n ∈z
n ∈z
x
y = x + 4π ; n = –2
y
y = x + 2π ; n = –1
y = x ;n = 0
4π
y = x – 2π ; n = 1
y = x – 4π ; n = 2
2π
–4π
O
–2π
x
2π
–2π
y = – x + 4π ; n = 2
y = –x + 2π ; n =1
y = –x ; n = 0
y = –x – 2π ; n = –1
y = –x – 4π ; n = –2
Fig. 1.98
Introduction of Graphs
i.e., two infinite set of perpendicular straight lines which could be shown as:
Thus, graph for cos y = cos x; represents two infinite set of perpendicular straight lines which have
infinite number of points of intersections; (So, if asked number of solutions then they are infinite).
EXAMPLE
SOLUTION
∴
12
Sketch the curve for sin y = sin x.
Here sin y = sin x
⇒
y = nπ + (– 1) n x; n ∈ z
sin y = sin x; represent two straight lines;
 nπ + x; n even integer
y =
 nπ – x; n odd integer
y = x + 4π ; n = 4
y
5π
y = x + 2π ; n = 2
4π
y = x ;n = 0
y = x – 2π ; n = –2
3π
y = x – 4π ; n = –4
2π
π
–5π –4π –3π –2π –π
0 π
–π
2π
3π
4π
5π
6π
y = 5π – x ; n = 5
–2π
–3π
–4π
–5π
x
y = 3π – x ; n = 3
y = π– x ;n = 1
y = –π – x ; n = –1
y = –3π – x ; n = –3
Fig. 1.99
43
Play with Graphs
i.e., two infinite set of perpendicular straight lines as shown in Fig. 1.99:
Thus, the graph for sin y = sin x.
EXAMPLE
13
Find the number of solutions for;
sin
πx 99x
=
.
2
500
πx
99x
and g( x) =
,
2
500
to find number of solutions; we shall plot both the curves as;
Let
SOLUTION
f ( x) = sin
y
g(x) = 99x
500
1
sin
–8
–6
–4
0
–2
1
2
4
6
πx = f (x)
2
x
–1
Fig. 1.100
Clearly, from the above figure, the number of solutions are 7.
EXAMPLE
14 Find the number of solutions for;
SOLUTION As, cos x = x
∴ to plot the curve for
y = cos x; y = x and find the
point of intersection as to
obtain number of solutions.
Here, the two curves intersect
at a point A.
So, cos x = x has only one
solution.
cos x = x
y
y=x
π/2
1
y = cos x
A
–2π –3π/2
–π
O
–π/2
π/2
π
3π/2
x
2π
–1
–π/2
Fig. 1.101
EXAMPLE
greatest
Find the number of solutions for; [ x] = { x}. where [ • ], { • } represents
integer and fractional part of x.
SOLUTION
∴ to plot
15
As,
y
[ x] = { x}
{x}
2
y = [ x]; y = { x}
and find point of intersection.
{x}
Here, the only point of intersection is x = 0 ,
–2
∴ only one solutions.
1
{x}
A
–1
0
1
–1
–2
44
Fig. 1.102
2
3
x
Find the number of solutions of
4 { x} = x + [ x]
where { ⋅ }, [ ⋅ ] represents fractional part and greatest integer function.
16
As we know, to find
number of solutions of two curves we
should find the point of intersection of
two curves.
∴
4 { x} = x + [ x]
⇒ 4( x – [ x]) = x + [ x]
{Q x = [ x] + { x}}
⇒
4x – x = 4 [ x] + [ x]
⇒
3x = 5 [ x]
3
…(i)
[ x] = x
⇒
5
∴ To plot the graph of both
3
y = [ x] and y = x.
5
Clearly, the two graphs intersects when
SOLUTION
y
y = [x]
y =3x
5
2
1
–2
0
1
–3/5
–1
–1
2
x
3
–2
Fig. 1.103
[ x] = 0 and [ x] = 1
…(ii)
5
x = [ x]
3
∴
x=
5
⋅0
3
[from Eqs. (i) and (ii)]
x=
and
Introduction of Graphs
EXAMPLE
5
(1)
3
∴ x = 0 and
x=
5
are the only two solutions.
3
EXAMPLE
17
Find the value of x graphically satisfying;[ x] – 1 + x 2 ≥ 0 ; where [ ⋅ ] denotes
the greatest integer function.
As,
[ x] – 1 + x 2 ≥ 0
⇒
x 2 – 1 ≥ – [ x]
2
Thus, to find the points for which f ( x) = x – 1 is greater than or equals to
SOLUTION
y
2
x –
1>
5
2
A
1
Thus, to find A and B.
It is clear that f ( x) and g( x) intersects when;
– [ x] = 2 .
∴
x2 – 1 = 2
x=± 3
(neglecting x = +
⇒
x=–
3
–4
–3
–2
0
1B
–1
–1
1>
]
3
– [x
]
– [x
4
x 2–
g( x) = – [ x].
where two functions f ( x) and g( x) could be
plotted as shown in Fig. 1.104;
From the adjoining figure; the solution set lies
when
or
x≤ A
x ≥ B.
2
3
4
x
x 2 –1< – [x]
Fig. 1.104
3 as A lies for x < 0 )
45
Play with Graphs
Thus, for A : x = –
3 and
∴ Solution set for which
x – 1 ≥ – [ x] holds.
⇒
Note
for B :
x = 1.
2
x ∈ (– ∞, –
3] ∪ [1, ∞ ).
The method discussed in previous example is very important as it reduces your
calculations, so students should practice these forms.
E X A M P L E 18 Find the values of x graphically which satisfy; – 1 ≤ [ x] – x 2 + 4 ≤ 2 ; where
[ ⋅ ] denotes the greatest integer function.
SOLUTION As, – 1 ≤ [ x] – x 2 + 4 ≤ 2
⇒
x 2 – 5 ≤ [ x] ≤ x 2 – 2
2
Thus, to find the points for which f ( x ) = x – 5 is less than or equal to g( x ) = [ x] and g( x) = [ x] is
less than or equal to h( x) = x 2 – 2 , where the three functions f ( x), g( x)and h( x)could be plotted
as;
y
h (x ) = x 2 – 2
f (x ) = x 2 – 5
5
4
3
g(x)
2
–4
–3
– 3
–1
A
B
h (x) < g (x) < f (x)
C
g (x )
1
0
–1
1
2 2
5 3
5
4
x
–2
–3
g(x)
–4
–5
Fig. 1.105
Thus, from the above graph;
x 2 – 5 ≤ [ x] ≤ x 2 – 2
when x ∈[ A , B] ∪ [C, D]
where A and D is the point of intersection if;
x2 – 5 = ± 2
⇒
x=–
3, 7
and C is point of intersection of
x2 – 2 = 1 ⇒ x =
∴
∴
A =–
3,
B = – 1,
C=
3.
3 and
D = 7.
– 1 ≤ [ x] – x 2 + 4 ≤ 2 is satisfied;
x ∈[–
when
3, – 1] ∪ [ 3, 7 ].
E X A M P L E 19 If 0 ≤ a ≤ 3, 0 ≤ b ≤ 3 and the equation; x 2 + 4 + 3 cos ( ax + b) = 2x has
atleast one solution then find the value of ( a + b).
SOLUTION
⇒
46
Here,
or
x 2 + 4 + 3 cos( ax + b) = 2x
x 2 – 2x + 4 = – 3 cos( ax + b)
2
( x – 1) + 3 = – 3 cos( ax + b)
y
cos ( a + b) = – 1
⇒
a + b = π, 3 π, 5 π, …
But
3π > 6
⇒
a + b = π as
EXAMPLE
20
f (x) = (x – 1)2 + 3
4
(1, – 3 cos(a + b))
g (x) = – 3 cos (ax + b)
3
From figure the two curves could atmost
touch at one point only when
– 3 cos( a + b) = 3
⇒
Introduction of Graphs
for above equation to have atleast one
solution; plot f ( x) = ( x – 1) 2 + 3 and
g( x) = – 3 cos( ax + b) in such a way that
they touch each other.
O
1
x
2
–3
Fig. 1.106
0 ≤ a, b ≤ 3.
If A + B + C = π and
A, B, C are angles of ∆; then show
3 3
.
2
sin A + sin B + sin C <
SOLUTION Here;
we
have
three
trigonometric ratios sin A , sin B, sin C.
∴ Let y = sin x, on which there are three
points x = A , x = B and x = C shown as;
As from the figure; In ∆PQR,
Centroid of ∆ formed by P( A , sin A )
Q( B, sin B) R(C, sin C ) is;
y
H
1
(A, sin A)
(B, sin B)
Q
G
R (C, sinC)
P
O A B I
 A + B + C sin A + sin B + sin C 
,
G=



3
3
π
2
where; G, H and I are collinear
C π
2π
x
Fig. 1.107
 A + B + C sin A + sin B + sin C 
A + B + C 
, 0 ; G 
,
I





3
3
3
and
A + B + C
 A + B + C 
H
, sin 




3
3
HI > GI
From figure;
i.e.,
ordinate of H > ordinate of G
⇒
 A + B + C  sin A + sin B + sin C
sin 
>


3
3
⇒
3 3
> sin A + sin B + sin C.
2
47
π
π
, then show A ( cosec A ) < .
6
3
SOLUTION Here, graph for y = sin x is shown as;
π
π
where P( A , sin A ) and Q  , sin 
6
6
Play with Graphs
EXAMPLE
21
If 0 < A <
From adjoining figure;
slope of OP > slope of OQ
π
sin – 0
sin A – 0
6
>
⇒
π
A–0
–0
6
sin A 3
A
π
or
⇒
>
<
A
π
sin A 3
y = sin x
π
sin
6
Q
P
sin A
O
π/6
A
EXAMPLE
SOLUTION
Fig. 1.108
π
3π
, then show that: A cosec A + B cosec B + C cosec C <
.
2
2
Here, graph for y = sin x is shown as;
22
If 0 < A , B, C <
y
π
π
P( B, sin B); Q(C, sin C ); R( A , sin A ); S  , sin  .
2
2
From figure;
⇒
⇒
x
π/2
π
A (cosec A ) < .
3
or
where
y
R
S
A
π/2
Q
slope of OP > slope of OS:
π
sin – 0
sin B – 0
2
>
π
B– 0
2
π
sin B 2
B
or
>
<
π
B
sin B 2
B cosec B <
or
P
O
B
C
x
Fig. 1.109
π
2
…(i)
Similarly, slope of OQ < slope of OS and slope of OR < slope of OS.
⇒
C cosec C < π/2
…(ii)
A cosec A < π/2
…(iii)
Adding Eqs. (i), (ii) and (iii), we get
A cosec A + B cosec B + C cosec C <
Note
48
3π
.
2
Students must practice above method in different questions of trigonometric
inequality as it saves time.
2x + 3 ; – 3 ≤ x < – 2

(ii) f (x ) = x + 1; – 2 ≤ x < 0
x + 2 ;
0 ≤x ≤1

2. Construct the graph of the function:
(i) f (x ) = | x – 1| + | x + 1|
3 x ;
– 1≤ x ≤ 1
(ii) f (x ) = 
x
4
1≤ x < 4
–
;

(iii) f (x ) = [x ] + | x – 1| ; – 1 ≤ x ≤ 3
(where [ ⋅ ] denotes greatest integer function)
x 4 ; x 2 < 1
(iv) f (x ) = 
2
x ; x ≥ 1
3. Is f (x ) = x 2 + x + 1 invertible? If not in which
4. If f is defined by y = f (x ) ; where x = 2 t – | t |,
y = t 2 + t | t |, t ∈ R. Then construct the graph
for f (x ).
5. Construct the graph for f (x ) = [[x ] – x ]; where
[ ⋅ ] denotes greatest integer function.
tan–1 α π
6. If 0 < α < 1 ; then show;
> .
α
4
7. Find
the
number
of
solutions
for;
cos–1(cos x ) = [x ] where [ ⋅ ] denotes the
greatest integer function.
8. Find
the
number
of
solutions
for;
[[x ] – x ] = sin x ; where [ ⋅ ] denotes the greatest
integer function.
9. Find the values of x graphically which satisfy
x2
≤ 1.
x –1
Introduction of Graphs
1. Construct the graph for;
x – 1; x < 0
 1
(i) f (x ) =  ;
x =0
4
2
x ; x > 0
10. Find the value of x for which x 3 – [x ] = 3, where
[ ⋅ ] denotes the greatest integer function.
region it is invertible.
ANSWERS
3. x ≥ –
1
2
7. 5 solutions.
8. infinite
 – 1– 5 – 1 + 5 
9. x ∈ 
,

2
2


10. x = 22 / 3 .
49
Play with Graphs
Myths About Jangent Lins
Remark 1
y
L
C
P
O
x
L meets C only at P but is not
tangent to C.
Fig. 1
Remark 2
y
C
L
P
x
O
L is tangent to C at P but meets C at
several points
Fig. 2
Remark 3
y
L
C
P
O
x
L is tangent to C at P but lies on two
sides of C, crossing C at P.
Fig. 3
50
2
CURVATURE AND
TRANSFORMATIONS
In this chapter we shall study:
➥ The bending of curves at different points.
➥ Transformations of curves.
2.1 CURVATURE
“The study of bending of curves at different points is known as curvature.” or “Rate at
which the curve curves”.
y-axis
Consider a curve and a point P on it and let Q be a point
near P. Let A be a point on the curve.
arc AP = s,
arc AQ = s + δs
∴
arc PQ = δs
A
Let ψ and ψ + δψ be angles which the tangents at P and
Q makes with x-axis.
∴ δψ is called the total curvature of the arc PQ.
δψ
is called the average curvature of the arc PQ.
δs
δψ dψ
lim
=
= curvature of the curve at P.
δs→ 0 δs
ds
Note
To study curvature we shall define
dψ
ds
or
P
ψ
Curvature and Transformations
er
t
ap
h
C
Q
ψ + δψ
O
x
Fig. 2.1
d2 y
.
dx2
2.2 CONCAVITY, CONVEXITY AND POINTS OF INFLEXION
(a) Concave upwards
If in the neighbourhood of a point P on a curve is above the tangent at P, it is said to be concave
upwards.
Mathematically
dy
d 2y
increases as x increases.
⇒
>0
dx
dx 2
51
Geometrically
y
y
Tangent
Play with Graphs
Tangent
P
P
x
O
x
O
Fig. 2.2
(b) Convex upwards or Concave downwards
If the curve is below the tangent at P, it is said to be convex upward or concave downward.
Mathematically
dy
decreases as x decreases.
dx
Geometrically
⇒
d 2y
dx 2
<0
y
y
Tangent
Tangent
P
P
x
O
x
O
Fig. 2.3
(c) Point of Inflexion
If at a point P, a curve changes its concavity from upwards to downwards or vice versa. Then P is
called point of inflexion.
Geometrically
y
y
P
O
P
x
O
x
Fig. 2.4
Mathematically
d 2y
(i)
= 0 at the point.
dx 2
d 2y
d 2y
d 3y
changes
its
sign
as
x
increases
through
the
value
at
which
;
i.e.,
(ii)
=
0
≠ 0.
dx 2
dx 2
dx 3
52
; d 2y
0
Concave up
Tangent
x
O
Here;
P
dx
dx 2 >
Concave
0
P
up
Tangent
; d 2
dx 2 y
>
Tangent
0
0
; d 2y
dx 2 < 0
P
Concave down
dy
>
dy
dx <
0
P
dy
dx >
dy
dx < 0
0
Tangent
Concave down
y
; d 2
dx 2 y
<
y
x
O
dy
< 0 for both curves.
dx
Here;
dy
> 0 for both curves.
dx
Fig. 2.5
2.3 PLOTTING OF ALGEBRAIC CURVES USING CONCAVITY
Here; if y = f ( x) = ( x – α )( x – β ). 0 < α < β.
Then we know it has roots α and β and would be
plotted as shown in Fig. 2.6.
From above discussion it becomes clear that to
plot curves we require;
(i) Point of intersection on x-axis. (i.e., y = 0)
(ii) Point of maximum and minimum value.
(iii) Interval for which function increases or
decreases.
(iv) Point at which concave up, concave down
and point of inflexion.
SOLUTION
(i) Put
1
Sketch
O
x
β
α
min
Concave up
when x < x0
Concave up
x = x 0 when x > x0
Fig. 2.6
y = ( x – 1)( x – 2).
Here; y = ( x – 1)( x – 2)
y =0 ⇒
x = 1, 2 .
y = x – 3x + 2
dy
d 2y
= 2x – 3 and
=2
⇒
dx
dx 2


d 2y
3
 as
∴ minimum at x =
> 0
2
2
dx


3
and decreases when
(iii) Increases when x >
2
3
x< .
2
(ii)
y = f (x) = (x – α)(x – β)
(for point of intersection on x-axis.)
2
y
3/2
x<
ng
asi
3/2
cre
x<
De
up
ve
nca
Co
O
1
3/2
–1/4
Inc
rea
sin
Co
nca g x >
3/2
ve
up
x>
3/2
EXAMPLE
y
Curvature and Transformations
In general we can represent concavity as;.
Note
x
2
minimum x = 3/2
Fig. 2.7
53
3
3
or x < .
2
2
∴ Graph is sketched as shown in Fig. 2.7.
EXAMPLE
2
Sketch the curve
y = ( x – 1)( x – 2)( x – 3).
Here; y = ( x – 1)( x – 2)( x – 3)
(i) Put y = 0
⇒
x = 1, 2 , 3.
(ii) y = x 3 – 6x 2 + 11x – 6
dy
d 2y
and
⇒
= 6x – 12
= 3x 2 – 12x + 11
dx
dx 2
dy
6± 3
when
=0
⇒
x=
dx
3
6– 3
d 2y
as
∴ maximum when;
=–2 3
x=
3
dx 2
6+ 3
d 2y
as
minimum when;
=2 3
x=
3
dx 2
SOLUTION
1 6 – √3 2
2
Concave
down for
x<2
–6
Fig. 2.8
SOLUTION
54
6 + √3
2
3
minimum
Sketch the curve y = ( x – 1) 2 ( x – 2).
Here; y = ( x – 1) 2 ( x – 2)
(i) Put y = 0
⇒
x = 1, 1, 2
3
2
(ii) y = x – 4x + 5x – 2
dy
and
= 3x 2 – 8x + 5
⇒
dx
dy
5
when
=0
⇒
x = 1,
dx
3
d 2y
= – 2.
∴ maximum when; x = 1 as
dx 2
eas
ing
O
ing
3
maximum
as
EXAMPLE
x<
cre
6– 3
3
6+ 3
or
x>
3
6– 3
6+ 3
decreases when;
< x<
3
3
(iv) Concave
upwards
when x > 2 and
concave down when x < 2 .
∴ Graph is sketched as shown in Fig. 2.8.
⇒ Increases when;
de

6 – 3 
6 + 3

 x –
= 3 x –
3  
3 

ing
y
Incr
dy
= 3x 2 – 12x + 11
dx
eas
(iii) Here;
Incr
Play with Graphs
(iv) Concave upwards for x >
d 2y
dx 2
= 6x – 8
Concave
up for
x>2
x
dx 2
= 2.
sin
g
d 2y
dy
5

= 3x 2 – 8x + 5 = 3( x – 1)  x – 

dx
3
5
⇒ Increases when;
x < 1 or x > .
3
5
Decreases when;
1< x< .
3
4
and
(iv) Concave up when x >
3
4
concave down when x < .
3
∴ Graph is sketched as shown in Fig. 2.9.
(iii) Here;
EXAMPLE
4
rea
5
as
3
maximum
at x = 1
O
1de
sin
g
4 5
3 3
In
cr
ea
cr
ea
s
in
g
Concave down
when x < 4/3
Inc
x=
2
minimum at
x = 5/3
Concave up
when x > 4/3
Fig. 2.9
Sketch the curve y = 3x 2 – 2x 3 .
Here; y = 3x 2 – 2x 3
3
when y = 0
…(i)
⇒
x = 0,
2
dy
also
= 6x – 6x 2 = 6x(1 – x)
dx
d 2y
and
= 6 – 12x = 6 (1 – 2x)
dx 2
d 2y
= – 6
⇒ maximum when;
x = 1 as
dx 2

 …(ii)
2
d y

and minimum when; x = 0 as
=
6

2
dx
y increases when;
0 < x < 1.

 …(iii)
y decreases when;
x < 0 and x > 1

SOLUTION
1/2
1 3/2
x
sin
ea
g
also;
minimum O
when x = 0
maximum
when x = 1
ing
cr
⇒
as
re
Inc
De
Here;
ng
SOLUTION
asi
5
cre
EXAMPLE
x<
De
1
2
1
Concave down when;
x>
2
∴ Graph is sketched as shown in Fig. 2.10.
Concave up when;
y
Concave up
Curvature and Transformations
minimum when;
Concave down
x = 1/2
Fig. 2.10


 …(iv)


Sketch the graph for the function:
f ( x) = |x + 3|( x + 1).
( x + 3)( x + 1); x ≥ – 3
y = |x + 3|( x + 1) = 
 – ( x + 3)( x + 1); x < – 3
x = – 1, – 3 when
dy  + 2x + 4; x ≥ – 3
=
dx  – 2x – 4; x < – 3
y =0
2
and
d y
dx
2
 2;
=
– 2 ;
…(i)
x≥ – 3
x< – 3
55
y
Inc
rea
sin
g
Play with Graphs
3
maximum at x = –3
–2
x
O
–1
1
2
asin
g
–3
Incr
e
minimum at x = –3
Concave up
Concave down
x = –3
Fig. 2.11
x < – 3 or x > – 2

decreasing when – 3 < x < – 2

maximum at x = – 3

minimum at x = – 2 
Increasing when
⇒
…(ii)
…(iii)
concave up when x < – 3


concave down when x > – 3
Note
…(iv)
Above example could also be solved by using transformations discussed in later part of
chapter.
EXAMPLE
6
Sketch the graph for: f ( x) =
Here; y =
SOLUTION
x+1
x2 + 3
x = – 1 when y = 0

1

when x = 0
y =
3

⇒
x2 + 3
x+1
…(i)
Increasing when; – 3 < x < 1

dy – x 2 – 2x + 3 – ( x + 3)( x – 1)
⇒
=
=

2
2
2
2
Decreasing when; x < – 3 or x > 1
dx
( x + 3)
( x + 3)
…(ii)
y
maximum
1/2
1/3
–5
–4
–3
–2
(–3, –1/6) minimum
–1
O
1
–1/6
Fig. 2.12
56
(1, 1/2)
2
3
4
5
x
d 2y
also;
dx 2
minimum at
x=– 3
as;
maximum at
x=1
as;
Note



3
2
2( x + 3x – 9x – 3)
–
+
decn –3 Incn
=
d 2y
–
1 decn
( x 2 + 3) 3
1
> 0
36
dx


d 2y
1

=
–
<
0

4
dx 2
2
=+
…(iii)
In above curve x-axis works as asymptote, i.e., the curve would never meet x-axis. For
detail refer chapter 3.
2.4 GRAPHICAL TRANSFORMATIONS
Here, we shall discuss the transformations as;
(i) f ( x) transforms to f ( x) ± a
(ii) f ( x) transforms to f ( x ± a).
(iii) f ( x) transforms to ( a f ( x))
(iv) f ( x) transforms to f ( ax).
(v) f ( x) transforms to f (– x).
(vi) f ( x) transforms to – f ( x).
(vii) f ( x) transforms to – f (– x)
(viii) f ( x) transforms to|f ( x)|.
(ix) f ( x) transforms to f (|x |).
(x) f ( x) transforms to|f (|x|)|.
(xi) y = f ( x) transforms to|y | = f ( x).
(xii) y = f ( x) transforms to|y| = |f ( x)|.
(xiii) y = f ( x) transforms to|y| = |f (|x|)|.



 Where| • | means modulus or absolute value function.



(xiv) y = f ( x) transforms to y = [ f ( x)].
(xv) y = f ( x) transforms to y = f ([ x]).
(xvi) y = f ( x) transforms to y = [ f ([ x])].
(xvii) y = f ( x) transforms to [ y ] = f ( x).
(xviii) y = f ( x) transforms to [ y ] = [ f ( x)].



 Where [ • ] denotes greatest integer less than or equal
 to x.

(xix) y = f ( x) transforms to y = f ({ x}).
(xx) y = f ( x) transforms to y = { f ( x)}.
(xxi) y = f ( x) transforms to y = { f ({ x})}.
(xxii) y = f ( x) transforms to { y } = f ( x).
(xxiii) y = f ( x) transforms to { y } = { f ( x)}.



 Where { • } denotes fractional part of x.


Curvature and Transformations

dy
,
 Using number line rule for
dx

(xxiv) y = f ( x) transforms to y = f –1 ( x), f –1 ( x) represents inverse of f ( x).
57
Now, we shall study the following cases as;
Play with Graphs
(i) When f (x), transforms to f (x) ± a. (where a is + ve)
i.e.,
f ( x) → f ( x) + a
shift the given graph of f ( x) upward through ‘a’ units
again,
f ( x) → f ( x) – a
shift the given graph of f ( x) downward through ‘a’ units.
Graphically it could be stated as:
y
y = f (x ) + a
a
a
a
aa
a
y = f (x )
a
a
a
O
x where a > 0
a
y = f (x ) – a
Fig. 2.13
EXAMPLE
1
y = e x + 1; y = e x – 1, with the help of y = e x .
Plot
We know; y = e x
SOLUTION
(exponential function) could be plotted as;
y
y = ex
(0, 1)
x
O
Fig. 2.14
⇒
y = e x + 1, is shifted upwards by 1
unit, shown as
y
Also y = e x – 1, is shifted downwards by
1 unit, shown as
y
y = ex + 1
y = ex
y = ex
y = ex – 1
1
2 1
1
1
1
1
O
Fig. 2.15
58
x
1
1
1
1
1 1
O
Fig. 2.16
x
SOLUTION
2
Plot
y = |x| + 2 and
y = |x| – 2 , with the help of y = |x|.
We know; y = |x|(modulus function) could be plotted as;
y
y = |x | = x ; x > 0
y = |x | = –x ; x < 0
2
1
–2
–1
1
O
x
2
Fig. 2.17
⇒ y = |x| + 2 is shifted upwards by 2 units.
y
y = |x | + 2 = –x + 2 ;
x<0
y = | x | = –x ;
x<0
y = |x | + 2 = x + 2 ; x > 0
y = |x | = x ; x > 0
2
2
2
2
2
2
x
O
Fig. 2.18
also y = |x| – 2 is shifted downwards by 2 units.
y
y = |x | = – x ;
x<0
y = |x | – 2 = –x – 2 ;
x<0
y = |x | = x ;
x>0
2
2
2
2
O
y = |x | – 2 = x – 2;
x>0
Curvature and Transformations
EXAMPLE
x
2
–2
Fig. 2.19
EXAMPLE
SOLUTION
3
Plot
y = sin –1 x; y = (sin –1 x) + 1 and
y = (sin –1 x) – 1.
We know, y = sin –1 x (Inverse trigonometric) could be plotted as;
y
π/2
O
–1
y = sin–1x ; where –1 ≤ x ≤ 1 and
–π/2 ≤ y ≤ π/2
x
1
–π/2
Fig. 2.20
59
⇒
and
y = sin –1 x + 1, is shifted upwards by 1 unit.
y = sin –1 x – 1, is shifted downwards by 1 unit.
Play with Graphs
y
–1
(1, π/2 + 1) y = (sin x ) + 1
π/2 + 1
(1, π/2) y = sin–1x
π/2
1
–1
(1, π/2 – 1) y = (sin x ) – 1
O
(–1, –π/2 + 1) –1
–1
(–1, –π/2)
1
x
–π/2
–π/2 – 1
(–π/2 – 1)
x = –1
x=1
Fig. 2.21
(ii) f (x) transforms to f (x – a)
i.e., f ( x) → f ( x – a); a is positive. Shift the graph of f ( x) through ‘a’ unit towards right f ( x )
transforms to f ( x + a ).
i.e., f ( x) → f ( x + a); a is positive. Shift the graph of f ( x) through ‘a’ units towards left.
Graphically it could be stated as
y
y = f (x + a )
y = f (x)
a
y = f (x – a)
a
O
–a
a
a
x
a
a
a
Fig. 2.22
EXAMPLE
SOLUTION
⇒
1
Plot
y = |x|,
y = |x – 2| and
y = |x + 2 |.
As discussed f ( x) → f ( x – a); shift towards right.
y = |x – 2 | is shifted ‘2’ units towards right.
y
y = |x| = –x ; x < 0
y = |x – 2| = –x + 2; x < 2
y = |x | = x ; x > 0
2
2
y = |x – 2| = (x – 2); x > 2
2
2
O
60
2
2
Fig. 2.23
x
y
y = |x| = –x ; x < 0
y = |x + 2 | = –(x + 2); x < –2
y = |x + 2 | = x + 2; x > –2
y = |x | = x ; x > 0
2
2
2
2
x
2 O
–2
Fig. 2.24
EXAMPLE
2
π

y = sin  x – 

2
Plot
π

y = sin  x +  .

2
and
As we know; y = sin x could be plotted as;
SOLUTION
y
π
2
–2π
y = sin(x + π/4)
–
π
2
–π
3π
2
–
π
2
π
2
O
π
2
y = sin(x – π/2)
π
π
2
x
2π
3π
2
π
2
y = sin x y = sin(x – π/4)
y=1
y = sin(x + π/4) y = sin x
y = –1
Curvature and Transformations
y = |x + 2 | is shifted ‘2’ units towards left.
also
Fig. 2.25
EXAMPLE
SOLUTION
3
y = sin –1 x;
Plot
y = sin –1 ( x – 1) and
y = sin –1 ( x + 1).
y
We know;
⇒
y = sin –1 ( x – 1) is shifted ‘1’ unit towards right.
and
y = sin –1 ( x + 1) is shifted ‘1’ unit towards left.
Shown as in Fig. 2.27.
(–1, π/2)
y = sin–1(x + 1)
y = sin–1x
π/2
1
1
1
–2
1
1
y = sin–1(x – 1)
1
O
–1
1
2
O
–1
y
(1, π/2)
π/2
y = sin –1 x could be plotted as shown in Fig. 2.26.
1
x
–π/2
Fig. 2.26
x
–π/2
Fig. 2.27
61
(iii) f (x) transforms to a f (x)
Play with Graphs
i.e.,
y
f ( x) → af ( x); a > 1
Stretch the graph of f ( x) ‘a’ times along y-axis.
f ( x) →
1
f ( x); a > 1.
a
x
Shrink the graph of f ( x) ‘a’ times along y-axis.
Graphically it could be stated as shown in
Fig. 2.28.
EXAMPLE
1
Plot y = x; y = 2x and
y =
Fig. 2.28
1
x.
2
y
SOLUTION
As we know graph for y = 2x.
y = 2x
(1, 2)
2
y = 2x; is stretch of f ( x) ‘2’ times
1
along y-axis and y = x; is shrink of f ( x) ‘2’
2
times along y-axis.
y=x
∴
1
(1, 1)
O
1
–1
(–1, –1/2)
(–1,–2)
–2
Fig. 2.29
EXAMPLE
2
Plot
y = sin x
y = 2 sin x.
and
We know; y = sin x and f ( x) → a f ( x)
⇒ Stretch the graph of f ( x) ‘a’ times along y-axis.
∴ y = 2 sin x ⇒ stretch the graph of sin x ‘2’ times along y-axis.
SOLUTION
y
y=2
2
y = 2sin x
y=1
1
–π
–π/2
y = sin x
O
–1
–2
π/2
y= 1x
2
(1, 1/2)
–1
Shown as in Fig. 2.29.
62
y = f ( x)
y = 1 f (x)
a
y = a f ( x)
x
π
y = –1
y = –2
Fig. 2.30
Above curve is plotted for the interval [– π, π] as periodic with period 2 π .
x
SOLUTION
3
Plot
y = sin x and y =
1
sin x.
2
As we know;
y
1
f ( x)
a
1
y =
⇒ shrink the graph of f ( x)‘a’
times along y-axis.
y =
∴
y = sin x
1/2
–π
–π/2
1
sin x
2
y = 12 sin x
O
π/2
π
x
–1/2
–1
⇒ shrink the graph of f ( x)
‘2’ times along y-axis.
(iv) f (x) transforms to f (ax)
i.e.,
f ( x) → f ( ax);
Fig. 2.31
y
a>1
y = f (ax) y = f (x) y = f ( 1 x)
a
Shrink (or contract) the graph of f ( x) ‘a’
times along x-axis.
again
1 
f ( x) → f  x ; a > 1
a 
1/a
–a
–1 –1/a
1
a
x
O
Stretch (or expand) the graph of f ( x) ‘a’
times along x-axis.
Graphically it could be stated as shown in
Fig. 2.32.
EXAMPLE
1
Plot
y = sin x and
Curvature and Transformations
EXAMPLE
Fig. 2.32
y = sin 2x.
SOLUTION Here; y = sin 2x, is to shrink (or contract) the graph of sin x by ‘2’ units along
x-axis. Shown as in Fig. 2.33.
y
y=1
–2π
–π
–
π
2
O
π
2
π
3π
2
x
y = –1
Fig. 2.33
From above figure sin x is periodic with period 2 π and sin 2 x with period π.
63
EXAMPLE
Play with Graphs
SOLUTION
2
y = sin x and
Plot
x
y = sin .
2
 x
Here; y = sin   , is to stretch (or expand) the graph of sin x ‘2’ times along
 2
x-axis. Shown as in Fig. 2.34.
y
y = sin (x /2)
y = sin x
–2π
–
3π
2
–π
–
π
2
O
π
2
π
3π
2
x
2π
Fig. 2.34
 x
From above figure sin x is periodic with period 2 π and sin   is periodic with period 4 π.
 2
EXAMPLE
3
y = sin –1 x and
Plot
y = sin –1 ( 2x).
y
Here; y = sin –1 ( 2x), is to shrink
(or contract) the graph of sin –1 x ‘2’ times
SOLUTION
π
2
along x-axis.
y = sin–1(2x)
y = sin–1(x)
Shown as in Fig. 2.35.
–1 – 1
2
1
2
O
–
π
2
Fig. 2.35
EXAMPLE
4
SOLUTION
To plot
Plot
x

y = sin –1  – 1 .
3

x

y = sin –1  – 1 . We should follow as;
3

(i) Plot y = sin –1 x
 x
(ii) Plot y = sin –1   ,
 3
i.e., stretch graph ‘3’ units along x-axis.
x

(iii) Plot y = sin –1  – 1 , i.e., shift the graph (ii) by ‘3’ unit towards right.
3

64
1
x
y
y
π
2
π
2
y = sin–1x
–1
–
x
1
O
y = sin–1(x /3)
–3
π
2
–
Fig. 2.36
x
3
O
π
2
Fig. 2.37

x
(iii) Plot y = sin –1  – 1 :

3
y
π
2
–3
–2
O
–1
3
3
–
1
2
3
4
5
6
x
π
2
Curvature and Transformations
 x
(ii) Plot y = sin –1   :
 3
(i) Plotting y = sin –1 x :
Fig. 2.38
(iv) f ( x ) transforms to f (– x )
f ( x) → f (– x)
i.e.,
To draw y = f (– x), take the image of the curve y = f ( x) in y-axis as plane mirror.
OR
“Turn the graph of f ( x ) by 180° about y-axis.”
Graphically it is stated as;
y = f (–x)
y
y
y = f ( x)
OR
O
y = f ( x)
x
x
O
y = f (–x)
Fig. 2.39
65
Play with Graphs
EXAMPLE
1
SOLUTION
As
Plot
y = e– x .
y = e x is known; then y = e – x is the image in y-axis as plane mirror for y = e x ;
shown as;
y
y = ex
y = e–x
x
O
Fig. 2.40
EXAMPLE
2
Plot the curve
y = log e (– x).
Here; y = log e (– x); is to take mirror image of y = log e x about y-axis. Shown as;
SOLUTION
y
y = loge x
y = loge (–x )
O
(–1, 0)
(1, 0)
x
Fig. 2.41
EXAMPLE
3
Plot the curve
y = sin –1 (– x).
y
Here; y = sin (– x);
is
the
y = sin –1 ( x) about y-axis.
Shown as in Fig. 2.42.
–1
SOLUTION
π
2
y = sin–1(–x )
y = sin–1x
mirror image of
x
1
–1
–
π
2
Fig. 2.42
(vi) f (x) transforms to – f (x)
i.e.,
To draw
f ( x) → – f ( x);
y = – f ( x) take image of y = f ( x) in the x-axis as plane mirror.
OR
“Turn the graph of f ( x ) by 180° about x-axis.”
66
1
Plot the curve
y = – ex .
y
As y = e x is known;
∴ y = – e x take image of y = e x in the x-axis as plane
1
mirror.
O
SOLUTION
y = ex
x
–1
y = –e x
Fig. 2.43
EXAMPLE
2
Plot the curve
y = – (log x).
y
y = log x
SOLUTION As y = log x is given then y = – log x is
the image of y = log x in the x-axis as plane
mirror.
x
O
y = – log x
Fig. 2.44
EXAMPLE
3
SOLUTION
As
∴
Plot the curve
y = – { x}; where { ⋅ } denotes the fractional part of x.
y = { x} is known;
y = – { x} is the image of y = { x} about x-axis as plane mirror.
Curvature and Transformations
EXAMPLE
y
1
y = {x }
–3
–2
–1
1
O
2
3
y = –{x }
x
–1
Fig. 2.45
(vii) f (x) transforms to – f (– x)
i.e.,
f ( x) → – f (– x);
to draw y = – f (– x)take image of f ( x)about y-axis to obtain f (– x)and then take image of f (– x)about
x-axis to obtain – f (– x).
∴
f ( x) → – f (– x)
67
⇒ (i) Image about y-axis.
Graphically it could be stated as;
(ii) Image about x-axis.
Play with Graphs
y
y
y = –f (–x)
OR
y = f (x)
y = –f (–x)
O
x
–a
–b
x
O
y = f (–x)
y = f (x)
y = –f (x)
Fig. 2.46
EXAMPLE
1
Plot the curve
y = – e– x .
y = e–x
SOLUTION
As
y = e x is known;
y
y = ex
1
(i) Take image about y-axis; for y = e .
–x
O
–1
(ii) Take image of y = e – x about x-axis; for y = – e – x .
Shown as in Fig. 2.47.
y = –e–x
Fig. 2.47
EXAMPLE
2
SOLUTION
As
Plot the curve y = – log(– x).
y = log x is known;
(i) Take image about y-axis,
for y = log(– x).
(ii) Take image of y = log(– x) about x-axis,
for y = – log(– x).
y
y = log x
y = log (–x)
O
y = –log (–x)
Fig. 2.48
68
x
x
3
Plot the curve
y = – {– x}. (where { ⋅ } denote fractional part).
As we know the curve for y = { x}.
To plot y = – {– x}
∴
(i) Take image about x-axis.
(ii) Take image about y-axis.
SOLUTION
y
1
y = {x }
y = {–x }
x
O
y = –{–x }
–1
Fig. 2.49
EXAMPLE
4
Plot the curve for y = – [– x]. (where [ ⋅ ] denotes the greatest integer function.)
As we know the curve for y = [ x].
to plot y = – [– x]
∴
(i) Take image about x-axis.
SOLUTION
y
3
Curvature and Transformations
EXAMPLE
y = [x]
2
1
–4
–3
–2
–1
O
1
2
3
4
5
x
–1
–2
–3
y = [–x]
–4
Fig. 2.50
69
(ii) Take image about y-axis.
y
y = –[–x]
Play with Graphs
3
2
1
–3
–4
–2
O
–1
1
2
3
x
4
–1
–2
–3
–4
Fig. 2.51
(viii) f (x) transforms to|f (x)|; (where| • |represents modulus function)
i.e.,
f ( x) → |f ( x)|
Here; y = |f ( x)|is drawn in two steps.
(a) In the I step, leave the positive part of f ( x), {i.e., the part of f ( x) above x-axis) as it is.
(b) In the II step, take the mirror image of negative part of f ( x). {i.e., the part of f ( x) below
x-axis} in the x-axis as plane mirror.
OR
Take the mirror image (in x-axis) of the portion of the graph of f ( x) which lies below x-axis.
OR
Turn the portion of the graph of f ( x) lying below x-axis by 180° about x-axis.
Graphically it could be stated as
Graph of f ( x ) :
Graph for| f ( x )| :
y
y
y = |f (x)|
–1
O
1
x
–1
O
y = f (x)
Fig. 2.52
70
Fig. 2.53
1
x
Above transformation of graph is very important as to discuss differentiability of f (x).
As from above example we could say y = f (x) is differentiable for all x ∈ R – {0 }.
But; y = | f (x)| is differentiable for all x ∈ R – {– 1, 0 , 1} as, “at sharp edges function is not
differentiable.”
EXAMPLE
Draw the graph for y = |log x|.
1
To draw graph for y = |log x|we have to follow two steps:
(i) Leave the ( + ve) part of y = log x, as it is
(ii) Take images of (– ve) part of y = log x, i.e., the part below x-axis in the x-axis as plane mirror.
Shown as:
SOLUTION
Graph for y = log x :
Graph for y =|log x|:
y
y
y = log x
y = | log x |
x
1
O
O
1
x
Fig. 2.54
Fig. 2.55
which is differentiable for all x ∈ ( 0, ∞ )
which is clearly differentiable for all
x ∈ ( 0, ∞ ) – {1}. “as at x = 1 their is a sharp
edge”.
EXAMPLE
2
Curvature and Transformations
Note
Draw the graph for y = |x 2 – 2x – 3|.
As we know the graph for y = x 2 – 2x – 3 = ( x – 3)( x + 1) is a parabola; so to
sketch y = | x 2 – 2x – 3 | we have to follow two steps.
(i) Leave the positive part of y = x 2 – 2x – 3, as it is.
(ii) Take the image of negative part of y = x 2 – 2x – 3, i.e., the part below x-axis in the x-axis
as plane mirror shown as in Fig. 2.56.
SOLUTION
Graph for y = x 2 – 2x – 3 = ( x – 3)( x + 1) :
y
3
2
1
–3
–2 –1
O
1
2
3
x
–1
–2
–3
–4
(1, –4)
Fig. 2.56
b
2a
D
and y = –
4a
minimum at x = –
71
Graph for y =| x 2 – 2x – 3| :
Play with Graphs
y
(1, 4)
4
3
–1 O
1
x
3
Fig. 2.57
Clearly above curve is differentiable for all x ∈ R – {– 1, 3}.
EXAMPLE
3
Sketch the graph for y = |sin x|.
Here; y = sin x is known.
∴ To draw y = |sin x |, we take the mirror image (in x-axis) of the portion of the graph of sin x
which lies below x-axis.
SOLUTION
y
y = |sin x|
–2π
–π
π
O
2π
3π
x
Image of portion below
x-axis
Fig. 2.58
From above figure it is clear;
y = |sin x|is differentiable for all x ∈ R – { nπ; n ∈ integer}.
(ix) f (x) transforms to f (|x|)
i.e.,
f ( x) → f (|x|).
If we know y = f ( x), then to plot y = f (|x|), we should follow two steps:
(i) Leave the graph lying right side of the y-axis as it is.
(ii) Take the image of f ( x) in the right of y-axis as the plane mirror and the graph of f ( x) lying
leftward of the y-axis (if it exists) is omitted.
OR
Neglect the curve for x < 0 and take the images of curve for x ≥ 0 about y-axis.
72
y
y
y = f (x )
y = f (|x |)
Neglected
x
O
x
O
Image of
f (x) about y-axis
when x > 0
Fig. 2.59
EXAMPLE
1
Sketch the curve
y = log|x|.
SOLUTION As we know, the curve y = log x.
∴ y = log|x|could be drawn in two steps:
(i) Leave the graph lying right side of y-axis as it is.
(ii) Take the image of f ( x) in the y-axis as plane mirror.
y
y
y = log x
O
y = log |x |
x
1
–1
O
x
1
Curvature and Transformations
Graphically shown as;
Fig. 2.60
EXAMPLE
SOLUTION
2
Plot the curve
y = e| x| .
As we know the curve for y = e x .
y
y
y = e|x |
y = ex
Image
for y = f (x); x ≥ 0
(1, 0)
neglect
(1, 0)
O
Fig. 2.61
∴ To plot
Fig. 2.62.
x
O
x
Fig 2.62
y = e , neglect the curve for x < 0and take image about y-axis for x ≥ 0. Shown as in
| x|
73
EXAMPLE
Plot the curve y = sin|x|.
3
SOLUTION
Play with Graphs
y
y
y = sin |x |
y = sin x
neglected
x
O
x
O
y = sin x
y = sin |x |
Fig. 2.63
EXAMPLE
Plot the curve
4
y = |x|2 – 2 |x| – 3 .
As we know, the curve for y = x 2 – 2x – 3 is plotted as shown in Fig. 2.64.
SOLUTION
y
y
ting
neglec
y = x 2 – 2x – 3
–1
O
x
3
1
–3
(1, –4)
–4
y = |x |2 – 2|x | – 3
–1 O
1
3
x
(1, –4)
Fig. 2.64
Fig. 2.65
∴ y = f (|x |), i.e., y = |x| – 2 |x| – 3 is to be plotted as shown in Fig. 2.65.
2
which shows y = |x|2 – 2 |x| – 3 is differentiable for all x ∈ R – { 0}.
(x) f (x) transforms to|f (|x |)|
f ( x) → |f (|x |)|
i.e.,
Here, plot the curve in two steps;
(i) f ( x) → |f ( x)|
y
(ii) |f ( x)| → |f (|x|)|
OR
y = f (x )
1
(i) f ( x) → f (|x |)
(ii) f (|x|) → |f (|x |)|.,
transformations.
i.e., (viii)
and
(ix)
O
1/2 1
–1
Graphically it could be stated as shown in Fig. 2.66.
Fig. 2.66
74
x
(ii) y =| f (| x|)|:
y
y
y = |f (x )|
1
SOLUTION
1
x
1/2
–1 –1/2 O 1/2 1
Fig. 2.67
Fig. 2.68
O
EXAMPLE
y = |f (|x |)|
1
Sketch the curve for
x
y = ||x |2 – 2 |x | – 3 |.
As we know the graph for y = x 2 – 2 x – 3, shown as;
y
y = x 2 – 2x – 3
–1
O
x
3
1
Curvature and Transformations
(i) y =| f ( x )|:
Fig. 2.69
(i) y = x 2 – 2x – 3 → y = |x|2 – 2 |x| – 3.
(ii) y =|x |2 –2 |x | – 3 → y = ||x|2 –2 |x |–3 |
y
y
(–1, 4)
2|x | –
3|
(1, 4)
–3
–1
O
1
3
y = || 2
x| –
3
x
–3
–1
O
1
3
x
–3
(–1, –4)
(1, –4)
Fig. 2.70
–3
Fig. 2.71
Clearly, above figure is differentiable for all
x ∈ R – {– 3, 0, 3}.
75
EXAMPLE
Play with Graphs
SOLUTION
Sketch the graph for y = e –|x| –
2
1
.
2
As we know the graph for y = e – x .
y
y = ex
1
x
O
Fig. 2.72
(i) y = e – x → y = e – x –
1
.
2
(ii) y = e – x –
1
1
→ y = e –|x| – .
2
2
y
y
1
1/2
1/2
–log 2
O
x
log 2
O
–1/2
–1/2
Fig. 2.73
(iii) y = e –|x| –
log 2
x
1
y = e–|x | –
2
Fig. 2.74
1
2
y
1/2
–log 2 O
log 2
1
y = e–|x | –
2
x
–1/2
Fig. 2.75
(xi) y = f (x) transforms to|y| = f (x)
Clearly|y| ≥ 0 ⇒ if f ( x) < 0; graph of|y| = f ( x) would not exist.
if f ( x) ≥ 0; |y| = f ( x) would be given as y = ± f ( x).
Hence, the graph of |y | = f ( x) exists only in the regions where f ( x) is non-negative and will be
reflected about x-axis only when f ( x) ≥ 0. “Region where f ( x) < 0 is neglected”.
76
(i) Remove (or neglect) the portion of the graph which lies below x-axis.
(ii) Plot the remaining portion of the graph, and also its mirror image in the x-axis.
Graphically it could be stated as shown in Fig. 2.76.
Graph for y = f ( x ) :
y
y
(1, 1)
1
y = f (x )
O
1/2
x
1/2 1
|y | = f (x )
(1, 1)
1
x
O
–1
neglecting
–2
(1, –1)
mirror image
about x-axis.
–2
Fig. 2.76
EXAMPLE
1
Sketch the curve |y| = ( x – 1)( x – 2).
SOLUTION
As
we
know
the
graph
for
y = ( x – 1)( x – 2), is shown in Fig. 2.77.
⇒ y = ( x – 1)( x – 2) → |y| = ( x – 1)( x – 2)
,
as shown in Fig. 2.78.
Curvature and Transformations
OR
y
y
2
1 3/2 2
O
3
x
(3/2, –1/4)
O
1
2
3
x
neglecting
Fig. 2.77
Image on x-axis,
when (x–1)(x–2) ≥ 0
Fig. 2.78
EXAMPLE
SOLUTION
2
Plot the curve |y| = sin x.
Here, we know the curve for y = sin x.
77
y
Play with Graphs
1
–2π
–π
π
O
2π
x
y = sin x
3π
–1
Fig. 2.79
∴
y = sin x → |y| = sin x
y
1
–2π
–π
π
O
3π
2π
π
neglected
x
2π
Image neglected –1 Image neglected Image
neglected
Fig. 2.80
EXAMPLE
SOLUTION
Sketch the curve |x| + |y | = 1.
3
As the graph for y = 1 – x is;
y
1
–1
O
x
y=1–x
1
–1
Fig. 2.81
(i) y = 1 – x → y = 1 – |x |.
(ii) y = 1 – |x| → |y | = 1 – |x |.
y
y
ne
gl
1
d
te
ec
1
|y | = 1 – |x | or |x | + |y | = 1
–1
1
x
ag
e
Im
d
te
ec
gl
ne
ag
Im
d
te
–1
ec
gl
Image
x
y = 1 – |x |
ne
1
e
O
ag
–1
Im
e
O
–1
Fig. 2.83
Fig. 2.82
78
Clearly above figure represents a square.
Curvature and Transformations
EXAMPLE
Sketch the curve|x| – |y | = 1 .
4
As the graph for y = x – 1 is known;
SOLUTION
y
y=x–1
x
O
1
–1
Fit. 2.84
(i) y = x – 1 → y = |x | – 1
(ii) y = |x | – 1 → |y | = |x | – 1
y
y
ag
Im
e
y = |x | – 1
O
x
–1
ne
gl
ec
te
d
1
e
d
te
–1
x
O
–1
ec
gl
ne
Im
ag
e
1
ag
Im
–1
Fig. 2.86
Fig. 2.85
(xii) y = f (x) transforms to|y| = |f (x)|
i.e.,
y = f ( x) → |y| = |f ( x)|; is plotted in two steps.
(i)
y = f ( x) → y = |f ( x)|
(ii)
y =|f ( x)| → |y| = |f ( x)|
Graphically it could be stated as;
y
y
y = f ( x)
O
(i)
x
y
y = |f (x)|
O
x
|y | = |f (x)|
O
y = f (x) → y = |f (x)|
y = |f (x)| → |y | = |f (x)|
(ii)
(iii)
x
Fig. 2.87
79
EXAMPLE
Plot the curve for|y| = |e – x|.
Here; curve for
SOLUTION
Play with Graphs
1
y = e – x is shown as;
y
y = e–x
1
x
O
Fig. 2.88
(i) y = e – x → y = |e – x|
(ii) y = |e – x| → |y| = |e – x|.
y
y
–x
y= e
1
1
remains same
O
x
O
x
–1
|y | = e–x
Fig. 2.89
Fig. 2.90
EXAMPLE
2
Plot the curve |y| = |e x – 1| .
As we know the curve for y = e x is shown as;
SOLUTION
Fig. 2.91
(i) y = e → y = e – 1.
x
(ii) y = e x – 1 → y = |e x – 1 |.
x
y
y
1
1
O
–1
80
y = |e x – 1|
Fig. 2.92
x
O
neglected
–1
Fig. 2.93
x
y
|y | = |e x – 1|
1
x
O
–1
Fig. 2.94
EXAMPLE
SOLUTION
3
Plot the curve |y| = sin x +
1
.
2
Here; we know the graph for y = sin x, is shown as
y
1
–2π
–π
π
O
2π
x
–1
Curvature and Transformations
(iii) y = |e x – 1 | → |y | = |e x – 1 |.
Fig. 2.95
1
(i) y = sin x → y = sin x + .
2
y
3/2
y = 3/2
1/2
–2π
–π
O
π
–1/2
2π
y = 1/2
x
y = –1/2
Fig. 2.96
(ii) y = sin x +
1
1
→ y = sin x +
2
2
81
y
Play with Graphs
3/2
y = 3/2
y = 1/2
1/2
–π
–2π
π
O
x
2π
y = –1/2
y = –3/2
Fig. 2.97
(iii) y = sin x +
1
1
→ |y | = sin x +
2
2
y
|y | = sin x + 1
2
y = 3/2
y = 1/2
x
O
y = –1/2
y = –3/2
Fig. 2.98
(xiii) y = f (x) transforms to|y| = |f (|x |)|
y = f ( x) → |y| = |f ( x)|.
i.e.,
The steps followed are:
(i)
y = f ( x) → y = |f ( x)|.
(ii)
y = |f ( x)| → y = |f (|x |)|
(iii)
y = |f (|x |)| → |y | = |f (|x |)|.
Graphically it could be stated as:
y
1
O
Fig. 2.99
82
x
(ii) y = |f ( x)| → y = |f (|x |)|
y
Curvature and Transformations
(i) y = f ( x) → y = |f ( x)|
y
1
1
x
x
O
ne
gl e
cte
d
O
Fig. 2.100
Fig. 2.101
(iii) y = |f (|x |)| → |y | = |f (|x |)|.
y
1
x
O
–1
Fig. 2.102
EXAMPLE
SOLUTION
1
Plot the|y| = e –|x| –
1
.
2
Here; we know the graph for y = e – x .
y
1
x
O
Fig. 2.103
1
(i) y = e – x → y = e – x –
2
(ii) y = e – x –
1
1
→ y = e –|x| – .
2
2
y
y
1/2
1/2
O log 2
–1/2
Fig. 2.104
log 2
–log 2
O
x
x
–1/2
y = –1/2
Fig. 2.105
83
(iii) y = e –|x| –
1
1
.
→ y = e –|x| –
2
2
(iv) y = e –|x| –
1
1
.
→ |y | = e –|x| –
2
2
Play with Graphs
y
y
1/2
–log 2 O
1/2
x
log 2
–1/2
Fig. 2.106
SOLUTION
2
Fig. 2.107
Plot |y| = |log|x ||.
Here; y = log x is plotted as;
y
y = logex
x
1
O
Fig. 2.108
(i) y = log x → y = log|x|
(ii) y = log|x | → y = |log|x ||
y
y
y = | log|x| |
–1
O
1
x
O
–1
neglected
(iii) y = |log|x || → |y | = |log|x ||
y
|y | = | log |x | |
–1 O
84
1
neglected
Fig. 2.110
Fig. 2.109
1
Fig. 2.111
x
1
2
x
O
–1/2
EXAMPLE
|y | = e–|x |–
x
SOLUTION
3
Sketch the curve |y | = ||x |2 – 3 |x | – 2| .
As we know the graph for y = x 2 – 3x – 2 .
y
–1 O
1
x
3
–3
Fig. 2.112
(i) y = x 2 – 3x – 2 → y = |x |2 – 3 |x | – 2
(ii)
y
y
3
–1 O
1
x
3
–3
–1 O
1
–3
3
x
Curvature and Transformations
EXAMPLE
Fig. 2.114
Fig. 2.113
y = |x |2 – 3 |x | – 2 → y = ||x |2 – 3 |x | – 2 |.
(iii) |y | = ||x|2 – 3|x| – 2|
y
3
–3
–1 O
|y | = ||x |2 – 3|x | – 2|
1
3
x
–3
Fig. 2.115
85
(xiv) y = f (x) transforms to y = [f (x)] ; (where [ • ] denotes the greatest integer function)
f ( x) → [ f ( x)]
Play with Graphs
i.e.,
Here; in order to draw y = [ f ( x)] mark the integer on y-axis. Draw the horizontal lines through
integers till they intersect the graph. Draw vertical dotted lines from these intersection points; finally
draw horizontal lines parallel to x-axis from any intersection point to the nearest vertical dotted line
with blank dot at right end in case f ( x) increase.
OR
Step 1.
Step 2.
Step 3.
Step 4.
Plot f ( x).
Mark the intervals of unit length with integers as end points on y-axis.
Mark the corresponding intervals {with the help of graph of f ( x)} on x-axis.
Plot the value of [ f ( x)] for each of the marked intervals.
Graphically it could be shown as:
3
2
y = f (x)
1
O
–1
–2
x1 x2
x3
x4
x5 x6 x7 x8
3
2
1
O
y = [f (x)]
–1
–2
x1 x2
x3
x4
x5 x6 x7 x8
Fig. 2.116
EXAMPLE
SOLUTION
86
1
Sketch the curve
y = [sin x].
Here, sketch for y = sin x is shown as in Fig. 2.117.
1
–2π –π
–3π
π
O
2π
x
3π
y = sin x
–1
Fig. 2.117
y = sin x → y = [sin x]
y
1
x
O
–1
–3π– 5π –2π – 3π –π x = – π
2
2
2
y = [sin x ]
x=
π π
2
3π 2π 5π 3π
2
2
Fig. 2.118
EXAMPLE
2
Sketch the curve y = [ x 2 – 1]. (where [ ⋅ ] denotes greatest integer function).
Curvature and Transformations
y
When – 2 ≤ x ≤ 2 .
Here y = x 2 – 1
could be plotted as shown in Fig. 2.119.
SOLUTION
(i) y = x 2 – 1 → y = [ x 2 – 1].
y
y
–√3
–2 –√2 –1
3
3
2
2
1
1
1 √2 √3 2
O
–1
Fig. 2.119
x
–√3
–2 –√2 –1
1 √2 √3 2
O
x
–1
y = [x 2 – 1]; –2 ≤ x ≤ 2
Fig. 2.120
87
EXAMPLE
3
Sketch the curve; y = [ 2 – x 2 ] ; where [ ⋅ ] denotes the greatest integer
Play with Graphs
function.
SOLUTION
We know, y =
2 – x2
y ≥ 0.
represents a circle for
Shown as in Fig. 2.121.
y
2
√2
1
– √2 –1
0
Thus, the graph for y = [ 2 – x2 ]
y
y = √2 – x 2
x
√2
1
√2
1
– √2 –1
0
y = √2 – x 2
x
√2
1
Fig. 2.121
(xv) y = f (x) transforms to y = f ([x])
Here, mark the integers on the x-axis. Draw vertical lines till they intersect the graph of f ( x).
From these intersection points draw horizontal lines (parallel to x-axis) to meet the nearest right
vertical line, with a black dot on each nearest right vertical line which can be shown as in Fig. 2.122.
y = f (x)
y = f ([ x])
y
–5
–4 –3
–2
–1
O
1
Fig. 2.122
88
2
3
4
5
x
y = f ([x])
–5
–4 –3
–2
O
–1
1
2
3
4
5
x
Fig. 2.123
OR
y = f ( x ) → y = f ([ x])
Step 1. Plot the straight lines parallel to y-axis for integral values of x
(say – 3, – 2, – 1, 0, 1, 2, 3, …)
Step 2. Now mark the points at which x = – 3, x = – 2, x = – 1, x = 0, x = 1, … on the curve.
Step 3. Take the lower marked point for x say if n < x < n + 1 then take the point at x = n and
draw a horizontal line to the nearest vertical line formed by x = n + 1, proceeding in this way we get
required curve.
EXAMPLE
SOLUTION
1
y = e[x ].
Plot the curve
Here the graph for y = e[x ] is shown as;
y = ex
y
Curvature and Transformations
y
y = e [x ]
–4
–3
–2
–1
O
1
2
3
x
Fig. 2.124
EXAMPLE
SOLUTION
2
Sketch the curve
y = sin[ x] when
–2 π ≤ x ≤ 2 π.
The curve for
y = sin[ x]; could be plotted as shown in Fig. 2.125.
89
Play with Graphs
y
–6
–5
–4 –3
–2
O 1
–1
2
3
4
5
x
6
y = sin x
y = sin [x ]
Fig. 2.125
EXAMPLE
SOLUTION
3
Sketch the curve for y = cos[ x] ; – π ≤ x ≤ π.
The curve for
y = cos[ x] could be plotted as;
y
y = cos [x ]
–π –3
–2
x
2 3 π
O 1
–1
Fig. 2.126
EXAMPLE
SOLUTION
4
Plot the curve
The curve for
y = [ x]2 ; –2 ≤ x ≤ 2 .
y = [ x]2 ; –2 ≤ x ≤ 2 could be plotted as;
y
y = x2
4
y = [x ] 2
1
–2
–1
O
1
Fig. 2.127
(xvi) y = f (x) transforms to y = [f ([x])]
Here, we should follow two steps;
90
2
x
EXAMPLE
Sketch the curve y = [sin[ x]]; where [ ⋅ ] denotes the greatest integral function
1
when 0 ≤ x ≤ π.
SOLUTION
Here; first we shall plot the curve for y = sin[ x], when x ∈[ 0, π].
y
y
y = sin [x ]; 0 ≤ x ≤ π
sin 2
sin 1
sin 3
O
sin 2
sin 1
sin 3
1
2
x
3 π
O
1
2
x
3 π
y = [sin [x]]; 0 ≤ x ≤ π
Fig. 2.128
From above figure we conclude that;
0 ≤ x ≤ π ⇒ y = sin[ x] ∈ [ 0, 1).
when
⇒
y = [sin[ x]] → 0 for all
EXAMPLE
SOLUTION
2
0 ≤ x ≤ π.
y = [ e[x ]] ; when – 4 < x < 2 .
Plot the curve
Here to sketch y = [ e[x ]], we should follow the steps as;
(i) y = e
(ii) y = e[x ]
(iii) y = [ e[x ]]
x
(i) y = ex
Curvature and Transformations
y = f ( x) → y = f ([ x])
y = f ([ x]) → y = [ f ([ x])]
(i)
(ii)
(ii) y = e[ x ]
y
y
(2, e2)
4
3
2
e 2.7
(1, e)
1 (–1, 1 )
)
(–2,
2
e
1
e
(0, 1)
(–3, 3 )
e
(1, e)
1 (–1, 1 ) 1
e
1 (–2, 2 )
(0, 1)
e
)
(–3, 3
e
–4
–3
–2
–1
O
Fig. 2.129
1
–4
2
–3
–2
–1
O
1
2
x
x
Fig. 2.130
91
(iii) y = [ e[ x ]]; from Fig. 2.130.
y
Play with Graphs
 0 < e[x ] < 1; x < 0

e[x ] =  e[x ] = 1
; 0≤ x< 1
 [x ]
 e = 2.7 ; 1 ≤ x < 2
Thus;
e
1
y = [ e[x ]]
x< 0
 0;

⇒ y = 1; 0 ≤ x < 1 shown as in
 2; 1 ≤ x < 2

Fig. 2.131.
Graph for;
y = [e [x] ]
2
y = [ e[x ]];
when
–4
–3
–2
–1
O
1
2
x
Fig. 2.131
x < 2.
(xvii) y = f (x) transforms to [y ] = f (x) / Graph of Candles
Here, to plot[ y ] = f ( x); we check only those points for which f ( x)∈ integers, as[ y ] ∈integers for
all x.
Thus; [ y ] = f ( x) represents only integral values of y. Here, domain of f ( x) are set of values of x
for which f ( x) ∈integers.
EXAMPLE
1
Sketch the curve;
[ y ] = sin x.
– 1 ≤ sin x ≤ 1 but since; [ y ] = sin x.
SOLUTION
As we know;
⇒
sin x = – 1, 0, 1 are only solutions;
x=
or
nπ
∈ Domain of [ y ] = sin x.
2
Thus, [ y ] = sin x is shown as in Fig. 2.132.
Graph for y = sin x :
y
1
–2π
– 3π
2
–π
O –π
2
–π
2
–1
Fig. 2.132
92
π
3π
2
2π
x
y
–
(–3π, 0)
(–2π, 0)
F′
2
3π , – 1
2
D′
E′
5π , – 1
2
–
C′
(–π, 0)
A
B′
5π , 1
2
π,1
2
1
O (0, 0)
A′
π , –1
2
–
(π, 0)
E
(2π, 0)
B
D
C
3π , – 1
2
x
[y ] = sin x
Fig. 2.133
From above figure, (the points marked O, A, B, C, D, E, A ′ , B′ , C ′ , D′ , E′ ,… is the graph for
candles), or graph for [ y ] = sin x.
EXAMPLE
SOLUTION
Sketch the curve [ y ] = sin –1 x.
2
As we know the graph for y = sin –1 x; shown as in Fig. 2.134 and 2.135.
Graph for y = sin –1 x;
Graph for [ y ] = sin –1 x;
y
y
2
π/2
π/2
C (sin 1, 1)
1
1
–1 sin(–1)
O sin 1
1
x
–1
–π/2
–1
B (0, 0)
–sin 1 O
A
x
sin 1 1
–1
(–sin 1, –1)
Fig. 2.134
Curvature and Transformations
Graph for [ y ] = sin x :
–π/2
Fig. 2.135
From above figure, [ y ] = sin x.
(xviii) y = f (x) transforms to [y ] = [f (x)]
As we have earlier discussed y = [ f ( x)], i.e., transformation (xiv), and we know [ y ] implies
only those values of x for which f ( x) ∈ integer.
EXAMPLE
SOLUTION
1
Sketch the curve; [ y ] = [sin x] .
To sketch the curve [ y ] = [sin x] we first plot y = [sin x].
93
(i) Graph for y = [sin x]
Play with Graphs
y
2
1
–2π –π
–3π
O
π
2π
x
3π
–1
Fig. 2.136
(ii) Graph for [ y ] = [sin x]
y
2
(–3π/2, 1)
(π/2, 1)
1
–2π –π
–3π
O
(5π/2, 1)
π
2π
4π
3π
x
–1
Fig. 2.137
From above figure it is clear [ y ] = [sin x] is periodic with period 2 π.
(xix) y = f (x) transforms to y = f ({x}); (where { ⋅ } denotes fractional part of x,
i.e., {x} = x – [x] )
f ( x) → f ({ x})
Graph of f ( x – [ x]) or f ({ x}) can be obtained from the graph of f ( x) by following rule.
“Retain the graph of f ( x) for values of x lying between interval [0, 1). Now it can be repeated for
rest of the points. (taking periodicity 1).
New obtained function is graph for y = f ({ x})”.
Graphically it could be stated as;
Graph for y = f ( x )
Graph for y = f ({ x})
y
y
y1
–3
94
–2
–1 O
neglecting
y1
1
2
Fig. 2.138
3
4
x
–4
–3
–2
–1 O
1
Fig. 2.139
2
3
4
x
SOLUTION
1
y = ({ x} – 1) 2 .
Sketch the curve
Here, we know the curve for y = ( x – 1) 2 shown as;
Graph for y = ( x – 1)2:
Graph for y = ({ x} – 1)2 :
y
y
y = (x – 1)2
1
1
y = ({x} – 1)2
0
–2 –1
1
x
2
–3
–1 O
–2
1
2
3
x
4
Fig. 2.141
Fig. 2.140
Now; to plot y = ({ x} – 1) retain the
graph for the interval x ∈[ 0, 1) and
repeat for length ‘one’.
2
EXAMPLE
2
Sketch the curve |y| = ({ x} – 1) 2 .
As discussed in above example y = ({ x} – 1) 2 .
Thus, |y| = ({ x} – 1) 2 is image of ({ x} – 1) 2 on x-axis whenever ({ x} – 1) 2 is positive.
SOLUTION
Graph of| y | = ({ x} – 1)2 :
y
1
–4
–3
–2
–1
O
1
2
3
x
4
Curvature and Transformations
EXAMPLE
–1
Fig. 2.142
EXAMPLE
3
Sketch the graph of y =
2x
2[x ]
.
As we know 2 x is exponential function and we want to transform it to 2 x –[x ], it retain
the graph for x ∈[ 0, 1) and repeat for rest points.
SOLUTION
Graph for y = 2x
Graph for y = 2{ x}
To retain graph between x ∈[ 0, 1).
y
y
y = 2x
2
2
y = 2{x}
1
O
Fig. 2.143
1
1
x
–3
–2
–1 O
1
Fig. 2.144
2
3
4
x
95
Play with Graphs
(xx) y = f (x) transforms to y = {f (x)}
Here, plot the horizontal lines for all integral values of y and for the point of intersection on
y = f ( x) plot draw vertical lines and translate the graph for boundary y = 0 and y = 1.
EXAMPLE
SOLUTION
1
Sketch the curve
y = {sin x}. (where { ⋅ } denotes the fractional part of x).
As we know the graph for y = sin x. Shown as.
Graph of sin x :
y
1
y = sin x
x
O
–1
Fig. 2.145
As to retain the curve when 0 ≤ y < 1; and shift other sections of graph between y = 0 to y = 1.
Graph for y = {sin x} :
y
1
As it is
–3π
–
5π
2
–2π
3π –π
–
2
O π π
2
–π
2
3π
2
2π 5π
2
3π
–1
Fig. 2.146
EXAMPLE
SOLUTION
2
Sketch the curve
y = { x 2 }.
As we know the curve y = x 2 , is shown as:
y
5
y = x2
4
3
2
1
–2 –√3 –√2 –1
O
1 √2 √3
96
Fig. 2.147
x
2
x
y
5
4
3
Translated
between
0 ≤ y <1
2
1
–2 –√3 –√2 –1
O
1 √2 √3
x
2
Fig. 2.148
EXAMPLE
SOLUTION
3
Sketch the curve
y = { e x }.
As we know the curve y = e x ;
Graph for y = { ex } :
shown as:
To retain the graph for 0 ≤ y < 1 and transform
the others to 0 ≤ y < 1.
3
y
2
3
1
O
2
log 2 log 3
Fig. 2.149
Curvature and Transformations
Now to sketch y = { x 2 }; retain the graph for 0 ≤ y < 1 and for other intervals transform the
graph between 0 ≤ y < 1.
Transformed
between
0≤y<1
1
Retained
y = {e x }
O
log 2 log 3
x
Fig. 2.150
(xxi) y = f (x) transforms to y = {f ({x})}
Here, we have to follow two steps :
(i) Draw the graph for y = f ({ x}).
(ii) f ({ x}) → { f ({ x})}
EXAMPLE
SOLUTION
1
Sketch the curve y = { e{ x} }.
As we know the curve for y = e x , is plotted as shown in Fig. 2.151.
97
Now to sketch y = e{ x} retain the graph for
y = e x between 0 ≤ x < 1 and repeat for entire
real x.
Graph for y = ex :
Play with Graphs
y
Graph for y = e{ x} :
y = ex
y
e
1
e
O
x
1
1
Fig. 2.151
–2
–1
O
1
2
x
3
Fig. 2.152
Graph for y = { e{ x} }:
y
e
2
1
y = {e {x}}
–3
–2
–1
–log 2 O
log 2 1
2
3
4
x
Fig. 2.153
Here, we know the graph for y = e{ x} , now to plot straight lines||to x-axis for integral values of y
and retain the graph for 0 ≤ y < 1 and transform the others between 0 ≤ y < 1.
EXAMPLE
SOLUTION
2
Sketch the curve
y = {sin{ x}}.
As we know the curve y = sin x; is plotted as;
Graph for y = sin x :
y
sin1
O
1
π
2
Fig. 2.154
98
π
2π
x
y
1
sin 1
–3
–2 –1 O
2
1
5
4
3
x
Fig. 2.155
Graph for y = {sin{ x}} :
From above figure
y = sin{ x} ⇒ 0 ≤ y < 1.
y
sin1
–4 –3
–2 –1 O
1
1
2
3
4
y = {sin {x }}
x
Fig. 2.156
Curvature and Transformations
Graph for y = sin{ x} :
So, the graph of y = sin{ x} and y = {sin{ x}} are same.
(xxii) y = f (x) transforms to {y} = f (x)
Here; retain the graph of y = f ( x) only when y = f ( x) lies between y ∈[ 0, 1) and neglect the
graph for other values.
Graphically it could be stated as;
y
neglected
Graph for y = f (x )
1
O
Graph for {y} = f (x )
x
Fgi. 2.157
99
As we know the curve for y = x 2 , is plotted as:
Graph for y = x 2:
Graph for { y } = x 2:
y
1
ect
y = x2
ed
y
ed
lect
neg
Play with Graphs
SOLUTION
Sketch the curve { y } = x 2 .
1
neg
l
EXAMPLE
1
{y } = x 2
–1
O
x
1
–1
Fig. 2.158
EXAMPLE
SOLUTION
2
x
O
1
Fig. 2.159
Sketch the curve { y } = sin x.
As we know the curve y = sin x, is plotted as shown in Fig. 2.160 and 2.161.
Graph for y = sin x :
y
–3π
–2π –π
π
O
2π
3π
x
Fig. 2.160
Now to sketch { y } = sin x: We retain the graph for 0 ≤ y < 1 and neglect the graph for other
values.
Graph for { y } = sin x :
y
{y } = sin x
x
neglected
Fig. 2.161
100
EXAMPLE
1
Sketch the curve { y } = { x}.
As we know the curve for y = x.
SOLUTION
Graph for y = { x} :
y
y
1
1
–2
O
–1
x
1
y = {x}
–2
–3
O
–1
1
2
x
–1
–2
Fig. 2.163
Fig. 2.162
Graph for { y } = { x} : From above figure we can see y = { x} attains all values between [0, 1).
Thus, graph remains same.
y
1
Curvature and Transformations
(xxiii) y = f (x) transforms to {y} = {f (x)}
As we have earlier discussed y = { f ( x)} (i.e., transformation ( xx), which shows y = { f ( x)}
belongs to [0, 1) ⇒ { y } = { f ( x)}. Thus, the graph of y = { f ( x)} and { y } = { f ( x)} are same.
{y } = { x }
–4
–2
–3
O
–1
1
2
x
3
Fig. 2.164
EXAMPLE
SOLUTION
2
Sketch the curve { y } = {cos x}.
As we know the curve y = {cos x} is plotted as shown in Fig. 2.165.
y
–
5π
2
–
3π
2
–π Oπ
2
2
3π
2
5π
2
x
Fig. 2.165
101
Play with Graphs
From above figure; y = {cos x} lies between [0, 1), which shows y = {cos x} and { y } = {cos x}
are same.
Thus, graph for { y } = {cos x}:
y
{y } = {cos x }
x
O
Fig. 2.166
(xxiv) y = f (x) transforms to y = f –1 (x)
As discussed in chapter 1. y = f –1 ( x) is the mirror image of y = f ( x) about y = x.
OR
“Interchange x and y-axis when function is bijective.”
Graphically it could be stated as :
y
y = f (x)
y =x
y = f –1(x)
O
x
Fig. 2.167
2.4 SKETCHING h( x ) = MAXIMUM { f ( x ), g ( x )} AND h( x ) = MINIMUM { f ( x ), g ( x )}
(i) h (x ) = maximum {f (x ), g (x )}
⇒
 f ( x); when f ( x) > g( x)
h( x) = 
g( x); when g( x) > f ( x)
∴ Sketch f ( x) when its graph is above the graph of g( x) and sketch g( x) when its graph is above
the graph of f ( x).
(ii) h (x ) = minimum {f (x ), g (x )}
⇒
 f ( x), when f ( x) < g( x)
h( x) = 
g( x), when g( x) < f ( x)
∴ Sketch f ( x) when its graph is lower and otherwise sketch g( x).
102
One must remember the formula we can write;
f (x) + g (x)
f (x) – g (x)
max {f (x), g (x)} =
+
2
2
min {f (x), g (x)} =
f (x) + g (x)
2
–
f (x) – g (x)
2
OR
“To
draw
the
graph
of
y = max { f ( x ), g( x )} or y = min { f ( x ), g( x )}.”
functions
of
the
form
We first draw the graphs of both the functions f ( x) and g( x) and their points of intersections.
Then we find any two consecutive points of intersection. In between these points either
f ( x) > g( x) or f ( x) < g( x), then, in order to max{ f ( x ), g( x )} we take those segments of f ( x ) for
which f ( x ) > g( x ), between any two consecutive points of intersection of f ( x) and g( x).
Similarly, in order to min { f ( x ), g( x )}, we take those segments of f ( x ) for which f ( x ) < g( x ),
between any two consecutive points of intersection of f ( x) and g( x).
EXAMPLE
1
3π 

x ∈  – π,
.

2
Sketch the graph of y = max {sin x, cos x}, ∀
First plot both y = sin x and
y = cos x by a dotted curve as can be seen
3π 

from the graph in the interval  – π,


2
y
SOLUTION
1
B
and then darken those dotted lines for
which f ( x) > g( x) or g( x) > f ( x).
–π
A
–
π
2
From adjacent figure the point of
intersections are A, B, C.
O
π
2
π
C
3π
2
x
–1
Curvature and Transformations
Note
Fig. 2.168
∴ Graph of max {sin x, cos x}
y
y = max. { sin x, cos x }
B
–π
–
π
2
O
π
2
A
π
3π
2
x
C
Fig. 2.169
EXAMPLE
2
Sketch the graph for y = min {tan x, cot x}.
SOLUTION First plot both f ( x) = tan x and g( x) = cot x by a dotted curves as can be seen from the
graph and then darken those dotted lines for which f ( x) < g( x) and g( x) < f ( x).
103
Play with Graphs
y
B
–π
3π
–
2
D
π
–
2
A
F
π
2
O
C
π
x
3π
2
E
Fig. 2.170
From above figure we obtain the graph of min{tan x, cot x}.
Graph of min {tan x, cot x} :
y
1
–
3π
2
–π
–
π
2
π
2
O
π
3π
2
x
–1
Fig. 2.171
EXAMPLE
3
Sketch the curve y = min {|x |,|x – 1 |,|x + 1 |}.
SOLUTION First plot the graph
for:
y = |x |, y = |x – 1 |, y =|x + 1|
by a dotted curve as can be seen
from the graph and then darken
those dotted lines for which
and
|x | < {|x – 1 |,|x + 1 |};
|x – 1 | < {|x |,|x + 1 |}
|x + 1 | < {|x |,|x – 1 |}.
Graph for
y
y = |x + 1|
y = |x |
3
y = |x – 1|
2
1
–3
–2
–1– 1
2
O 1
2
x
1
y = | x|, y =| x – 1|, y = | x + 1| .
Fig. 2.172
104
2
3
|x
1;
x
=
≥
y
1
y
–
|x
x
1;
≤
–1
O 1
2
1
2
=
–
–1– 1
2
y =1–x
y
–x
–2
1|
=
x
=
1|
–
+
–3
1
2
x +1 –x x
x
3
Fig. 2.173
From above figure;
 –( x + 1) ;

( x + 1) ;


;
 –( x)

min {|x – 1 |,|x |,|x + 1 |} = 
( x)
;


(1 – x) ;

( x – 1) ;
x≤ –1
– 1≤ x≤ –
–
1
2
1
≤ x≤ 0
2
1
0≤ x≤
2
1
≤ x≤1
2
x≥1
2.5 WHEN f ( x ), g ( x ) → f ( x ) + g ( x ) = h ( x )
Curvature and Transformations
As from the above curve graph for y = min {|x – 1|,|x|,|x + 1|} is plotted as;
There is no direct approach; but we can use following steps if minimum or maximum value of any
one is known.
Step 1. Find maximum and minimum value of g( x) say; a ≤ g( x) ≤ b.
Step 2. Plot the curve h( x) = f ( x) + g( x) between f ( x) + a to f ( x) + b.
i.e.,
f ( x) + a ≤ h( x) ≤ f ( x) + b
Step 3.
Step 4.
Step 5.
Check g( x) = 0 ⇒ h( x) = f ( x).
When g( x) > 0 ⇒ h( x) > f ( x).
When g( x) < 0 ⇒ h( x) < f ( x).
EXAMPLE
1
Plot the curve y = x + sin x.
Here, y = x + sin x = f ( x) + g( x)
as we know;
g( x) = sin x ∈ [– 1, 1]
…(i)
∴
x – 1≤ y ≤ x + 1
⇒ To sketch the curve between two parallel lines y = x + 1and y = x – 1(called bounded limits)
also;
…(ii)
g( x ) = 0 ⇒ y = x
…(iii)
g( x) > 0 ⇒ y = x + sin x > x
SOLUTION
105
g( x) < 0 ⇒ y = x + sin x < x
Using Eqs. (i), (ii), (iii) and (iv), we get
…(iv)
y=x–1
Play with Graphs
y=x
y=x+1
x
(π, π)
(π/2, π/2 + 1)
y
+ sin
y=x
(0, 0) 1
π/2
(3π/2, 3π/2 – 1)
1
–3π/2
–π/2
–π
–1
π
x
3π/2
–1
(–π/2, –π/2 – 1)
(–3π/2, –3π/2 + 1)
(–π, –π)
Fig. 2.174
2.6 WHEN f ( x ), g ( x ) → f ( x ) ⋅ g ( x ) = h ( x )
There is no direct approach but we can use the following steps if minimum and maximum of any
one is known.
Step 1. Find the minimum and maximum of any one of them say a ≤ g( x) ≤ b.
Step 2.
From step 1;
Step 3.
Check
EXAMPLE
1
Sketch the curve;
SOLUTION
Here; y = x sin x;
where,
– 1 ≤ sin x ≤ 1
⇒
– x≤ y ≤ x
y = x sin x.
y
y = –x
y=x
y = x sin x
…(i)
also;
sin x = 0
⇒
x = – 2 π, – π, 0, π, 2 π, …
∴ y = 0 when
x = – 2 π, – π, 0, π, …
106
a ⋅ f ( x) ≤ h ( x) ≤ b ⋅ f ( x)
g( x) = 0 ⇒ h( x) = 0 .
–2π
–3π/2
–π –π/2
O
π/2
Fig. 2.175
π
3π/2
2π
x
y =x
when
y =–x
when
π 5π
,
,…
2 2
π 3π
x=– ,
,…
2 2
x=
sin x
.
x
sin x
1
1
lies between – to .
SOLUTION As we know; –1 ≤ sin x ≤ 1 ⇒ y =
x
x
x
1
1
or
to sketch the curve where – ≤ y ≤
x
x
Here at x = 0; y is not defined but as;
sin x
x → 0 ⇒ y =
→ 1
x
also;
y = 0 at x = nπ ; n ∈ z – { 0}
Using Eqs. (i), (ii) and (iii), the curve could be plotted as;
EXAMPLE
2
Draw the graph of the function y =
…(i)
…(ii)
…(iii)
y
y = 1/x
y = –1/x
–3π –5π/2 –2π
–3π/2
–π
–π/2
π/2
O
π
3π/2
2π
5π/2
x
3π
y=
sin x
x
Curvature and Transformations
and
y = –1/x
y = 1/x
Fig. 2.176
EXAMPLE
SOLUTION
3
Draw the curve: y = e – x sin x.
As we know;
y
y = e–x
–1 ≤ sin x ≤ 1
∴
y = e x sin x
⇒
– e– x ≤ y ≤ e– x
y = e–xsin x
…(i)
–2π
–π
Thus, y is bounded between
y = – e– x
and
O
π
2π
x
3π
4π
y = e– x .
Shown as in Fig. 2.177;
y = –e–x
Fig. 2.177
107
Play with Graphs
SOME MORE SOLVED EXAMPLES
EXAMPLE
Sketch the curves of the following :
(ii) y = [ x] +
(i) y = x – [ x]
1
(iii) y = |[ x] +
SOLUTION
x – [ x]
x – [ x] |; where [ • ] represent greatest integer function
(i) As we know that;
Also, for any
∴
0 ≤ x – [ x] < 1 for all x ∈ R.
x ∈[ 0, 1), we have
x2 ≤ x ≤ x
x – [ x] ≤ x – [ x]
 x + 1 , when – 1 ≤ x < 0

, when 0 ≤ x < 1
 x
Now,
y = x – [ x] = 
 x – 1 , when 1 ≤ x < 2

 x – 2 , when 2 ≤ x < 3 … and so on.
Thus, from above we have to plot;
when 0 ≤ x < 1
y = x,
y =
∴ the curve for y =
x – 1, when 1 ≤ x < 2
(i.e., same as shifting x towards right by 1 unit.)
y = x – 2, when 2 ≤ x < 3
(i.e., same as shifting x towards right by 2 units)
x – [ x] is periodic with period ‘1’. Shown as in Fig. 2.178.
y
1
–3
√x + 3
√x + 2
√x + 1
–2
–1
O
√x
1
√x – 1
√x – 2
2
3
Fig. 2.178
(ii) As we know;
⇒
108
y = [ x] +
x – [ x]
 –1 + x + 1, when – 1 ≤ x < 0

, when 0 ≤ x < 1
 x
y =
1 + x – 1 , when 1 ≤ x < 2

 2 + x – 2 , when 2 ≤ x < 3 … and so on.
√x – 3
x
y = [ x] +
Thus, the curve for
x – [ x]
y
4
3 + √x – 3
3
2 + √x – 2
2
1 + √x – 1
1
√x
–4
–3
–2
O
1
–1 + √x + 1
–1
–1
2
3
4
x
5
–2 + √x + 2
–3 + √x + 3
–2
–3
Fig. 2.179
From above curve we could discuss that;
y = [ x] + x – [ x] is continuous and differentiable for all x.
(iii) The graph for y = [ x] +
the graph of y = [ x] +
x – [ x] is obtained by reflecting the portion lying below x-axis of
x – [ x] about x-axis and keeping the portion lying above x-axis (as it
is).
Thus, the graph for y = [ x] +
Curvature and Transformations
Thus, the graph for y = [ x] + x – [ x] is obtained by the graph of y = x – [ x] by translating
it by [x] units in upward or downward direction according as [ x] > 0 or [ x] < 0.
x – [ x] .
y
–4
–3
–2
–1
O
1
2
3
4
x
Fig. 2.180
109
EXAMPLE
2
Sketch the curve
y = (1 – x 2/ 3 ) 3/ 2 .
Here;
(a) f (– x) = f ( x) ∴ even function or symmetric about y-axis
x=0 ⇒ y =1 
(b) When

y = 0 ⇒ x = ± 1
Domain ∈ [– 1, 1] 
(c)

Range ∈ [ 0, 1] 
Play with Graphs
SOLUTION
…(i)
…(ii)
…(iii)
dy

> 0 when x ∈ [– 1, 0] 

dx

dy
< 0 when x ∈ [ 0, 1] 

dx
2

d y
> 0 when x ∈ [– 1, 1] 
2
dx

(d)
(e)
…(iv)
…(v)
From above;
y
1
y = (1 – x 2/3 )3/2
–1
O
1
x
Fig. 2.181
EXAMPLE
3
Sketch the graph for y = ( x – 1)x 2/ 3 .
In y = ( x – 1)x 2/ 3
(a) Not symmetric about any axis or origin.
SOLUTION
110
…(i)
(b)
as x = 0 ⇒ y = 0 

as y = 0 ⇒ x = 0, 1
…(ii)
(c)
Domain ∈ R

Range ∈ R 
…(iii)
(d)
2( x – 1) 5x – 2
dy
= x 2/ 3 +
=
dx
3x1/ 3
3x1/ 3
⇒
dy
2
> 0, when x < 0 or x > 
5
dx

dy
2

< 0, when 0 < x <

dx
5
…(iv)
y
=
10x + 2
O
9x 4 / 3
–2/5

2
or x > 0
5

⇒ dx

2
d y
2

< 0, when – < x < 0

5
dx 2
d 2y
2
y = (x –1)x 2/3
2/5
x
1
> 0, when x < –
…(v
Fig. 2.182
)
Thus, curve for y = ( x – 1) x 2/ 3 is shown in Fig. 2.182.
EXAMPLE
Draw the graph for:
(i) y = | 1 – |x – 1 ||
4
(ii) |y | = |1 – |x – 1||
As we know the graph for y = x – 1
SOLUTION
y
y=x–1
O
1
x
–1
Curvature and Transformations
(e)
1
 1/ 3

x ⋅ 5 – ( 5x – 2) ( x) –2/ 3

1
3
= 

2
2/ 3
3
dx
x



d 2y
Fig. 2.183
(a) y = ( x – 1) → y = |x – 1|
(b) y = |x – 1 | → y = – |x – 1 |
y
y
Im
e
ag
W
y = |x – 1|
y
0
1
is
ax
x-
<
t
ou
ab
n
he
1
ut
–1
bo
la
al
1
x
Sm
O
x
O
ax
x-
–1
is
Fig. 2.185
Fig. 2.184
111
Play with Graphs
(c) y = – |x – 1 | → y = 1 –|x – 1 |
(d) y = 1 – |x – 1 | → y = 1 – |x – 1 |
y
y
1
1
O
1
y = |1 – |x – 1||
x
2
O
y = 1 – |x – 1|
Fig. 2.186
1
x
2
Fig. 2.187
(e) y = 1 – |x – 1 | → |y | = 1 – |x – 1 |
y
(1, 1)
1
|y | = |1 – |x – 1||
1
O
–1
x
2
(1, –1)
Fig. 2.188
EXAMPLE
SOLUTION
5
Draw the graph for:
1
1
(i) y = 2 –
(ii) y = 2 –
|x – 1 |
|x – 1 |
We know the graph for y =
(iii) |y | = 2 –
1
is shown as;
x
y
y=
(1, 1)
1
–1 O
(–1, –1)
1
–1
Fig. 2.189
112
1
x
x
1
|x – 1 |
1
1
→ y =
x
x–1
(b) y =
1
1
→ y =
x–1
|x – 1|
Curvature and Transformations
(a) y =
y
y
(2, 1)
O
1
(2, 1)
1
x
2
O
x
1
(0, –1)
x=1
x=1
Fig. 2.190
(c) y =
Fig. 2.191
1
1
→ y = –
|x – 1 |
|x – 1 |
(d) y = –
1
1
→ y = 2 –
|x – 1|
|x – 1|
y
y
y=2–
1
O
1
x
2
(0, –1)
(2, 1)
x
3 2
2
O 1
2
–1
(2, –1)
1
|x – 1|
x=1
Fig. 2.193
x=1
Fig. 2.192
(e) y = 2 –
1
1
→ y = 2 –
|x – 1|
|x – 1|
y
(f) y = 2 –
1
1
→ |y |= 2 –
|x – 1 |
|x – 1 |
y
y= 2–
|y | = 2 –
1
|x – 1|
1
|x – 1|
1
1
x
O
O
1
2
3
2
x
–1
1
2
3
2
x=1
x=1
Fig. 2.194
Fig. 2.195
113
Play with Graphs
EXAMPLE
6
Sketch the graph for:
(i) y = e|x| – e – x
(ii) y = |1 + e|x| – e – x|
(iii) |y| = |1 + e|x| – e – x|
(iv) |y | ≤ | 1 + e|x| – e – x |
SOLUTION
(i) Here
y = e |x| – e –x
ex – e– x ; x ≥ 0
y =
–x
–x
e – e ; x < 0
ex – e– x ; x ≥ 0
y =
; x< 0
0
⇒
To discuss;
y = ex – e– x
(i) when x = 0 ⇒ y = 0 (it passes through origin)
(ii) when y = 0 ⇒ e 2x – 1 = 0 ⇒ x = 0
(iii) f (– x) = – f ( x)
as
y = f ( x) = e x – e – x
⇒
f (– x) = e – x – e x = – f ( x);
it shows y = f ( x) = e x – e – x is odd function, i.e., symmetric about origin.
(iv)
y = ex – e– x
dy
e 2x + 1
⇒
= ex + e– x =
> 0 for all x ∈ R.
dx
ex
∴ y is increasing for all x.
e 2x – 1
d 2y
x
–x
(v)
=
e
–
e
=
ex
dx 2
d 2y
> 0 when x > 0; concave up and increasing .
⇒
dx 2
d 2y
also
< 0 when x < 0; concave down and increasing from above discussion
dx 2
y = ex – e– x ; is plotted as shown in Fig. 2.196.
Recall : Increasing
n
ow
ed
v
a
nc
Co
e–
1
e
Inc
con reasi
ng
cav
ed
ow
n
–1
Inc
r
co easin
nc
ave g
up
y
n
x
O
1
1
–e
e
y = e x – e –x ; x ∈ R
Fig. 2.196
114
Co
u
ve
ca
p
y =e
|x|
–e
–x
y = ex – e–x; x ≥ 0
ex – e– x ; x ≥ 0
=
; x < 0.
0
∴ y = e|x | – e–x
y = 0; x < 0
Thus from Fig. 2.197.
x
(0, 0)
Fig. 2.197
(ii) Plotting of y = | 1 + e |x| – e –x |
y
y
y = 1+ ex – e–x; x ≥ 0
y = 1; x < 0
y = |1+ e|x | – e–x|
1
1
x
O
x
O
Fig. 2.198
Fig. 2.199
From above Fig. 2.198.
y = e|x| – e – x → y = 1 + e|x| – e – x
⇒
1 + e x – e – x ; x ≥ 0
y =
; x< 0
1
Thus, the graph for;
y = |1 + e|x| – e – x|
is same as;
y = 1 + e|x| – e – x
∴ Graph for y = |1 + e
| x|
(iii) Plotting of | y | = | 1 + e
|x|
Curvature and Transformations
y
Now;
{as y ≥ 1 for all x ∈ R}
–x
– e | is shown in Fig. 2.199.
– e –x |
y
y = 1 + ex – e–x; x ≥ 0
y = 1; x < 0
1
x
O
–1
y = –1; x < 0
y = –(1 + ex – e–x); x ≥ 0
Fig. 2.200
115
Play with Graphs
(iv) Plotting of | y | ≤ |1 + e |x| – e –x |
y
y = 1 + ex – e–x; x ≥ 0
From above figure check any point say
(0, 0)
⇒ 0 ≤ | 1 | (True,
towards (0, 0).
thus
to
shade
y = 1; x < 0
1
|y | ≤ |1 + e|x | – e–x|
x
area
From given figure shade part represents
the area bounded between two curves.
y = –1; x < 0
y = –(1 + ex – e–x); x ≥ 0
Fig. 2.201
EXAMPLE
7
Sketch the curve of the following:
(i) |x + y| ≥ 1
(ii) |x – y | ≤ 1
(iii) |x| + |y | ≤ 2
(iv) |x| – |y | ≥ 1
SOLUTION
(i) Here to sketch, | x + y | ≥ 1
we know; |x + y| ≥ 1
⇒
x + y ≥ 1

x + y ≤ – 1
or
x+y≥1
and
x+ y ≤ –1
y
First we shall draw the graph for
x + y = 1 and x + y = – 1, now we
shall consider any fixed point say
(0, 0) and check x + y ≥ 1 and
x + y ≤ – 1 holds or not.
As;
1
–1
x
1
x + y ≤ –1
–1
x + y ≥ 1 ⇒ 0 ≥ 1 (false).
∴ shaded part is away from origin.
again as;
⇒
x+y≤–1
0≤ – 1
(false)
∴ Shaded part is away from origin
shown as in Fig. 2.202.
(ii) To sketch | x – y | ≤ 1
we know
116
|x – y| ≤ 1
⇒
–1 ≤ x – y ≤ 1
⇒
–1 – x ≤ – y ≤ 1 – x
Fig. 2.202
1
+
y
=
x
Thus, to plot y between ( x – 1) to ( x + 1).
In figure shaded parts is towards origin as;
putting any fixed point say (0, 0); in|x – y| ≤ 1.
⇒
| 0 – 0| ≤ 1
⇒
0≤ 1
=
x
–
1
1
y
–1
x
O
(true)
1
–1
|x – y | ≤ 1
(iii) To sketch | x | + | y | ≤ 2
Fig. 2.203
Here; |y | = 2 – |x | is plotted as;
y
y
y
2
2
2
|y | = 2 – | x |
1
–2 –1 O 1
–1
x
2
O
–2
y=2–x
2
x
O
–2
x
2
y = 2 – |x |
–2
–2
(i)
–2
(ii)
(iii)
Fig. 2.204
y
From above graph of|y | = 2 – |x |, we can check shading
2
of
|y | ≤ 2 – |x |
as at (0, 0)
⇒ 0≤ 2
or
Curvature and Transformations
y
( x – 1) ≤ y ≤ ( x + 1)
or
|x | + |y | ≤ 2 .
–2
(true)
∴ shading towards origin; shown as in Fig. 2.205.
2
–2
(iv) Sketching of | x | – | y | ≥ 1
x
|x | + |y | ≤ 2
Fig. 2.205
To sketch |x | – |y | ≥ 1; we proceed as:
y = ( x – 1) → y = |x | – 1 → | y |= |x | – 1 → |x | – |y | ≥ 1
y
y
y=x–1
O
1
–1
(i)
y = |x | – 1
x
x
–1 O
1
–1
(ii)
117
y
y
|x | – |y | ≥ 1
Play with Graphs
|y | = |x | – 1
–1 O
x
1
–1 O
x
1
(iv)
(iii)
Fig. 2.206
EXAMPLE
SOLUTION
8
Sketch the curve;
(i) |x + y | + |x – y | ≤ 4
(ii) |x + y | + |x – y | ≥ 4
(i) As we know;
y
|x | – |y |; |x | ≥ |y |
|x – y | = 
 –(|x | – |y |); |x | ≤ |y |
2
Thus; |x + y| + |x – y| ≤ 4
⇒
 x + y + x – y ≤ 4; |x | ≥ |y |

 x + y – x + y ≤ 4; |x | < |y |
⇒
 2x ≤ 4; |x | ≥ |y |

 2y ≤ 4; |x | < |y |
or
|x + y | + |x – y | ≤ 4
⇒
|x | ≤ 2
and
–2
2
x
|x + y | + |x – y | ≤ 4
–2
Fig. 2.207
|y | ≤ 2
Thus, to shade the portion when –2 ≤ x ≤ 2 and –2 ≤ y ≤ 2 . Shown as in Fig. 2.207.
and
(ii) Again;
|y | ≥ 2
|x + y| + |x – y| ≥ 4
⇒
|x | ≥ 2
Thus, to shade the portion when x ≤ –2 or x ≥ 2 and y ≤ – 2 or y ≥ 2 . Shown as in
Fig. 2.208.
y
2
–2
O
2
–2
|x + y | + |x – y | ≥ 4
Fig. 2.208
118
x
Sketch the curve for;
9
(ii) 2|x| |y| + 2|x|–1 ≤ 1; |x | ≤
(i) 2| x||y | + 2| x|–1 ≤ 1
SOLUTION
2|x||y| + 2|x|–1 ≤ 1
(i) Here
⇒
1

2| x| |y| +  ≤ 1
2

⇒
|y| +
1
1
and |y | ≤ .
2
2
1
≤ 2 –|x|
2
1
≤ 2 –|x| , we proceed as;
2
1
1
1
 – x 1
y =  2 –  → y = 2 –|x| – → |y| = 2 –|x| – → |y| ≤ 2 –|x| –

2
2
2
2
(iv)
(ii)
(iii)
(i)
Thus, to plot |y| +
Shown as;
y
(0, 1)
y = 2–x
O
x
Fig. 2.209
y
y
y = 2–x–1/2
neglecting
1/2
1/2
Image
O
x
1
–1
O
x
1
–1/2
–1/2
(i) y = 2 – x –
1
2
(ii) y = 2 –|x | –
y
1
2
y
1/2
–1
Curvature and Transformations
EXAMPLE
O
1/2
x
1
–1/2
(iii) | y | = 2 –|x| –
–1
x
O
1
–1/2
1
2
(iv) | y | ≤ 2 –|x | –
1
2
Fig. 2.210
119
Here, figure (iv) is shaded towards origin as putting
Play with Graphs
⇒ 0≤ 1 –
1
or
2
(ii) Plotting of
x = 0, y = 0 in |y| ≤ 2 –|x| –
1
0 ≤ , which is true, therefore, shaded towards origin.
2
2|x||y | + 2|x|–1 ≤ 1;
|x | ≤
1
.
2
y
1
1
and |y | ≤ .
2
2
y = 1/2
1/2
from above figure;
2|x||y | + 2|x|–1 ≤ 1
1
1
and |x | ≤ , |y | ≤
2
2
Fig. 2.211.
EXAMPLE
2|x ||y | + 2|x | –1 ≤ 1;
1 and |y | ≤ 1
|x | ≤
2 x = –1/2
2
shown as in
x
O
–1
1
–1/2
y = –1/2
x = 1/2
Fig. 2.211
Sketch the graph of the function:
f ( x) = log 2 (1 – x 2 )
SOLUTION Here,
log 2 (1 – x 2 ) exists when, –1 < x < 1 

i. e.,
doman ∈ (– 1, 1)


and
range ∈ (– ∞, 0]

as
x → ± 1 ⇒ y → – ∞
differentiating
y = log 2 (1 – x 2 ),
dy
(–2x)
x
we get
=
⋅ (log 2 e) = 2 log 2 e 2
2
dx 1 – x
( x – 1)
10
Using number line rule;
–
+
–1
–
0
…(i)
…(ii)
+
1
dy

> 0, when – 1 < x < 0

dx

dy
< 0, when 0 < x < 1 

dx
=
(1 – x 2 ) 2
< 0 for all
y
x ∈ (– 1, 1)
from above results we can draw
y = log 2 (1 – x 2 ) as shown in Fig. 2.212.
O
–1
n
dow
cave
g
Con
easin
decr
∴ y = log(1 – x 2 ) is concave down for x ∈ (–1, 1)} …(iv)
cave
d
increa own
sing
dx 2
–2 log 2 e(1 + x 2 )
Con
also,
d 2y
…(iii)
1
x
y = log2 (1 – x 2 )
Fig. 2.212
120
11
SOLUTION
Here;
y = log(sin x) .
Sketch the graph of
(i) y = log(sin x) is defined, when sin x > 0
⇒ x ∈ ( 2nπ, ( 2n + 1)π ).
(ii) Since, sin x is periodic with period 2 π .
∴ to discuss y = log(sin x), when x ∈ ( 0, π ) as exists when x ∈ ( 0, π ) and then plotted for
entire number line.
(iii)
as x → 0
⇒
y = log( 0) = – ∞.
π
as y → 0
⇒
sin x → 1
⇒ x= .
2
dy
1
(iv)
=
⋅ cos x = cot x
dx sin x
dy
 π
⇒
> 0 when x ∈  0, 
 2
dx
dy
π 
< 0, when x ∈  , π
2 
dx
(v)
d 2y
= – cosec 2 x < 0 for all x ∈ ( 0, π )
dx 2
Thus, increasing and concave down x ∈( 0, π / 2)
x ∈( π / 2, π ).
decreasing and concave down
and to plot the curve only when x ∈ ( 2nπ, ( 2n + 1) π ). Shown as in Fig. 2.213.
–π
–
π
2
O
π
2
π
3π
2
2π
5π
2
3π
x
g and
ve dow
conca
sing a
nd con
cave d
3π
2
n
increa
–
asin
decre
–2π
own
y
Curvature and Transformations
EXAMPLE
Fig. 2.213
EXAMPLE
SOLUTION
Sketch the graph for
y = – log sin x e.
–1
is defined when,
(i) Here, y = – log sin x e =
log e sin x
12
sin x > 0 and
i.e.,
sin x ≠ 1.
x ∈ ( 2nπ, ( 2n + 1) π ) – { 2nπ + π / 2}; n ∈ N
…(i)
121
–1
is periodic with period 2 π,
log e sin x
∴ to discuss the graph for x ∈ ( 0, π ) – { π / 2} and sketch for the entire number line.
dy
dy
1
1
cot x
(iii) ∴
=+
⋅
⋅ cos x
⇒
=+
2
dx
dx
(log sin x) sin x
(log sin x) 2
Play with Graphs
(ii) y =
dy
π

> 0 when x ∈  0, 

dx
2
dy
π 
< 0 when x ∈  , π
2 
dx
or
 (log sin x) 2 ⋅ (– cosec 2 x) – cot x ⋅ 2 (log sin x) ⋅ cot x
=
+


dx 2
(log sin x) 4


{(log sin x) 2 cosec 2 x + 2 (log sin x) cot 2 x}
=–
(log sin x) 2
d 2y
(iv)
⇒
Thus,
 π
x ∈ ( 0, π ) –  
 2
dx

π 
 increasing and concave up x ∈  2 , π
1

y =
⇒ 
log sin x
 decreasing and concave up x ∈  0, π 

 2
d 2y
> 0 for all
2
y
y=–
–2π
–
3π
2
–π
–
π
2
π
2
O
π
3π
2
2π
5π
2
1
logesin x
3π
x
y = – logsin x e
Fig. 2.214
EXAMPLE
13
Sketch the curve:
(i) [ y ] = cos x; x ∈ [–2 π, 2 π]
(ii) [ y ] = [cos x]; x ∈ [–2 π, 2 π] ; where [ ⋅ ] denotes greatest integer function.
SOLUTION
As we know; y = cos x could be plotted as;
y
1
–2π – 3π
2
–π
–
π O
2
π
2
–1
122
Fig. 2.215
π
3π
2
2π
x
From above curve
x = – 2 π, –
cos x = – 1, 0, 1 when
3π
3π
π
π
,– π, – , 0, , π,
, 2π
2
2
2
2
y
⇒ [ y ] = cos x is possible only
2
3π
π 

when x ∈  ±2 π, ±
, ± π, ± , 0
2
2 

1
Thus,
–2π
cos x = – 1 ⇒ –1 ≤ y < 0
– 3π
2
–π
–π
2
π
2
O
cos x = 0
⇒ 0≤ y < 0
–1
cos x = 1
⇒ 1≤ y < 2
–2
Thus, [ y ] = cos x is plotted as
π
[y] = cos x
shown in Fig. 2.216.
Fig. 2.216
(ii) Sketching of [ y ] = [cos x ]
First to plot y = [cos x]. Shown as;
y
(–2π, 1)
–2π
– 3π
2
(2π, 1)
(0, 1)
–π
2
π
2
O
3π
2
2π
x
3π
2
2π
x
Curvature and Transformations
(i) Sketching of [ y ] = cos x
Fig. 2.217
y = [cos x]
From above figure
⇒
Thus,
3π
π

 –1 ; 2 < |x | < 2

3π
π

and
y =  0 ; 0 < |x | ≤
≤ |x | < 2 π
2
2

 1 ; |x | = 0, 2 π


when
y =0
⇒
[ y ] ∈ [ 0, 1)
y =1
⇒
[ y ] ∈ [1, 2)
y = –1
⇒
[ y ] ∈ [– 1, 0)
123
graph for [ y ] = [cos x]
Thus,
Play with Graphs
y
2
(2π, 1)
1
(–2π, 1)
–2π < x ≤ –
3π
2
–
–
π
π
≤x<0<x≤
2
2
3π
π
<x<–
2
2
–1
x
3π
≤ x < 2π
2
2π < x <
3π
2
Fig. 2.218
EXAMPLE
14
Sketch the curves
(i) y = 4 – [ x]
(ii) [ y ] = 4 – [ x]
(iii) [|y |] = 4 – [|x |]
SOLUTION
As we know;
(i) y = 4 – [ x] could be plotted with in two steps;
y = 4 – x → y = 4 – [ x]
y
y
5
4
3
4
3
y=4–x
2
2
1
–4 –3 –2 –1 O
–1
1
1
2
3
4
5
x
–4 –3 –2 –1 O
–1
–2
–2
–3
–3
(i)
(ii)
Fig. 2.219
(ii) Since, we know from above curve
y = 4 – [ x]
⇒ y ∈integer, thus, on taking integral part, say
[ y ] = I ⇒ I ≤ y < ( I + 1)
124
1
2
3
y = 4 – [x ]
4
5
x
Graph for [ y ] = 4 – [ x]
y
6
5
Included
4
Excluded
3
2
1
–3 –2 –1 O
–1
1
2
3
x
5
4
–2
and so on
–3
[y ] = 4 – [x ]
Fig. 2.220
In above curve lower boundary are included and upper are excluded.
(iii) To sketch [| y |] = 4 – [| x |]
From above figure we can say;
[ y ] = 4 – [ x] → [ y ] = 4 – [|x |] → [|y |] = 4 – [|x |]
y
5
4
3
Curvature and Transformations
∴
2
1
–5
–4 –3 –2 –1 O
–1
1
2
3
–2
4
5
x
[|y |] = 4 – [|x |]
–3
–4
–5
[|y |] + [|x |] = 4
Fig. 2.221
EXAMPLE
SOLUTION
Shade the region whose co-ordinates x and y satisfy the equation.
cos x – cos y > 0
 y – x
 x + y
Here, cos x – cos y > 0 can be written as, 2 sin 
>0
 sin 
 2 
 2 
15
125
 y – x
 x + y
sin 
>0
 sin 
 2 
 2 
Play with Graphs
⇒
This inequality holds true for all points,
 y – x
 x + y
when sin 
 have same sign.
 and sin 
 2 
 2 
i.e.,
 x + y
sin 
 > 0 and
 2 
 y – x
sin 
>0
 2 
or
 x + y
sin 
 < 0 and
 2 
 y – x
sin 
<0
 2 
⇒
or
x + y > 2nπ and y – x > 2nπ
x + y < 2nπ and y – x < 2nπ
 y > 2nπ – x
 y < 2nπ – x
or 
⇒

 y > 2nπ + x
 y < 2nπ + x where n ∈ z
Here, the equation x + y = 2kπ represents a system of parallel straight lines corresponding to
different values of k.
Say
k = 0, 1 ⇒ y = – x and y = – x + 2 π.
i.e., the set of points whose coordinates satisfy the inequality
0 < x + y < 2π
Similarly, for general points:
…(i)
2kπ < x + y < ( 2k + 1) π
x + y
 > 0.
 2 
Now, the set of points which satisfy sin 
y
8π
6π
4π
2π
–8π
–6π
–4π
–2π
O
y – x = 2n π
x
2π
4π
–2π
–4π
–6π
–8π
x + y = 2k π
Fig. 2.222
126
6π
8π
cos x > cos y
2nπ <
x+y
< ( 2n + 1) π
2
⇒
2 ( 2nπ ) < x + y < 2 ( 2n + 1) π
⇒ only those points of x, y which satisfy Eq. (ii).
Similarly,
2 ( 2p – 1) π < y – x < 2 ( 2p ) π
From above results, graph for cos x > cos y is shown in Fig. 2.222.
EXAMPLE
16
Sketch the curve
…(ii)
…(iii)
x 2/ 3 + y 2/ 3 = a 2/ 3 .
As to sketch, x 2/ 3 + y 2/ 3 = a 2/ 3 .
(i) Curve is symmetric about y-axis (as when x is replaced by (– x) curve remains same)
(ii) Curve is symmetric about x-axis (as when y is replaced by (– y) curve remains same).
(iii) Curve is symmetric about origin (as when x is replaced by y and y by x curve remains same).
(iv) When,
x=0 ⇒ y =±a
when,
y = 0 ⇒ x = ± a.
(v)
y 2/ 3 = ( a 2/ 3 – x 2/ 3 ),
or
y 2 = ( a 2/ 3 – x 2/ 3 ) 3 ;
SOLUTION
differentiating both sides,
2y
dy

 2
= 3 (a 2 / 3 – x 2 / 3 )2 ⋅  – x –1 / 3 

 3
dx
dy
< 0 when x > 0, y > 0 [to discuss 0 < x ≤ a and to take symmetry for rest of graph
dx
using Eqs. (i), (ii) and (iii)].
dy
(vi) From above
= – x –1/ 3 ( a 2/ 3 – x 2/ 3 ) 2
y
dx
Differentiating again, we get;
⇒
d 2y
2
 dy 
 1
 2
+   = – x –1/ 3 ⋅ 2( a 2/ 3 – x 2/ 3 ) ⋅  – x –1/ 3  + x –4 / 3 ⋅ ( a 2/ 3 – x 2/ 3 ) 2
2
 3



3
dx
dx
4
1
= x –2/ 3 ( a 2/ 3 – x 2/ 3 ) + x –4 / 3 ( a 2/ 3 – x 2/ 3 ) 2
3
3
1 –2/ 3 2/ 3
y
(a
– x 2/ 3 ){ 4 + x –2/ 3 ( a 2/ 3 – x 2/ 3 )}
= x
3
a
d 2y
whenever;
and
=
+
ve
0
<
x
≤
a
y
>
0.
decreasing and
dx 2
concave up
x
Thus; when 0 < x ≤ a and y > 0
O
–a
a
2
dy
d y
x2/3 + y2/3 = a2/3
< 0 and
> 0,
⇒
dx
dx 2
y
Curvature and Transformations
⇒
–a
i.e., decreasing and concave up.
∴ Graph for x 2/ 3 + y 2/ 3 = a 2/ 3
Note
Fig. 2.223
Above curve is known as Astroid represented by the following parametric equations:
x = a cos3 t 
 0 ≤ t ≤ 2π , a > 0
y = a sin3 t 
127
Play with Graphs
Eliminating cos t and sin t from above equations, we get x 2/ 3 + y 2/ 3 = a 2/ 3 .
where the quantities x and y are defined for all values of t. But since, the functions cos 3 t and sin 3 t are
periodic with period 2 π, it is sufficient to discuss the curve for t ∈[ 0, 2 π].
Domain of function ∈ [ 0, 2π]
…(i)
Thus,

Range of function ∈ [– a, a] 
Now,
Here,
∴
dx

= – 3a cos 2 t sin t

dt

dy
= + 3a sin 2 t cos t

dt
dy
π
3π
dx
, 2 π.
= 0,
= 0 ⇒ t = 0, , 8 π,
2
2
dt
dx
dy

= – tan t
dx

…(ii)
…(iii)


dx
3a cos t sin t 
From Eqs. (i), (ii), (iii) and (iv), we construct a table;
d 2y
2
Range of t
0< t <
π
2
π
< t< π
2
3π
π< t<
2
3π
< t < 2π
2
1
=
…(iv)
4
Sign of
dy
dx
Sign of
Domain (x)
Range (y)
0< x< a
0< y < a
–
+
– a< x< 0
0< y < a
+
+
– a< x< 0
– a< y < 0
–
–
0< x< a
–a < y < 0
+
–
d 2y
dx 2
y
On the basis of above information we can sketch
x = a cos 3 t
a
y = b sin 3 t
Students are adviced to convert the cartesian into
parametric if possible.
–a
O
Astroid : x2/3 + y2/3 = a2/3
or x = a cos3t
y = a sin3t
x
a
–a
Fig. 2.224
EXAMPLE
SOLUTION
Sketch the curves:
ex + e– x
ex – e– x
(i) y =
(ii) y =
2
2
17
e x + e –x
2
∴ even or symmetric about y-axis
ex – e– x
ex + e– x
(iv) y =
ex + e– x
ex – e– x
(i) Sketching of f (x ) = y =
(a) f (– x) = f ( x),
128
(iii) y =
…(i)
dx
=
2
ex + e– x
e 2x + 1
=
2
2 ex
ex – e– x
=
2
dx 2

> 0 when x > 0 

dx

2
d y

when
<
0
x
<
0
2

dx
d 2y
⇒
(iii) Sketching of y =
(b)
(c)
(d)
ex + e–x
2
1
x
O
Fig. 2.225
y
y=
O
e x – e– x
2
x
2
…(iv)
From Eqs. (i), (ii), (iii) and (iv) graph of y =
(a)
y=
In
co crea
nc
av sing
ed
ow
n
(d)
d 2y
y
ing
as p
u
cre
de cave
n
co
dy

> 0 whenever x > 0

dx
…(iii)
From above;

dy

< 0 whenever x < 0

dx

d 2y
…(iv)
>
0
x
for
all

dx 2

Thus, from Eqs. (i), (ii), (iii) and (iv) graph of
ex + e– x
is;
y =
2
e x – e –x
(ii) Sketching y =
2
(a) f (– x) = – f ( x),
∴ odd function or symmetric about origin …(i)
(b)
…(ii)
x=0 ⇒ y =0
x
–x
dy e + e
(c)
…(iii)
=
> 0 for all x
dx
2
inc
co reas
nc
av ing
eu
p
d 2y
⇒
…(ii)
Inc
co reas
nc
av ing
eu
p
Differentiating, we get;
x=0 ⇒y =1
dy e x – e – x
e 2x – 1
=
=
dx
2
2 ex
Curvature and Transformations
(b) When
Fig. 2.226
ex – e– x
.
2
e x – e –x
e x + e –x
f (– x) = – f ( x), ∴ odd function or symmetric about origin
…(i)
x=0
…(ii)
as
⇒
y =0
Domain ∈ R


Range ∈ (– 1, 1)
dy
> 0 for all
dx
x ∈R
…(iii)
…(iv)
129

< 0 for x > 0
2

dx

2
d y

>
0
x
<
0
for
2

dx
y
d 2y
Play with Graphs
(e)
…(v)
1
y=
O
From Eqs. (i), (ii), (iii), (iv) and (v) graph of
ex – e– x
is shown as in Fig. 2.227.
y = x
e + e– x
(iv) Sketching y =
e –e
Fig. 2.227
–x
(a) f (– x) = – f ( x), ∴ odd function or symmetric about origin
…(i)
as x → 0; y → ∞
y

…(ii)
(b)
as y → 1; x → ∞ 
ex + e–x

y= x
y → – 1; x → – ∞
e – e–x
1
Domain ∈ R – { 0} 
(c)
…(iii)

Range ∈ R – [– 1, 1]
x
O
dy
…(iv)
(d)
< 0 for all x ∈ R – { 0}
–1
dx
2

d y
> 0 for x > 0
2

dx
(e)
…(v)

2
d y
Fig. 2.228

< 0 for x < 0
2

dx

ex + e– x
From above information, graph for y = x
as shown in Fig. 2.228.
e – e– x
Note
In many applications we come across exponential functions of the form
1 x
(e – e – x ) and
2
1 x
(e + e – x ) known as “hyperbolic functions” represented by;
2
ex – e– x 

2
–x 
x
e +e 
cos hx =

2
called hyperbolic sine and hyperbolic cosine.
ex – e– x 
tan hx = x
e + e – x 
also,

ex + e– x 
cot hx = x
e – e – x 
sin hx =
…(i)
…(ii)
called hyperbolic tangent and hyperbolic cotangent. Where if; x = cos ht , y = sin ht
{Using Eq. (i)}
⇒
x 2 – y 2 = cos2 ht – sin2 ht = 1
which is equation of hyperbola.
ex – e– x
ex + e– x
Thus, sin hx =
and cos hx =
2
2
130
x
–1
e x + e –x
x
ex – e–x
e x + e –x
are hyperbolic functions.
SOLUTION Since, the function is periodic with period 2 π, it is sufficient to investigate the
function in the interval [0, 2 π].
dy
(a)
= 2 cos x – 2 sin 2x = 2 (cos x – 2 sin x cos x)
dx
dy
∴
= 2 cos x(1 – 2 sin x)
dx
dy
π π 5π 3π
Here,
=0 ⇒ x= , ,
,
dx
6 2 6 2
d 2y
(b)
= – 2 sin x – 4 cos 2x
dx 2
 d 2y 
3
where y =
 2
= – 3< 0
2
π
 dx  at x =
6
∴
at
x=
 d 2y 
 2
= 2 > 0;
 dx  at x = π
π
π
is maximum at x = .
6
6
2
∴ at
π
π
x = is minimum at x = , where y = 1
2
2
at
x=
5π
6
we have,
and
y =
3
2
(maximum)
at
and
x=
d 2y
dx
2
y
2
3/2
1
= – 3< 0
O
3π
, we have,
2
d 2y
dx
–1
= 6> 0
2
π
6
π
2
5π
6
π
3π
2
2π
x
Curvature and Transformations
y = 2 sin x + cos 2x.
E X A M P L E 18 Sketch the curve
–2
y = – 3 (minimum)
–3
Thus, curve for y = 2 sin x + cos 2x;
y = 2 sin x + cos 2x
Fig. 2.229
EXAMPLE
SOLUTION
19
Sketch the curve
y = e– x
As the curve y = f ( x) = e – x
(a) Symmetric about y-axis
2
(Gaussian curve)
2
{as f (– x) = f ( x)}
…(i)
(b)
as x → 0 ⇒ y → 1 

as y → 0 ⇒ x → ± ∞
…(ii)
(c)
dy

< 0 ⇒x> 0
2
dy

= e – x (–2x) ⇒ dx

dy
dx
> 0 ⇒ x < 0

dx
…(iii)
131
d 2y
Play with Graphs
(d)
2
2
= – 2e – x – 2x(– 2xe – x ) = 2e x ( 2x 2 – 1)
2
> 0 ⇒x< –
d y
1
dx
d 2y
dx
⇒
2
2
2
dx 2
<0 ⇒–
2
1
or x >
2
1
< x<
2
1

2



…(iv)
y
2
(e) y = e – x assumes maximum at
x=0 ⇒ y =1
and
Domain ∈ R
e up
cav
Con asing
e
incr
…(v)
Range ∈ ( 0, 1]
C
n 1 on
ow
de cav
d
e ng
cr e d
i
ea
av
o
nc reas
sin wn
o
C inc
g
e–1/2
O
–1/√2
1/√2
y=1
Conc
ave u
p
decr
easin
g
x
From above discussion;
Fig. 2.230
EXAMPLE
SOLUTION
20
Sketch the curve
y =
x
1 + x2
.
Domain ∈ R


Range ∈ [– 0.5, 0.5]
Here; (a)
…(i)
(b) f (– x) = – f ( x), hence, f ( x) is odd, symmetric about origin
(c) When
x=0
⇒
…(ii)
y =0
…(iii)
–
2
(d)
dy
1– x
=
dx (1 + x 2 ) 2
or
dy

> 0 when – 1 < x < 1

dx

dy
< 0 when x < – 1 or x > 1

dx
+
–
–1
1
Fig. 2.231
…(iv)
From above maximum at x = 1 and minimum at x = – 1}
2
d y
(e)
dx
2
d y
⇒
2
dx
d 2y
dx 2
=
–
2 ( x – 3)
(1 + x 2 ) 3
< 0, when x ∈ (– ∞, –
y =
…(v)
2
> 0, when x ∈ (– 3, 0) and ( 3, ∞ )
From above conditions graph for
132
2
x
x
1 + x2
3 ) and ( 0,
+
–√3




3 )

–
0
+
–√3
Fig. 2.232
…(vi)
Curvature and Transformations
y
1
(1, 0.5)
0.5
decr
easin (√3, √3/4)
g
ng
i
as
–√3
decr
easin
g
in
–1
re
c
O
1
(–√3, –√3/4)
–0.5
(–1, –0.5)
Concave down
x
√3
Concave up
Concave down
Concave up
Fig. 2.233
In above figure :
dy/dx
d 2y / dx 2
3
–
–
– 3< x< –1
–
+
– 1< x< 0
+
+
0< x< 1
+
–
1< x<
3
–
–
3< x< ∞
–
+
x
–∞ < x < –
EXAMPLE
Sketch the curve
21
y 2 = x 3 . (Semicubical parabola).
To plot the curve we discuss;
(i) Symmetric about x-axis (as y → – y curve remains same).
SOLUTION
(ii) Domain ∈ [ 0, ∞ )
(iii) Range ∈ R.
y 2 = x3
Here, to plot
⇒ y =
x3
and
y =–
x 3 , we draw y =
x 3 and take image
about x-axis for y = – x 3 . (i.e., to discuss curve when x, y ≥ 0).
dy 3 1/ 2 3
(iv)
= x =
x > 0 for all x, y > 0.
dx 2
2
y
d 2y
3
(v)
=
> 0 for all x, y > 0.
dx 2
4 x
⇒ Increasing and concave up when x > 0.
From Eqs. (i), (ii), (iii), (iv) and (v).
x
y
dy / dx
O
y = (x 3/2 )
x
d 2y / dx 2
y = –(x 3/2 )
+
+
+
+
+
–
–
–
Fig. 2.234
133
Note
In above curve y 2 = x 3 (semicubical parabola). For x = 0 we have y = 0 and y′ = 0, thus, the
Play with Graphs
branch of the curve has a tangent at y = 0 at origin. The second branch of y = –
x3 also
passes through origin and has the same tangent y = 0. Thus, two different branches of the
curve meet at origin, have the same tangent, and situated at different sides of origin. This
is known as cusp of first kind.
EXAMPLE
22
SOLUTION
Here;
y 2 = x 4 – x 6.
Sketch the curve:
y = ± x2 1 – x2
Thus, to plot the curve for y = x 2 1 – x 2
and take image about x-axis.
(i) Symmetric about x and y-axis.
(ii) Domain ∈[– 1, 1]
 2
2 
(iii) Range ∈ –
,
.
 3 3 3 3
(iv) When
x=0 ⇒ y =0
y = 0 ⇒ x = 0, ± 1.
dy
(v) 2y
= 4x 3 – 6x 5 = 2x 3 ( 2 – 3x 2 ) = 2x 3 ( 2 – 3x)( 2 + 3x)
dx
dy
2
and y > 0.
> 0 when; 0 < x <
⇒
dx
3
dy
2
< x < 1 and y > 0.
< 0 when
3
dx
Here, we are sketching the curve only when x, y ≥ 0and then take image about x and y-axis.
(vi)
d 2y
dx
2
> 0 when
2
5
0< x<
d 2y
dx
2
y > 0.
and
< 0 when
2
< x < 1 and
5
y > 0.
From above discussion.
y
By symmetry
–1
O
–
√2 – √2
5
3
Concave up and increasing
Concave downward increasing
2
Concave down and decreasing
3√3
1
x
– 2
By symmetry
3√3
√2 √2
5 3
Fig. 2.235
Note
134
At the origin (as the singular point) the two branches of the curve corresponding to plus
and minus in front of the radical sign are mutually tangent. Known as point of osculation
or tacnode or double cusp.
1. Sketch the curves; (where [ ⋅ ] denotes the
greatest integer function).
(i) y = |2 – | x – 1||
4
(ii) y = 2 –
| x – 1|
4
(iii) | y | = 2 –
| x – 1|
(iv) y = 2 –
(vi) y = 2sin x
(vii) y = log2 (| sin x | )
(viii) | y | = log2 | sin x |
 1
(ix) y = logsin x  
 2
 1
(x) | y | = logsin x  
 2
4
| x – 1|
1– x 2
3. Sketch the curve y = sin–1 

1+ x 2
4
| x – 1|
4. Sketch the curve y =
(v) | y | = 2 –
(vi) y = | e| x | – 2 |
5. Sketch the curves;
(i) y = x 2 – 2 | x |
(vii) | y | = | e| x | – 2 |
(viii) y = x – [x ]
(ix) y = x – [x ]
(x) y = (x – [x ])
(iii) y = e
(v) y = x – x
3
2
(ii) y = e –| x |
(iv) | y | = x
(vi) y 2 = x – 1
6. Construct the graph of the function
(xi) | y | = x – [x ]
(xii) | y | = (x – [x ])
|x|
2x
x +1
2
2
(xiii) y = | x – 1| + | x + 1|
(xiv) | y | = | x – 1| + | x + 1|
(xv) y = [| x – 1|]
(xvi) | y | = [| x – 1|]
(xvii) y = x + [x ]
(xviii) y = | x | + [| x |]
(xix) | y | = x + [x ]
(xx) | y | = | x | + [| x |]
2. Sketch the curves;
(i) y = sin x
(ii) | y | = sin x
(iii) y = | sin x | + | cos x |
(iv) | y | = cos x + | cos x |
(v) y = sin2 x – 2 sin x
y = f (x – 1) + f (x + 1)
1 – | x |, when | x | ≤ 1
where f (x ) = 
when | x | > 1
0,
7. Sketch the curves;
1
(i) y =
x –2
(iii) y =
1
|x | – 2
(ii) y =
Curvature and Transformations
EXERCISE
1
|x | – 2
(iv) | y | =
1
|x | – 2
8. Find number of solutions of 2 cos x = | sin x |.
When x ∈[ 0, 4π ].
9. Find the number of solutions of;
sin π x = |log| x ||.
10. Sketch the curve
sin 2x
y =
+ cos x .
2
ANSWERS
8. 2 solutions,
9. 6 solutions.
135
Play with Graphs
Graph Jells Us
Remark 1
Absolute maximum
No greater value of f any where.
Also a local maximum
Local minima
No greater value of f near by
y = f (x)
Local minimum
No smaller value of f near by
Local minima
No smaller value of f near by
Absolute minima
No smaller value of f any where.
Also a local minimum.
a
c
e
d
Local vs absolute (Global) Extrema
b
x-axis
Fig. 1
Remark 2
Absolute maximum
f ′undefined
Local maximum
f ′= 0
y = f (x )
f ′< 0
f ′> 0
f′ > 0
No extremum
f ′= 0
f ′> 0
c1
f ′< 0
Local minimum
f ′= 0
Local minimum
Absolute minimum
a
No extremum
f ′= 0
c2
c3
c4
c5
b
x-axis
Local vs absolute (Global) Extrema
Fig. 2
Remark 3
The First derivative Test for Local Extreme Value
(i) If f ′ changes from positive to negative at C (f ′ > 0 for x < c and f ′ < 0 for x > c), then f has a
local maximum value at c.
136
f ′> 0
f ′< 0
f ′> 0
f ′< 0
x=c
x=c
x-axis
x-axis
(b) f ′(c) undefined
(a) f ′(c) = 0
Fig. 3
(ii) If f ′ changes from negative to positive at c (f ′ < 0 for x < c and f ′ > 0 for x > c), then f has a
local minimum at x = c.
f ′< 0
f ′< 0
f ′> 0
f ′> 0
Local minimum
Local minimum
x=c
Graph Tells Us
Local maximum
Local maximum
x=c
x-axis
(a) f ′(c) = 0
x-axis
(b) f ′(c) undefined
Fig. 4
(iii) If f ′ does not change sign at c (f ′ has the same sign on both sides of c), then f has no local
extreme value at c.
no extreme
no extreme
f ′> 0
f ′> 0
x=c
x=c
x-axis
(a) f ′(c) = 0
x-axis
(b) f ′(c) undefined
Fig. 5
If f ′ < 0
(f ′ > 0) for x > a, then f has local
(iv) At a left end point ‘a’:
Local maximum
Local minimum
f ′< 0
maximum (minimum) value at
x = a.
x=a
x-axis
x=a
f ′> 0
x-axis
Fig. 6
(v) At a right end point ‘b’:
If f ′ < 0 ( f ′ > 0) for x < b, then f has local minimum
(maximum) at x = b.
Local maximum
Local minimum
x=b
x=b
x-axis
x-axis
Fig. 7
137
Play with Graphs
er
t
ap
h
C
ASYMPTOTES,
SINGULAR POINTS AND
CURVE TRACING
3
In this chapter we shall study:
➥ Plotting of Rational and Irrational functions.
➥ Intersection of Curve and Straight line at Infinity.
3.1 ASYMPTOTES
A straight line, at a finite distance from origin, is said to be an asymptote of the curve y = f ( x), if
the perpendicular distance of the point P on the curve from the line tends to zero when x or y both
tends to infinity.
OR
A straight line A is called an asymptote to a curve, if the distance δ from the variable point M of
the curve to this straight line approaches zero as the point M tends to infinity. Shown as:
y
y
ym
As
e
ot δ
M
pt
M
e
t
o
pt
m
sy
A
Curve
x
O
O
x
Fig. 3.1
Mathematically
Let y = f ( x) be a curve and let (x, y) be a point on it.
Tanget at (x, y) is given by;
Y–y =
Y=
138
dy
( X – x)
dx
dy 
dy

⋅ X + Y – x 

dx 
dx
…(i)
Now we shall discuss the following cases
(i) Asymptote parallel to x-axis.
(ii) Asymptote parallel to y-axis.
(iii) Asymptote of algebraic curves or oblique asymptotes.
(iv) Asymptote by inspection.
(v) Intersection of curve and its Asymptotes.
(vi) Asymptote by Expansion.
(vii) The position of the curve with respect to asymptote.
3.1 (i) Asymptote parallel to x-axis
Let the equation of curve be,
( a 0 x n + a1 x n – 1 y + a 2 x n – 2 y 2 + … + a n y n ) + ( b1 x n – 1 + b 2 x n – 2 y + … + b n y n – 1 )
…(i)
+ ( c 2 x n – 2 + c 3 x n – 2 y + … + c n y n– 2 ) + … + … = 0
then it can be arranged in descending powers of x as follows:
…(ii)
a 0 x n + ( a1 y + b1 )x n – 1 + ( a 2 y 2 + b 2 y + c 2 )x n – 2 + … = 0
n–1
n
Now, if a 0 = 0, i.e., the term consisting x is absent, then a1 y + b1 = 0, i.e., coefficient of x = 0
will make two roots of Eq. (i) infinite as coefficients of both x n and x n–1 are zero.
Hence, a1y + b1 = 0 is an asymptote parallel to x-axis.
Again if; both x n and x n–1 are absent, then a 2 y 2 + b 2 y + c 2 = 0, i.e., coefficient of x n– 2 being
zero will make three roots of Eq. (ii) infinite hence, a 2y 2 + b2y + c2 = 0 will give two asymptote
parallel to x-axis.
Method to find asymptote parallel to x-axis
To find the asymptote parallel to x-axis equate the coefficient of highest power of x to zero.
If the coefficient is constant, then there is no asymptote parallel to x-axis (horizontal).
3.1 (ii) Asymptote parallel to y-axis
From above article, if we need an asymptote parallel to y-axis, equate the coefficient of highest
power of y to zero.
If this coefficient is constant, then there is no asymptote parallel to y-axis (vertical).
EXAMPLE
1
Sketch the curve
SOLUTION Here;
y ( x – 5) = 1
∴ Asymptote parallel to x-axis.
⇒
y =
1
x–5
y =0
(equating highest power of x = 0)
x=5
(equating highest power of y = 0)
Asymptotes, Singular Points and Curve Tracing
Now, if asymptote exists, then x → ∞
dy
dy 

and  y – x  → finite limit say m and c
⇒

dx
dx 
dy
x dy
and
Say
→m
→c
y –
dx
dx
∴ Eq. (i) reduces to, Y = mX + c is asymptote of equation.
Asymptote parallel to y-axis.
⇒
139
Thus, x = 5 and y-axis are asymptotes shown as in figure.
Asymptote
Play with Graphs
y
y=
1
x–5
Asymptote
O
x y=0
x=5
Fig. 3.2
EXAMPLE
Show the curve y = tan x has an infinite number of vertical asymptote.
2
y = tan x
SOLUTION
here
y → ± ∞ as
or
tan x → ∞ as
π
3π
,±
,…
2
2
π
3π
,…
x→ ± , ±
2
2
x→ ±
i.e., equating highest power of y = 0.
(as y = tan x ⇒ y cot x = 1, where
Shown as:
cot x → 0).
y
O
–π/2
π/2
π
x
3π/2
Asymptote
Asymptote
–π
Asymptote
Asymptote
–3π/2
Fig. 3.3
EXAMPLE
3
SOLUTION
Here
⇒
or
140
Show the curve y = e1/ x has a vertical and horizontal asymptote.
y = e1/ x
y ⋅ e –1/ x = 0
e –1/ x → 0
as
x→ 0
(Since, lim e –1/ x → 0)
x→ 0
y = e1/ x
1
= log y
x
⇒
x=
Asymptote y =1
1
log y
1
O
which shows x(log y ) = 1 has an asymptote
parallel to x-axis as
log y = 0 ⇒ y = 1.
Thus, y = e
1/ x
has two asymptote
x = 0 and
x
Asymptote
⇒
Fig. 3.4
y = 1.
3.1 (iii) Asymptote of algebraic curves or oblique asymptote
An asymptote which is not parallel to y-axis is called an oblique asymptote. Let y = mx + c be an
asymptote of y = f ( x), then
y
and
c = lim ( y – mx)
m = lim
x→ ∞
x→ ∞
x
or x→ – ∞
or x→ – ∞
Method to find oblique asymptote
Suppose y = mx + c is an asymptote of the curve. Put y = mx + c in the equation of the curve and
arrange it in descending powers of x. Equate to zero the coefficients of two highest degree terms.
Solve these two equations, find m and c. Put them in y = mx + c to get asymptotes.
Note
1. Here, we will find non-parallel or non repeated asymptote only.
2. Neglect all imaginary values of m.
EXAMPLE
SOLUTION
1
Find the asymptotes to the curve
Here, the given curve y = x +
⇒
or
y =x+
1
and then sketch.
x
1
x
x y = x2 + 1
x 2 – xy + 1 = 0
(i) Asymptote parallel to x-axis
Equating highest power coefficient of x to zero in x 2 – xy + 1 = 0
⇒
1= 0
Asymptotes, Singular Points and Curve Tracing
y
From adjoining figure
(which is not true)
∴ no asymptote parallel to x-axis.
(ii) Asymptote parallel to y-axis
Equating highest power coefficient of y to zero in
141
x 2 – xy + 1 = 0
Play with Graphs
⇒
– x=0
(i.e., y-axis) is asymptote for y = x +
x=0
or
1
x
(iii) Oblique asymptote
Let
y = mx + c in
x 2 – xy + 1 = 0
i.e.,
x 2 – mx 2 – xc + 1 = 0
⇒
x 2 (1 – m ) – ( c)x + 1 = 0
Equating highest and second highest power of x to zero
i.e.,
1 – m = 0 and
∴
m = 1 and
or
y =x
is oblique asymptote to
c=0
c=0
1
y =x+ .
x
Now to trace the curve;
(iv) Symmetric about origin (as odd function)
(v) Domain ∈ R – { 0}.
(vi) Range ∈ (– ∞, – 2] ∪ [ 2 , ∞ )
(vii)
+
dy
x2 – 1
1
{using
number
line
rule,
=1– 2 =
–1
dx
x
x2
dy
> 0, when x < – 1 or x > 1
dx
dy
< 0, when – 1 < x < 1 – { 0}
dx
–
+
1
}
y max at x = – 1
which shows
y min at x = 1
d 2y
(viii) Also,
⇒
dx
d 2y
dx 2
=
2
x3
> 0, when
d 2y
dx 2
142
2
< 0, when
x> 0
(concave up)
x< 0
(concave down)
1
as:
x
up
ve
a
c
on y = x
y
d
Asymptote
an
1
y = x + , x > 0 ing
x
s
ea
cr
In
–1
c
2
Local minimum at x = 1
Asymptote
x
1
O
Local maximum
–2
y=x+
1 ,x<0
x
Fig. 3.5
EXAMPLE
SOLUTION
2
Find the asymptotes of the curve
Here, the curve y =
y =
x 2 + 2x – 1
and hence, sketch.
x
x 2 + 2x – 1
could be written as;
x
x 2 + 2x – yx – 1 = 0
(i) No asymptote parallel to x-axis.
(ii) Asymptote parallel to y-axis.
⇒ x = 0.
(iii) Oblique asymptote
Let
∴
y = mx + c be oblique asymptote
x 2 + 2x – x( mx + c) – 1 = 0
x 2 – mx 2 + 2x – cx – 1 = 0
⇒
x 2 (1 – m ) + x( 2 – c) – 1 = 0
…(i)
Asymptotes, Singular Points and Curve Tracing
Using above information we can trace y = x +
For oblique asymptote equate highest power and second highest power of x to zero.
i.e.,
∴
Coefficient of
x2 = 0 ⇒ m = 1
Coefficient of x = 0 ⇒ c = 2
1
y = x + 2 is oblique asymptote to y = x – + 2.
x
143
Play with Graphs
(iv) Neither symmetric about axis nor about origin.
(v) Domain ∈ R – { 0}.
(vi) Range ∈ R.
dy
1
(vii)
=1+ 2
dx
x
dy
⇒
> 0, for all x ∈ R – { 0}.
dx
d 2y
2
(viii)
=– 3
2
dx
x
2
d y
> 0, when x < 0
⇒
dx 2
d 2y
< 0, when x > 0
dx 2
1
Using above information, we can plot the curve y = x – + 2 as;
x
(concave down)
(concave up)
y
3
y=
2
x 2 + 2x – 1
x
1
–2
–1
x
O
1
2
Asymptote x = 0
pt
+
ym
x
=
y
As
2
ot
e
–1
–2
–3
Fig. 3.6
3.1 (iv) Asymptote by inspection
If the equation of the curve be of the form Fn + Fn – 2 = 0, where Fn and Fn– 2 are expressions in x
and y such that degree of Fn = n and degree of Fn– 2 ≤ n – 2, then every linear factor equated to zero
will give an asymptote if no two straight lines represented by any other factor of Fn is parallel or
coincident with it.
144
SOLUTION
1
Find the asymptote of the curve x 2 y + xy 2 = a 3 .
Here, the given curve is.
or
x 2 y + xy 2 = a 3
x 2 y + xy 2 – a 3 = 0
This equation is of the form
Fn + Fn – 2 = 0
Here,
and
F3 = x 2 y + xy 2
F0 = – a 3
∴ By inspection the asymptotes are given by
x 2 y + xy 2 = 0
∴ The asymptotes are
EXAMPLE
2
x = 0,
y = 0,
or
xy ( x + y ) = 0
x + y = 0.
Find the asymptote of the curve
y =x+
1
(by Inspection).
x
Here, the given curve is x 2 – xy + 1 = 0
This equation is of the form
Fn + Fn – 2 = 0
Here
F2 = x 2 – xy
F0 = 1
By inspection the asymptotes are given by
∴
or
x 2 – xy = 0
x( x – y ) = 0
The asymptotes are x = 0 and x – y = 0.
∴
SOLUTION
3.1 (v) Intersection of curve and its asymptote
An asymptote of curve of nth degree cut the curve in ( n – 2) points provided the asymptote is not
parallel to any asymptote.
Hence, if there be N asymptotes of the curve, then they cut the curve in N( n – 2) points.
Note
The number of asymptotes of an algebraic curve of nth degree can not be more than n.
EXAMPLE
points.
SOLUTION
1
Show the asymptote of the curve xy ( x 2 – y 2 ) + x 2 + y 2 – 1 = 0 cut at 8
The equation of the curve is,
xy ( x 2 – y 2 ) + x 2 + y 2 – 1 = 0
Here
…(i)
n=4
This equation is of the type
Fn + Fn – 2 = 0
Hence,
and
Fn = xy ( x 2 – y 2 ) = xy ( x – y ) ( x + y )
Asymptotes, Singular Points and Curve Tracing
EXAMPLE
Fn– 2 = x 2 + y 2 – 1
∴
Fn = 0
⇒ x = 0, y = 0, x – y = 0 and
x + y = 0 are the equations of asymptotes.
The combined equation of the asymptotes is,
xy ( x – y )( x + y ) = 0
…(ii)
145
Play with Graphs
Subtracting Eq. (ii) from (i), we get
x2 + y 2 – 1 = 0
Thus, intersection of curve and asymptotes lie on this curve since, there are 4 asymptotes, i.e.,
N = 4.
∴ Point of intersection of curve and asymptotes = 4( 4 – 2) = 8.
3.1 (vi) Asymptote by expansion
If the equation of the curve is of the form
y = mx + c +
A
B
C
+
+ 3 +…
x x2
x
y = mx + c will be an asymptote of the given curve.
Then
EXAMPLE
SOLUTION
1
Find the asymptote of the curve y 3 = x 2 ( x – a).
a

y 3 = x 2 ( x – a) = x 3 1 – 

x
1/ 3

a
1 a 1 a2 

or
y = x 1 – 
y = x 1 –
…
–

x
3 x 9 x2 

The curve is,
⇒
a 1 a2
–
…
3 9 x
A
B
y = mx + c + + 2 + …
x x
y =x–
or
⇒
y = mx + c
Hence,
which is of the form
is asymptote
a
y = x – is asymptote of the given curve.
3
EXAMPLE
SOLUTION
2
Find the asymptote for
Here; y = x +
1
x
1
y =x+ .
x
is of the form,
y = mx + c +
A
B
+ 2 +…
x x
⇒ y = x is asymptote of the curve
1
y =x+ .
x
Note
Above method is useful to find oblique asymptote. Thus, students are adviced to find
vertical and horizontal asymptote (i.e., asymptote parallel to x-axis and y-axis).
3.1 (vii) The position of the curve with respect to an asymptote
Let the equation of the curve is of the form;
A
B
C
y = mx + c +
+
+ 3 + …, then
x x2
x
146
or
or
(ii) A = 0, B > 0
(iii) A = 0, B = 0, C ≠ 0 and C and x have same signs and
(b) The curve lies below the asymptote if
(i) A ≠ 0 and, A and x have opposite signs.
or
or
(ii) A = 0, B < 0
(iii) A = 0, B = 0, C ≠ 0 and C and x have opposite signs.
EXAMPLE
For the curve y 5 = x 5 + 2x 4 ; show;
1
(i) The curve lies above the asymptote
(ii) The curve lies below the asymptote
SOLUTION
The given curve is,
2
, if x < 0.
5
2
y = x + , if x > 0.
5
y =x+
y 5 = x 5 + 2x 4
2

y 5 = x 5 1 + 

x
or
2

y = x 1 + 

x
1/ 5
2
8
2 1
8 1 

y = x 1 + ⋅ –
⋅
… = x + –
+…

5 x 25 x 2 
5 25x
∴ The asymptote is
(i) Now if
y =x+
2
;
5
8
and x have same sign
25
⇒ x < 0. Then the curve lie above the
8
and x have opposite sign
25
⇒ x > 0. Then the curve lie below the
A =–
asymptote.
(ii) Now if A = –
asymptote.
EXAMPLE
2
For the curve y = x +
1
x
show,
y = x, if
x> 0
(ii) The curve lies below the asymptote y = x, if
x< 0
(i) The curve lies above the asymptote
SOLUTION
Asymptotes, Singular Points and Curve Tracing
(a) The curve lies above the asymptote if
(i) A ≠ 0 and, A and x have same signs
1
The given curve is, y = x + , is of the form
x
A
B
C
y = mx + c +
+ 2 + 3…
x x
x
147
1
y =x+ .
x
A = 1 and x have same sign ⇒ x > 0, then the curve lies above the
Thus, y = x is the asymptote to
Play with Graphs
(i) Now if
asymptote.
A = 1 and x have opposite sign
(ii) Now if
⇒ x < 0, then the curve lies below the
asymptote.
3.2 SINGULAR POINTS
Here, we shall discuss the following
(i) Multiple points
(ii) Double points
Types of double points :
(a) Node
(b) Cusp
(c) Isolated point
(iii) Tangent at the origin.
(iv) Necessary conditions for existence of double points.
(v) Types of cusps.
3.2 (i) Multiple points
A point on a curve is said to be a multiple point of order r, if r branches of the curve pass through
this point.
If P is the multiple point of order r, then there will be r tangents at P, one of each of the r
branches. These r tangents may be real, imaginary, distinct, coincident.
3.2 (ii) Double points
A point on a curve is said to be a double point of the curve, if two branches of the curve pass
through this point.
Double points have two tangents, they may be real, imaginary, distinct or coincident.
Types of Double points
(a) Node
If the two branches of a curve pass through the double point and the tangents to them at the point
are real and distinct, then the double point is called a node as shown in Fig. 3.7.
y
Tangent
P
Curve
y
Tangent
nt
n
Ta
ge
Curve
Node
x
O
Fig. 3.7
Cusp
O
x
Fig. 3.8
(b) Cusp
If the two branches of the curve pass through the double point and the tangent to them are the
point is real and coincident, then the double point is called cusp as shown in Fig. 3.8.
148
x→ c+
x → c–
lim f ′ ( x) = ∞
2. lim f ′ ( x) = – ∞
and
1. lim f ′ ( x) = ∞
and
x→ c–
x→ c–
OR
x→ c+
shown as:
lim f ′ ( x) = – ∞
x→ c+
2. lim f ′ ( x) = – ∞ and lim f ′ ( x) = ∞
y-axis
x→ c+
x→ c–
y-axis
y = f ( x)
Cusp
lim f ′(x) = ∞
x→c–
lim f ′(x) = –∞
lim f ′(x) = ∞
x→c+
lim –f ′(x) = –∞
x→c+
x→c
Cusp
O
x-axis
c
y = f (x)
c
O
Fig. 3.9
Note
x-axis
Fig. 3.10
A cusp can either be a local maximum (1) or a local minima as in (2).
(c) Isolated point
If there are no real point on the curve in the neighbourhood of a point P is called an isolated or a
conjugate point.
y
P
isolated point
x
O
Fig. 3.11
3.2 (iii) Tangent at the origin
If an algebraic curve passes through the origin, the equation of tangent or tangents at the origin is
obtained by equating to zero the lowest degree terms in the equation of the curve.
EXAMPLE
SOLUTION
1
Show that the curve
Asymptotes, Singular Points and Curve Tracing
Cusps :
The graph of a continuous function y = f ( x) has a cusp at a point x = c if the concavity is
same on the both side of c and either.
1. lim f ′ ( x) = ∞ and lim f ′( x) = – ∞
y 2 = 4x 2 + 9x 4 has a node at origin and hence, sketch.
The equation of the curve is,
y 2 = 4x 2 + 9x 4
…(i)
149
Play with Graphs
It passes through the origin.
Now, equating to zero the lowest degree terms of the given curve, i.e.,
y 2 – 4x 2 = 0
⇒
y = 2x and y = – 2x
There are two real and distinct tangents y = 2x and y = – 2x. Thus, two branches of curve
passes through origin (0, 0).
…(ii)
∴ origin is node.
Now to sketch;
(iii) Symmetric about x-axis, y-axis and origin.
(iv) As x → 0 ⇒ y → 0
(v) Domain ∈ R.
(vi) Range ∈ R.
Here, we shall discuss the behaviour of y = x 4 + 9x 2 , x ≥ 0 and use symmetry to
construct y 2 = x 2 ( 4 + 9x 2 ).
dy
(vii) 2y
= 8x + 36x 3 = 4x( 2 + 9x 2 )
dx
dy
⇒
> 0 for all x, y > 0
dx
(viii) Also,
2
d 2y
 dy 
= 4 + 54x 2
  +y
 dx 
dx 2
⇒
2
y
node
O
> 0 for all x
2
Fig. 3.12
Show origin is a conjugate point for
x 4 + y 3 + 2x 2 + 3y 2 = 0
SOLUTION
The given curve is,
x 4 + y 3 + 2x 2 + 3y 2 = 0
It passes through origin.
∴ To find equation of tangent at origin equating the lowest degree term to zero.
2x 2 + 3y 2 = 0
i.e.,
⇒
y =±i
2
x
3
which are imaginary tangents.
Hence, origin is a conjugate point of the curve.
3.2 (iv) Necessary conditions for the existence of double points
Let (x, y) be a point on the given curve f ( x, y ) = 0.
The necessary and sufficient conditions for (x, y) to be a double points are:
f = 0,
150
nt
y = –2x
d y
2
x
Ta
n
ge
dx
Thus, the graph for y 2 = x 2 ( 4 + 9x 2 )
EXAMPLE
nt y = 2x
ge
n
Ta
∂f
= 0,
∂x
∂f
= 0 at ( x, y )
∂y
are not all zero, then,
 ∂ 2f 


 ∂x ∂y 
2
 ∂ 2f   ∂ 2f 
–  2  2 > 0
 ∂x   ∂y 
2
– f xx f yy > 0
f xy
or
(ii) The double point will be an isolated point, if
2
– f xx f yy < 0
f xy
(iii) The double point will be a cusp if
2
– f xx f yy = 0.
f xy
Here, if f xx = f xy = f y y = 0at ( x, y ), then it will be a multiple point of order greater than 2.
EXAMPLE
1
For the curve x 3 + x 2 + y 2 – x – 4y + 3 = 0 , find the double point and hence,
whether the point is node or isolated point.
SOLUTION
Let
f ( x, y ) = x 3 + x 2 + y 2 – x – 4y + 3 = 0
∴
f x = 3x 2 + 2x – 1
f y = 2y – 4
∴
for a double point f x = 0,
f x = 0 ⇒ 3x + 2x – 1 = 0
1
x= , –1
3
2
or
f y = 0 ⇒ 2y – 4 = 0 ⇒ y = 2.
1 
∴ Possible double points are  , 2 , (– 1, 2)
3 
1 
Q
f  , 2 ≠ 0 and f(– 1, 2) = 0
3 
∴ f(– 1, 2) is a double point.
f xx = 6x + 2
⇒
f xx
at (– 1, 2) = – 4
f xy = 0
⇒
f xy
at (– 1, 2) = 0
f yy = 2
⇒
f yy
at (– 1, 2) = 2
∴
f xy – f xx f yy = 0 – (– 4)( 2) = 8 > 0
∴ (– 1, 2) may be node.
fy = 0
Asymptotes, Singular Points and Curve Tracing
∂ 2f
∂ 2f
∂ 2f
and
,
∂x ∂y ∂ x 2
∂y 2
(i) Double point will be a node if
Now, if
For shifting origin to (– 1, 2), substitute x = X – 1, y = Y + 2 in the given equation,
we get,
or
X 3 – 2X 2 + Y 2 = 0
Y = ± X 2– X
For numerically small values of X, Y is real.
∴ (–1, 2) is a node on the given curve.
151
EXAMPLE
2
For the curve x 3 + 2x 2 + 2xy – y 2 + 5x – 2y = 0, find the double point and
Play with Graphs
hence, check whether node, cusp or isolated point.
SOLUTION
Let
f ( x, y ) = x 3 + 2x 2 + 2xy – y 2 + 5x – 2y = 0
∂f
fx =
= 3x 2 + 4x + 2y + 5
∂x
∂f
= 2x – 2y – 2
fy =
∂y
f xx =
…(i)
∂ 2f
= 6x + 4
∂x 2
∂ 2f
f xy =
=2
∂x ∂y
f yy =
∂ 2f
=–2
∂y ∂x
fx = fy = f = 0
f x = 0 ⇒ 3x 2 + 4x + 2y + 5 = 0
f y = 0 ⇒ 2x – 2y – 2 = 0
or
2y = 2x – 2
2
Solving Eqs. (ii) and (iii), we get 3x + 4x + 2x – 2 + 5 = 0
⇒
x=–1
also
x = – 1, y = – 2
∴ (– 1, – 2) is a double point.
At (– 1, – 2),
f xx = 6 (– 1) + 4 = – 2
f xy = 2 , f yy = – 2
For double points
∴
2
f xy
– f xx f yy
…(ii)
…(iii)
satisfies the given equation.
at (– 1, – 2) = 0
∴ (–1, – 2) may be a cusp.
For shifting the origin to (– 1, – 2) substitute x = X – 1, y = Y – 2 in the given equation.
( X – 1) 3 + 2( X – 1) 2 + 2( X – 1)( Y – 2) – ( Y – 2) 2 + 5( X – 1) – 2( Y – 2) = 0
or
…(iv)
X 3 – X 2 + 2XY – Y 2 = 0
Y=X±X X
∴
Y is real for all positive value of X.
∴ Two branches of (iv) pass through origin.
∴ Two branches of (i) pass through (– 1, – 2).
⇒ (– 1, – 2) is a cusp.
EXAMPLE
3
SOLUTION
Let
Find for the curve y 2 = x sin x origin is node, cusp or isolated point.
f ( x, y ) = y 2 – x sin x
f x = – sin x – x cos x
f y = 2y
f xx = – cos x + x sin x – cos x
152
∴ origin is node.
3.2 (v) Types of cusps
When two branches of a curve pass through a cusp and the tangents at cusp are coincident.
Therefore, normal to the branches at a cusp would also be coincident.
Cusp can be of five kinds
(a) Single cusp
If the branches of the curve lie on the same side of the common normal, then the cusp is called a
single cusp.
y
y
Ta
n
ge
nt
nt
ge
n
Ta
Normal
O
Normal
Single cusp of first species
x
x
O Single cusp of second species
Fig. 3.13
(b) Double cusp
If the branches of the curve lie on the both sides of the common normal, then the cusp is called
double cusp.
Here, both the branches of the curve lie on the both sides of common tangent, then the cusp is of
first kind.
Also if, the branches of the curve lie on the same side of the common tangent, then the cusp is
called cusp of second species or Ramphoid cusp.
y
y
Normal
nt
Normal
e
ng
Ta
nt
ge
n
Ta
O
Double cusp of first species
Fig. 3.14
x
Asymptotes, Singular Points and Curve Tracing
f xy = 0
f yy = 2
∴ at x = 0 : f xx = – 2, f xy = 0, f yy = 2
2
at
(0, 0)
– f xx f yy
∴
f xy
⇒
0 + 2( 2) = 4 > 0
2
f xy
– f xx f yy > 0
x
O
Fig. 3.15
153
(c) Point of oscu-inflexion
A double cusp of both the species is called a point of oscu-inflexion.
Play with Graphs
y
Normal
nt
ge
n
Ta
x
O
Fig. 3.16
EXAMPLE
SOLUTION
1
Sketch the curve y 2 ( a + x) = x 2 ( a – x).
Here, the curve is
y 2 ( a + x) = x 2 ( a – x)
…(i)
1. The curve is symmetrical about x-axis.
2. Curve passes through origin and cuts the x-axis at a point (a, 0).
3. Equating to zero the lowest degree terms of Eq. (i), we get the tangents at origin.
∴
y 2 = x 2 or y = ± x
∴ origin is node.
4. Equating to zero the highest degree term, we get the asymptote
x + a = 0, i.e., x = – a.
y =±x
5. From Eq. (i)
a– x
a+x
∴ y exists when – a < x ≤ a ⇒ Domain ∈ (– a, a].
6. As x increases from 0 to a ⇒ y increases upto a point then decreases to zero.
y increases when x ∈[ 0, a]
⇒
y decreases when x ∈ (– a, 0]
Thus, y 2 ( a + x) = x 2 ( a – x) could be plotted as;
y
(–a, 0)
asym ptote
node
nt
ge
y=x
n
Ta
(a, 0)
O
Ta
n
ge
nt
y = –x
x = –a
154
Fig. 3.17
x
1. Check symmetry
(a) A curve is symmetrical about x-axis, i.e., y is replaced by – y and curve remains same.
(b) A curve is symmetrical about y-axis, i.e., f (– x) = f ( x).
(c) A curve is symmetrical about y = x, i.e., on interchanging x and y curve remains same.
(d) A curve is symmetrical about y = – x. i.e., on interchanging x by – y and y by – x curve
remains same.
(e) The curve is symmetrical in opposite quadrants, i.e., f (– x) = – f ( x).
2. Check for origin
Find whether origin lies on the curve or not.
If yes, check for multiple points (See Art. 3.2).
3. Point of intersection with x-axis and y-axis
Put
x = 0 and find y, put y = 0 and find x. Also obtain the tangents at such points.
4. Asymptotes
Find the point at which asymptote meets the curve and equation of asymptote (see Art. 3.1)
5. Domain and range
To check in which part the curve lies.
6. Monotonicity and maxima minima
dy
and check the interval in which y increases or decreases and the point at which it
dx
attains maximum or minimum.
Find
7. Concavity and convexity
The interval in which,
d 2y
and
dx 2
d 2y
>0
<0
dx 2
Using all the above results we can sketch the curve
y = f ( x ).
Asymptotes, Singular Points and Curve Tracing
3.3 REMEMBER FOR TRACING CARTESIAN EQUATION
155
Play with Graphs
SOME MORE SOLVED EXAMPLES
EXAMPLE
SOLUTION
1.
2.
3.
4.
5.
1
Sketch the curve
y 2 (a 2 + x 2 ) = x 2 (a 2 – x 2 )
y2 =
Here, the curve is
(a 2 + x 2 )
The curve is symmetric about x-axis and y-axis {as on replacing y by – y curve remains same
and on replacing x by – x curve remains same thus, symmetric about x and y-axis
respectively).
It passes through origin and y = ± x are two tangents at origin. Thus, the origin is node.
It meets x-axis at (a, 0), (0, 0) and (– a, 0) and meets y-axis at (0, 0) only.
The tangents at (a, 0) and (– a, 0) are x = a and x = – a respectively.
The curve has no asymptote.
a2 – x2
Here,
y =±x 2
a + x2
∴
Domain ∈[– a, a]
dy
a 4 – 2a 2 x 2 – x 4
= 2
dx ( a + x 2 ) 3/ 2 ( a 2 – x 2 )1/ 2
6.
dy
→∞
dx
i.e.,
when
a 4 – 2a 2 x 2 – x 4 = 0
dy
=
dx ( a 2 + x 2 ) 3/ 2 ( a 2 – x 2 )1/ 2
=
=
=
⇒
x→ ± a
as
dy
=0
dx
a 4 – 2a 2 x 2 – x 4
Also
156
x 2 (a 2 – x 2 )
– { x 4 + 2a 2 x 2 + a 4 – 2a 4 }
( a 2 + x 2 ) 3/ 2 ( a 2 – x 2 )1/ 2
– {( x 2 + a 2 ) 2 – ( 2a 2 ) 2 }
( a 2 + x 2 ) 3/ 2 ( a 2 – x 2 )1/ 2
– { x – (– 1 +
 0;

dy 
= + ve;
dx 
 – ve;

2 )a} { x + (–1 +
2 )a} { x 2 + (1 +
2 )a 2 }
( a 2 + x 2 ) 3/ 2 ( a 2 – x 2 )1/ 2
x = ± (– 1 +
2) a
x ∈ (–
2 ) a, (– 1 +
(– 1 +
x ∈ (– a, –
x ∈ (–
(– 1 +
(– 1 +
2 ) a)
2 ) a) or ( (–1 +
2) a,
i.e.,
y increasing when
and
y decreases when x ∈ (– a, – (– 1 +
(– 1 +
2 ) a, a)
2 ) a)
2 ) a) or ( (– 1 +
2 ) a, a)
–1+
where
2 = ( 0.6) approx
y (a + x ) = x 2 (a 2 – x 2 )
2
Thus, the curve for
2
2
y
Ta
n
rig
ge
n
ta
n
Ta
rig
in
(–a, 0)
nt
ge
to
(–0.6a, 0)
O
in
o
at
(0.6a, 0)
(a, 0)
x
Fig. 3.18
EXAMPLE
SOLUTION
2
y 2 ( x – a) = x 2 ( a + x).
Sketch the curve
Here, the curve is given by y 2 =
x 2 ( a + x)
( x – a)
1. Symmetrical about x-axis only.
2. It passes through origin and y 2 + x 2 = 0, i.e., y = ± ix are two imaginary tangents at origin.
Thus, origin is isolated point.
3. It meets x-axis at (– a, 0), (0, 0) and y-axis at (0, 0).
The tangent at (– a, 0) is x = – a.
x = a are three asymptote.
x+a
x 2 ( x + a)
5.
⇒ y =±x
y2 =
( x – a)
x– a
x+a
Thus, for domain;
≥ 0 and x ≠ a
x– a
4. y = ± ( x – a) and
x ≤ – a and
i.e.,
or
6.
⇒
+
–
–a
x> a
Domain ∈ (– ∞, – a] ∪ ( a, ∞ ) ∪ { 0}
1
1
 

 x – (1 + 5 )a  x – (1 –
2
2
dy
x – ax – a
2
2
 
=±
=± 
3/ 2
1/ 2
3/ 2
dx
( x – a) ( x + a)
( x – a) ( x + a)1/ 2
dy
> 0,
dx
when
1
x ∈ (– ∞, – a] ∪  (1 +
2
+
a
Asymptotes, Singular Points and Curve Tracing
OR
dy
> 0, when x ∈ (– ( 0.6) a, ( 0.6) a)
dx
dy
< 0, when x ∈ (– a, – ( 0.6) a) or (( 0.6) a, a)
dx

5 )a


5 )a, ∞

157
dy
< 0, when
dx
 1
x ∈  a, (1 +
 2

5 ) a

Play with Graphs
Thus, the curve;
y 2=
y
x 2(a + x)
( x – a) x + a = 0
(a, 2a)
ym
pto
(0, a)
te
x=a
asym ptote
as
x
(–a, 0) O (a, 0)
te
pto
ym
as
x 2(x + a)
y =
( x – a)
(0, –a)
2
(a, –2a)
y 2=
x 2 ( x + a)
(x – a)
y = –(x + a)
Fig. 3.19
EXAMPLE
3
Sketch the curve
y 2 = ( x – 1)( x – 2)( x – 3).
Here, y 2 = ( x – 1)( x – 2)( x – 3)
1. Symmetrical about x-axis.
2. It does not pass through origin.
3. It meets x-axis at (1, 0) (2, 0) and (3, 0) but it does not meet y-axis.
4. No asymptote.
5. For domain:
( x – 1) ( x – 2) ( x – 3) ≥ 0
–
SOLUTION
⇒
Domain ∈[1, 2] ∪ [ 3, ∞ )
1
–
2
+
3
y = ± ( x – 1)( x – 2)( x – 3)
6.
∴
dy
( 3x – 12x + 11)
=±
=±
dx
2 ( x – 1)( x – 2)( x – 3)
2
=±
⇒
158
+
3( x – 1.42)( x – 2.5)
2 ( x – 1)( x – 2)( x – 3)

6 – 3 
6 + 3
3 x –
 x –

3  
3 

2 ( x – 1)( x – 2) ( x – 3)

 6– 3
6+ 3
= 1.42 and
= 2.5/approx
 as
3
3


dy
> 0, when x ∈ (1, 1.42) ∪ ( 3, ∞ )
dx
dy
< 0, when x ∈ (1.42 , 2)
dx
y
O
1
2
3
x=1
x=2
x=3
x
Fig. 3.20
EXAMPLE
4
SOLUTION
Here,
y 2x 2 = x 2 – a2.
Sketch the curve
y2=
x2 – a2
x2
1. Symmetrical about both the axis.
2. It does not pass through origin.
3. x-intercepts are (a, 0) and (– a, 0)
The tangent at (a, 0) is x = a
the tangent at (– a, 0) is x = – a.
and
4. y = ± 1 are the two asymptotes.
5.
y =±
x2 – a2
⇒
x
a2
dy
=±
dx
x2 x2 – a2
Thus, the curve for y 2 =
x2 – a2
x2
Domain ∈ (– ∞, – a] ∪ [ a, ∞ )
⇒
dy
> 0,
dx
x ∈ (– ∞, – a) ∪ ( a, ∞ )
when
is,
y
asymptote
(0, 1)
y=1
2
y 2= x –2a
x
(–a, 0)
O
(a, 0)
(0, –1)
asymptote
x = –a
2
x
Asymptotes, Singular Points and Curve Tracing
y 2 = ( x – 1)( x – 2)( x – 3)
Thus, the curve
y = –1
x=a
Fig. 3.21
159
Play with Graphs
EXAMPLE
5
SOLUTION
Here,
Sketch the curve
y2 =
y 2 ( x 2 – 1) = 2x – 1.
2x – 1
x2 – 1
1. Symmetrical about x-axis.
2. It does not pass through origin.
1
2


3. It meets x-axis in  , 0
1 
The tangent at  , 0
2 
and y-axis in (0, 1) and (0, – 1) respectively.
is
1
x= .
2
4. x = 1, x = – 1 and
5. y 2 =
y = 0 are three asymptotes.

1
⇒ Domain ∈  – 1,  ∪ (1, ∞ )

2
2x – 1
x2 – 1
6. y = ±
2x – 1
⇒
x2 – 1


–x2 + x + 1
dy

=±
1/ 2
2
3/ 2
dx
 ( 2x – 1) ( x – 1) 
1

x ∈  –1,  ∪ (1, ∞ )

2
dy
< 0 when
dx
⇒
∴ y is decreasing in its domain.
2x – 1
Thus, the graph for y 2 = 2
is,
x –1
y
y 2=
2x – 1
x 2– 1
(0, 1)
x = –1
1
2
O
( , 0) (1, 0)
(0, –1)
asymptote
asymptote
(–1, 0)
x = 1/2 x = 1
Fig. 3.22
EXAMPLE
SOLUTION
160
6
Sketch the curve :
Here,
( x 2 + y 2 ) x – a( x 2 – y 2 ) = 0; ( a > 0)
 a – x
y 2 = x2 

 a + x
y=0
(asymptote)
x
4. x = – a is the only asymptote.
y =±x
5.
y
dy
a 2 – ax – x 2
=±
dx
( a + x) a 2 – x 2
dy
> 0 , when
dx
nt
e
ng
Ta
asymptote
6.
⇒
Domain ∈ (– a, a]

–1 + 5 
x ∈  – a,
a
2


.
(–a, 0)
⇒
dy
< 0 , when
dx
(a, 0)
O

 –1 + 5
x ∈
a, a .
2


Tangent
∴
a– x
a+x
x
Ta
n
ge
nt
Thus, the graph for
x = –a
 a – x
y 2 = x2 
 as shown in Fig. 3.23.
 a + x
EXAMPLE
7
x=a
Fig. 3.23
Sketch the curve x 3 + y 3 = 3ax 2 ( a > 0).
Here, x 3 + y 3 = 3ax 2
1. No line of symmetry.
2. Origin is cusp and x = 0 is tangent.
3. x-intercept, (0, 0) (3a, 0)
The tangent at (3a, 0) is x = 3a.
SOLUTION
 a 2a
4. y = a – x is asymptote and the curve meets asymptote at  ,  .
3 3 
5. Here;
x 3 + y 3 = 3ax 2
⇒
3ax 2 > 0
∴
x +y >0
6.
⇒
3
(as, a > 0)
3
i.e., x and y both cannot be negative (thus, curve would not lie in third quadrant).
dy
= x ( 2a – x)
y2
dx
dy
> 0, when x ∈ ( 0, 2a)
dx
dy
< 0, when x ∈ (– ∞, 0) ∪ ( 2a, ∞ )
dx
Asymptotes, Singular Points and Curve Tracing
1. Symmetric about x-axis.
2. Origin lies on the curve and y = ± x are two tangents at origin. So, origin is node.
3. x-intercept are (0, 0) and (a, 0). The tangent at (a, 0) is x = a.
161
Thus, the curve y 3 + x 3 = 3ax 2 is,
y=
a
Tangent
Play with Graphs
x+
Tangent
y
(0, a)
O
(a, 0)
(2a, 0)
x
(3a, 0)
as x = 3a
ym
pt
ot
e
x+
y=
a
Fig. 3.24
EXAMPLE
8
Sketch the curve with parametric equation θ.
x = a(θ + sin θ), y = a(1 + cos θ); x ∈ (– π, π ).
SOLUTION
Here, x = a(θ + sin θ) and
y = a(1 + cos θ) gives the following table for x and y
with θ.
θ
– π
0
π
x
– aπ
0
aπ
y
0
2a
0
So, that we have,
⇒
⇒
Also
and
Now,
162
– π≤θ≤ 0
(x, y) starting from (– aπ, 0) moves to the right and upwards to (0, 2a).
0≤ θ≤ π
the point (x, y) starting from (0, 2a) moves to the right and downward to ( aπ, 0).
dx
= a(1 + cos θ)
dθ
dy
= – a sin θ
dθ
dx
= 0 if θ = π, – π
dθ
tan θ
dy
,
=–
2
dx
Also,
Thus,
y
x
π
O
–π
Fig. 3.25
EXAMPLE
9
SOLUTION
Here;
x 5 + y 5 = 5a 2 xy 2 .
Sketch the curve :
1. The curve is symmetrical in opposite quadrants.
2. The curve passes through origin and x = 0, y = 0 are tangents. Thus, origin is node.
3. It meets coordinate axis at origin.
4. x + y = 0 is an asymptote.
5. On transfering to polar coordinates, we get.
r2=
5a 2 cos θ sin θ
cos 5 θ + sin 5 θ
π
when, θ = 0, r = 0 when, θ = , r = 0
2
3π 2
π
As θ increases from to
, r is negative and hence, r is imaginary.
2
4
∴ no portion of the curve lies in this region.
3π
3π
to π ⇒ r decreases from ∞ to 0.
At θ =
, r = ∞ as θ increases from
4
4
∴ Curve x 5 + y 5 = 5a 2 xy 2
y
x 5+ y 5= 5a 2xy 2
Asymptotes, Singular Points and Curve Tracing
dx
= 0.
dθ
tangent at θ = π and θ = – π are x = π and x = – π.
the curve for x = a (θ + sin θ) and y = a (1 + cos θ).
except for the values + π of θ for which
x
O
as
y
m
pt
ot
e
x+y=0
Fig. 3.26
163
EXAMPLE
10
Sketch the curve
y 4 – x 4 + xy = 0.
Here,
y 4 – x 4 + xy = 0
1. No line of symmetry.
2. It passes through origin two tangents at (0, 0) as x = 0 and y = 0,
∴ origin is node.
3. It cuts the coordinate axes at the origin only.
4. y = x, y = – x are its asymptotes.
5. Converting into polar coordinates,
1
r 2 = tan 2 θ
2
π
π
6. When,
or 0 < 2 θ <
0< θ<
⇒ r 2 increases from 0 to ∞.
4
2
π
π
π
or
When,
<θ<
< 2θ < π ⇒ r 2 is negative,
4
2
2
π
π
∴ no curve when < θ < .
4
2
Hence, the curve
Play with Graphs
SOLUTION
y
y=x
te
to
mp
y
as
x
O
as
ym
pto
te
y = –x
Fig. 3.27
164
Plot the Curves :
1 4
x .
2
2. y = (x + 1)(x – 2)2
2
1
1 5
3. y = x – x 3 +
x .
5
2
10
4. y = (1 – x 2 )–1
1. y = 1 + x 2 –
5. y =
x4
(1 + x )3
24. y = (x + 1)2 / 3 + (x – 1)2 / 3
25. y 2 = 8x 2 – x 4
26. y 2 = (x – 1)(x – 2)(x – 3)
x –1
27. y 2 =
x +1
x 2 (1 – x )
2
28. y =
(1 + x )2
29. y 2 = x 4 (x + 1)
(1 + x )4
6. y =
(1 – x )4
30. x 2 (y – 2)2 + 2xy – y 2 = 0
1
1
31. x = (t + 1)2 , y = (t – 1)2
4
4
1
t2
32. x =
, y =
1+ t2
1– t2
x 2 (x – 1)
(x + 1)2
x
8. y =
(1– x 2 )2
7. y =
33. x =
1
9. y = 2x – 1 +
(x + 1)
t2
t
, y = 2
t –1
t –1
34. x = – 5 t 2 + 2 t 5 , y = – 3 t 2 + 2 t 3
x2 + 1
10. y = 2
x –1
35. x =
a 2x
a + x2
t2+ 1
t
, y =
t +1
4(1 – t )
36. x =
(t + 2)2
(t – 2)2
, y =
t –1
(t + 1)
11. y =
2
a + x 
12. y 2 = x 2 

b – x 
8a 3
x + 4a 2
cos x
14. y =
cos 2x
1– x 2
1– x 2
15. y = arc cos 
= cos–1 

2
1+ x 
1+ x 2
13. y =
2
16. y = arc sin (sin x ) = sin (sin x )
–1
17. y = sin (arc sin x )
18. y = arc tan (tan x )
 1
19. y = arc tan  
x 
20. y = (x + 2) e1/ x
1
21. y = ( x 2 + x + 1 – x 2 – x + 1)
2
22. y = x 2 + 1 – x 2 – 1
23. y = (x + 2)2 / 3 – (x – 2)2 / 3
t – t2
t2– t3
, y =
2
1+ t
1+ t 2
3
3
38. x + y = 3 axy , where a > 0.
37. x =
39. (x – a )2 (x 2 + y 2 ) = b 2x 2 ,
40. x
2/ 3
+y
2/ 3
=a
2/ 3
41. x + 2x y = y
6
3
,
where a, b > 0.
where a > 0
3
42. 4y 2 = 4x 2y + x 5
43. x 4 + 2y 3 = 4x 2y
44. x 3 – 2 x 2y – y 2 = 0
45. x 2y 2 + y = 1
46. x 3 + y 3 = 3x 2
Asymptotes, Singular Points and Curve Tracing
EXERCISE
47. y 5 + x 4 = xy 2
48. x 4 – y 4 + xy = 0
49. x 5 + y 5 = xy 2
50. x = a sin 2 θ (1 + cos 2 θ ),
y = a cos 2 θ (1– cos 2 θ )
165
INTRODUCTION OF GRAPHS
y
1. (i)
y
3
2
1
(ii)
y = x 2; x > 0
y = 1/4
x
–3 –2
–1
y
–
=x
1;
x
<0
y
2. (i)
y
4
–2
x
–2
y=
2
y=2
–1
2x
3
y=
3x
y=
(ii)
x
0 1
–1
–2
–3
–1
y=
Play with Graphs
Hints and Solutions
0
1
x
2
4
–
x
1
–2
–1
x
0
1
0
1
2
3
x < –1
–2x ;

Here; f(x ) =  2; –1 ≤ x ≤ 1
 2x ;
x >1

y
(iii)
y
(iv)
5
1
4
3
–1
2
x
–1
1
–1
0
1
2
3
x
Here,
Here,
166
–1≤x <0
– x ;
– x + 1 ; 0 ≤ x < 1

1≤x<2
f(x ) = x ;
x + 1 ;
2≤ x < 3

5;
x=3
x 4 ;
–1 < x <1
f(x ) = 
x
;
x
≤
–
1
or x ≥ 1

4
is shown as;
y
neg
y
f (x)
lec
ting
(0, 1)
3/4
(1,0)
0 3/4
–1/2
x
0
–1/2
i.e., inverse for f(x ) = x 2 + x + 1 would exists only when
1
x≥– .
2
Here,
2x 2 ; x ≥ 0
f(x ) = 
x≤0
0;
Here,
f(x ) = [[ x] – x] = [ x] + [– x]
y = 0; x < 0
–1
1
x≥– .
2
>0
2x 2
;x
y=
2
–2
x
neglecting
From above figures it is clear that the graph would exists only when;
y
f –1 (x)
Graph for f –1(x)
Graph for f (x)
4.
Hints and Solutions
y = x2 + x + 1
3. We know,
0
5.
x
1
y
x – x;
=
[ x] + (– 1 – [ x]);
1
–4
–3
–2
0
–1
1
2
3
x
4
From figure;
y
π/2
x ∉ integer
 0; x ∈ integer
=
– 1; x ∉ integer
–1
6.
x ∈ integer
(α, tan–1)
π/4
P
(1, π/4)
y = tan–1x
Q
O
Slope of OP < Slope of OQ.
⇒
α
1
x
⇒
π
–0
tan –1 α – 0
4
<
1–0
α–0
tan –1 α π
> .
α
4
167
y
(π, π)
Play with Graphs
4
π
3
y = cos–1 (cos x)
2
1
0 1
–1
–3 –2 –1
2
3 π4
5
x
2π
–2
–3
–4
Clearly, the above curves intersect at 5 points;
number of solutions = 5.
∴
y
8.
1
π
0 1 2 3
x
4 3π
2
–1
From above figure, number of solutions is infinite.
y
x2
≤1
x –1
or
x 2 ≤| x – 1|
1I
y
3
4
3
g(x ) = 3 + [ x]
2
1
Clearly; from figure the two curves f(x )and g(x )intersects
when g(x ) = 4.
168
x
√5–1 1
2
O
–√5–1
2
x 3 = 3 + [ x]
∴ to sketch f(x ) = x
and
Ix–
10. Here,
1
2 <
⇒
 –1 – 5 –1 + 5 
.
x ∈
,
2
2 

x
9. Clearly;
∴
f(x ) = 4
⇒
x3 = 4
or
x = 22 / 3 .
–4
–3
–2
–1
–1
0
1
2
x
Hints and Solutions
CURVATURE AND TRANSFORMATIONS
1. (i)
y
(ii)
2
2
y = |2 – |x – 1||
–1
1
O
–1
3
x
3
1
–2
y=2–
4
|x – 1|
x=1
y
(iii)
y
(iv)
2
–1 O
x
3
–1 O
–2
|y | = 2 –
4
|x – 1|
y = 2–
y
(vi)
2
–1
4
|x – 1|
x=1
y
(v)
x
3
1 2
1
O
– log 2
x
1
3
|y | = 2 –
O
x
log 2
y = e |x | – 2
4
|x – 1|
–2
x=1
y
(vii)
(viii)
y
1
–log 2
O
log 2
x
O
–1
1
2
3
4
x
y = x – [x]
| y | = e |x | – 2
169
y
Play with Graphs
(ix)
y
(x)
1
1
O
1
2
3
x
4
O
1
2
y
y
(xii)
1
1
–1
x
4
y = (x – [x])2
y = √x – [x]
(xi)
3
O
1
2
3
4
x
–1
O
1
2
3
x
4
–1
–1
|y | = (x – [x])2
|y | = √x – [x]
y
(xiii)
y = –2x
y
(xiv)
2
y = 2x
2
y=2
x
1
y = |x – 1| + |x + 1|
–1 O
–1
O
x
1
|y | = |x – 1| + |x + 1|
–2
(xv)
y
y
(xvi)
4
4
3
3
2
2
1
O
1
1
2
3
4
x
–4 –3 –2
y = [|x – 1|]
–1
O
1
–1
–2
–3
–4
|y | = [|x – 1|]
170
2
3
4
x
y
(xviii)
5
5
x+2
4
3
3
x+1
2
–x + 1
x+1
2
1
–2
–1
1
x
1
O
x+2
4
–x + 2
2
3
4
x
1
–x
x
–3
–2
–1
O
2
x
3
–1
x–1
–2
y = |x| + [|x|]
|y | = x + [x]
y
(xix)
Hints and Solutions
y
(xvii)
y
(xx)
5
5
x+2
4
x+2
4
3
3
x+1
2
x+1
2
1
x
1
O
1
2
x
3
–4
–3
–2
–1
O
x
1
–1
–1
–2
–2
–3
–3
–4
–4
–5
–5
|y | = x + [x]
2
3
4
x
|y | = |x | + [|x |]
y
2. (i)
y=1
1
sin x
–2π
–π
O
π
2π
3π
x
y = √sin x
171
y
Play with Graphs
(ii)
–2π
–π
π
O
2π
x
3π
|y| = √sin x
(iii)
y
√2
1
–
3π
2
–π
–
π
2
O
π
2
π
x
3π
2
y = |sin x| + |cos x|
y
(iv)
y
(v)
2
2
1
–
3π – π
–
2
2
5π
2
O
1
π
2
–1
–2
x
3π
2
O
|y | = cos x + |cos x|
–1
10.
y
(vi)
2
3/2
1
1/2
–3π/2
–π
–π/2
O
π/2
π
y = 2sin x
172
x
3π/2
2π
π/2
π
x
3π/2
y = sin2x – 2 sin x
2π
y
(viii)
–2π 3π –π – π O
–
2
2
π 3π
2
π
2
2π 5π
2
3π
x
–
3π
2
y
y = log2|sin x|
(ix)
–π
–2π – 3π
2
O
Hints and Solutions
y
(vii)
–
π
2
π
2
O
x
3π
2
it only represents points
π
2
π
2π
π
2π
5π
2
3π
x
y = logsin x1/2
y
(x)
–2π
–
3π
2
–π
O
π
2
5π
2
3π
x
|y| = logsin x1/2
y
3.
y
4.
π/2
1
–3
–1
O
–π/2
1
–2
x
y = sin–1
1 – x2
1 + x2
O
–1
1
2
y=
2x
1 + x2
–1
3
x
173
y
5. (i)
y
(ii)
Play with Graphs
1
O
–2
x
2
x
O
–|x |
y=e
y = x 2 – 2|x |
y
(iii)
y
(iv)
x
O
x
O
y = e |x |
|y | = x
y
(v)
(vi)
O
–1
x
O
x
1
y
x = y 2+ 1
3
y=x –x
y
6.
1
–2
7. (i)
O
–1
1
2
x
y = f (x –1) + f (x + 1)
y
(ii)
y
x
O
y=
1
x–2
–2
–2
174
x
2
O
y=
1
|x | – 2
y
(iv)
2
2
–2
O
x
2
–2
O
–2
y=
|y | =
1
|x | – 2
y
8.
x
2
1
|x | – 2
y
9.
2
|log|x ||
π
O
2π
Hints and Solutions
y
(iii)
x
O
x
sin πx
–2
Since, 2 cos x and|sin x |, intersects at two points for
x ∈[ 0, 2π ].
∴ number of solutions are four when x ∈[ 0, 4π ].
π/6
π/2
5π / 6
x2
y′
0
–2
0
9/8
y′′
–3 3
2
0
3 3
2
0
x
0
From above figure we have six solutions.
3π
2
3π 


 x 2,

2
0
–
1
0
– 3√3
4
2π
x4
9/8
0
+
– 3√15
16
0
3√3
4
y

 3π
, x 4


 2
0
–
3√15
16
1
y
O π
6
–1
1
sin 2 x + cos x,
2
periodic with period 2 π.
y=
1
π
2
5π π
6
3π
2
2π
x
175
Play with Graphs
ASYMPTOTES, SINGULAR POINTS AND CURVE TRACING
y
1.
2.
y=1+x2 –
–1 – 1 O
√3
1
√3
x
y
3.
O
(1, 0)
1
2
(2, 0)
x
y
2 x – 1x3 + 1 x5
5
2
10
(–1, 0)
x
y
5.
(–1, 0) O
4.
y=
–1
y = (x + 1) (x – 2)2
1 4
x
2
(√1 + √3 , 0)
1
y
(1, 0)
x
y
6.
y=x–3
–4
–1
1
x
–1
176
O
1
x
O
–3 –√17
2
y
8.
–3 +√17
2
x
–1
O
–1
x = –1
9.
y
–1 + 1
√2
–1
1
–
2
10.
y = 2x – 1
O
O
–1
x
1
–1
x = –1
y
(–√3a, 0)
y=1
1
–1
11.
x=1
y
x
1
2
x
1
Hints and Solutions
y
7.
O (a, 0)
y
12.
(–a, 0)
(√3a, 0)
x=1
x
(–a, 0)
O
(b, 0)
x
x=b
177
y
13.
(0 , 2a)
Play with Graphs
y
14.
y = 2a
–
x
O
y
15.
3π
4
–
π
4
π
4
O
x
3π
4
y
16.
π
2
π
–π
–
2
O
O
–
x
1
y
17.
π
2
π
2
x
π
π
2
y
18.
1
O
–1
1
x
–
π
2
x
π
2
–1
y
19.
y
20.
y=x+3
π
2
O
–
x
π
2
–2
178
–1
O
2
x
y
22.
1/2
x
O
x
O
–1
1
–1/2
y
23.
–2
x
2
O
–1
y
25.
– 8
2
8
O
x
x
O
2–
y
27.
–1
O
28.
1
x
1
y
26.
O
–2
y
24.
Hints and Solutions
y
21.
1
3
y
x
x
O
–3 + √17
2
179
y
Play with Graphs
29.
30.
y
10
3
–
4
5
x
O
4 2
– +
5 5
2
y
31.
O
–1
2
3
1
5/4
x
y
32.
1
1
1/2
O
x
x
1
–1
y
33.
y
34.
–
1/2
x
y= – 3
2 4
x
4
y
35.
19
16
O
63
16
x
y
36.
1
4
1/2
O
180
1/4
x
–4
– 16
3
16
3
x
y
38.
O
–1
O
√2 – 1
2
x
3√2a 3√4a
y+
a
√2 – 1
2
x
x+
–
Hints and Solutions
y
37.
=
0
39.
y
y
40.
a
a–b
O
a
a+b
x
O
–a
x
a
–a
y
41.
x
O
43.
y
42.
–1 – 24
25
y
x
O
y
44.
y=
1 4
√2 3√3
x – 1
2 8
x
–
9
8
–1
x
181
y
45.
y
46.
Play with Graphs
1
–1/4
x
O 1/4
O
1
2
y=
47.
y
x
3
–x
+1
y
48.
y=x
x
O
x
y=
49.
y
–x
y
50.
–3√3a , a
4 4
3 √3a , a
4 4
x
O
x
O
y=
–x
√3a , – 3a
4
4
– √3a , – 3a
4
4
(0, 2a)
182