Relations & functions
1. Discuss the commutavity and associativity of the binary operation  on R defined by
a b =ab/4  a,b R
2. Show that binary operation a * b  a  b - 4 , a, b  R is commutative and associative. Also
find its Identity element and inverse of an element.
3. Examine which of the following is a binary operation.
𝑎+𝑏
i)
a∗ 𝑏 = 2 , a,b N
𝑎+𝑏
a∗ 𝑏 =
ii)
2
, a,b ∈ Q, for the binary operation check the commutative and
associative property.
4. Consider the binary operations : R×R Rand o :R×R Rdefined as a
b= | a – b | and a
o b = a for all a, b R. Show that ′ ∗ ′ is commutative but notassociative, 'o' is associative
but not commutative
5.
Show the function f:R→R defined by f(x)=
2𝑥−1
3
, x∈ R is one-one and onto function . Also
find the inverse of the function f.
Solutions: Relation & Functions
1. Commutativity:
𝑎𝑏
a*b = 4
𝑏𝑎
=4
=b*a
* is commutative.
Associativity:
(a*b)*c =(𝑎𝑏
)*c
4
=(𝑎𝑏
)/4 =
4
a*(b*c) =a *
𝑎𝑏𝑐
𝑏𝑐
4
16
=a*
𝑏𝑐
𝑎𝑏𝑐
/4 = 16
4
* is associative.
2. Commutativity:
a*b=a+b-4=b+a-4=b*a
∴ * is commutative.
Associativity:
(a*b)*c=(a+b-4)*c=a+b-4+c-4=a+b+c-8
a*(b*c)=a* (b+c-4)*c=a+b-4+c-4=a+b+c-8
∴ * is associative.
Identity Element:
Let a,e∈ R , a*e=a=e*a
a+e-4=a∴ e=4& a=e*a∴ a=e+a-4
e=4 is the identity element.
Inverse Element:
a*b=4∴ a+b-4=4 b=8-a
b*a=4, b+a-4=4 b=8-a
∴ Inverse of a is 8-a
3. (i)a N, b N
a*b=
𝑎+𝑏
2
∉N
is not binary.
(ii) a ∈ Q, b ∈ 𝑄
a*b=
𝑎+𝑏
2
∈Q
is binary operation.
𝑎+𝑏

a*b=

(a*b)*c=(

2
=
𝑏+𝑎
2
𝑎+𝑏
2
a*(b*c)=a*(
= 𝑏 ∗ 𝑎 𝐶𝑜𝑚.
𝑎+𝑏
)∗𝑐 =
𝑏+𝑐
2
2
(
) = a*
+ 𝑐/4
𝑏+𝑐
)
2
2
=
𝑎+𝑏+2𝑐
=
4
2𝑎+𝑏+2𝑐
4
∴ (a*b)*c a*(b*c)
* is not associative.
4. (i)
a*b= |𝑎 − 𝑏|=|𝑏 − 𝑎|=b*a
* is commutative.
a*(b*c)=a*(|𝑏 − 𝑐| = |𝑎 − |𝑏 − 𝑐|| = |𝑎 − 𝑏 + 𝑐|
(a*b)*c=|𝑎 − 𝑏|*c = |𝑎 − 𝑏||−𝑐|= |𝑎 − 𝑏 − 𝑐|
∴ 𝑎 ∗ (𝑏 ∗ 𝑐) ≠ (𝑎 ∗ 𝑏) ∗ 𝑐
* is not associative.
(ii)
a o b =a
b o a=b
aob
boa
‘o’ is not commutative.
 (a o b) o c = a o c =a
a o (b o c) = a o b =a
‘o’ is associative.
5. One- One
x1=x2
⇒ 2𝑥1=2𝑥 2
⇒ 2𝑥1-1=2𝑥 2-1
x2
⇒ (2𝑥1-1)/3=(2𝑥2-1)/3
f(x1)=f(x2)
f is one-one.
Onti:For some y∈ R (Co-domain)
2𝑥−1
Y=f(x) =
3
⇒ 3𝑦 = 2𝑥 − 1
2x= 3y + 1
⇒𝑥=
3𝑦+1
2
∈R
Now, f(x)= (2(y+1)-1)/3 =3y/3 =y
∴ 𝑓 𝑖𝑠 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑒.
F-1(x)=
3𝑥∓1
2
INVERSE TRIGNOMETRY
1
1
1. Prove the following:2𝑡𝑎𝑛−1 (3) + 𝑡𝑎𝑛−1 (7) =
𝜋
4
2.Provethat:tan−1 1 + tan−1 2 + tan−1 3 = 𝜋.
3. Prove that
tan 1
1
1
1 
 tan 1  tan 1 
2
5
8 4
𝟏
𝟐
𝟏
𝟒
4. Prove the following:𝒕𝒂𝒏−𝟏 (𝟒) + 𝒕𝒂𝒏−𝟏 (𝟗) = 𝟐 𝐬𝐢𝐧−𝟏 (𝟓)
𝟔𝟐
𝟓
𝟑
5. Provethat:𝒕𝒂𝒏−𝟏 (𝟏𝟔) = 𝐬𝐢𝐧−𝟏 𝟏𝟑 + 𝐜𝐨𝐬 −𝟏 𝟓
SOLUTION
INVERSE TRIGNOMETRY
1. 𝑡𝑎𝑛−1
2/3
+𝑡𝑎𝑛−1
1−1/32
3
𝑡𝑎𝑛−1 4 +𝑡𝑎𝑛−1
1
7
1
7
𝜋
→ 𝑡𝑎𝑛−1 1 = 4
3
2. LHS= 𝑡𝑎𝑛−1 (−1) +𝑡𝑎𝑛−1 3
−3+3
= 𝑡𝑎𝑛−1 ( 1+9 ) == 𝑡𝑎𝑛−1 (0) = 𝜋
7
1
3. tan−1 9 +𝑡𝑎𝑛−1
4 . 2 𝑡𝑎𝑛−1
1 2
+
4 9
1.2
1−
4.9
5 . RHS= . 𝑡𝑎𝑛−1
8
=2. 𝑡𝑎𝑛−1
5
12
56+9
= 𝑡𝑎𝑛−1
72−7
= 𝑡𝑎𝑛−1 1 =
17
𝜋
4
1
=2. 𝑡𝑎𝑛−1 2 = . 𝑡𝑎𝑛−1
34
4
+. 𝑡𝑎𝑛−1 3 = . 𝑡𝑎𝑛−1
15+48
36−20
1
1
1−
4
= . 𝑠𝑖𝑛−1
4
5
=LHS
MATRICES
3
1. if A= [1
0
1
2. . If A=[2
2
2 0
4 0] , show that 𝐴2 − 7𝐴 + 10𝐼 = 02.
0 5
2 2
1 2], prove that 𝐴2 − 4𝐴 − 5𝐼 = 0. Hence find 𝐴−1
2 1
3 −5
) , 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝐴2 − 5𝐴 − 14𝐼 = 0. ℎ𝑒𝑛𝑐𝑒 𝑓𝑖𝑛𝑑 𝐴−1 .
−4 2
3. If𝐴 = (
3 2
4. Express the matrix A=[4 1
0 6
1 0
3 −2
5. A=[
]and I=[
]
4
−2
0
1
5
3] as the sum of a symmetric and skew –symmetric matrix.
7
, find k so that 𝐴2 = 𝑘𝐴 − 2𝐼
DETERMINANTS
1.
Using properties of determinants, prove that
𝑎 + 𝑏 + 2𝑐
|
𝑐
𝑐
𝑥
2. Ifx, yand z are different and|𝑦
𝑧
𝑎
𝑏 + 𝑐 + 2𝑎
𝑎
𝑥2
𝑦2
𝑧2
𝑏
| = 2(𝑎 + 𝑏 + 𝑐)3
𝑏
𝑐 + 𝑎 + 2𝑏
1 + 𝑥3
1 + 𝑦 3 | = 0 , show that xyz=-1.
1 + 𝑧3
3. Using properties of determinants, prove the following:
𝑎2 + 1
𝑎𝑏
𝑎𝑐
2
| 𝑎𝑏
𝑏 +1
𝑏𝑐 | = 1 + 𝑎2 + 𝑏 2 + 𝑐 2 .
𝑐𝑎
𝑐𝑏
𝑐2 + 1
4. Using properties of determinants prove the following:
1+𝑎
1
1
| 1
1+𝑏
1 | = abc + bc + ca + ab.
1
1
1+𝑐
5. Using properties of determinants prove the following:
1
|2
3
1+𝑝
3 + 2𝑝
6 + 3𝑝
1+𝑝+𝑞
1 + 3𝑝 + 2𝑞 | = 1
1 + 6𝑝 + 3𝑞
𝑥
2. Ifx, yand z are different and|𝑦
𝑧
ANS.2
𝑥2
𝑦2
𝑧2
1 + 𝑥3
1 + 𝑦 3 | = 0 , show that xyz=-1.
1 + 𝑧3
3. Using properties of determinants, prove the following:
𝑎2 + 1
𝑎𝑏
𝑎𝑐
2
| 𝑎𝑏
𝑏 +1
𝑏𝑐 | = 1 + 𝑎2 + 𝑏 2 + 𝑐 2 .
𝑐𝑎
𝑐𝑏
𝑐2 + 1
ANS.3
4. Using properties of determinants prove the following:
1+𝑎
1
1
| 1
1+𝑏
1 | = abc + bc + ca + ab.
1
1
1+𝑐
ANS.4
5. Using properties of determinants prove the following:
1
|2
3
ANS.5
1+𝑝
3 + 2𝑝
6 + 3𝑝
1+𝑝+𝑞
1 + 3𝑝 + 2𝑞 | = 1
1 + 6𝑝 + 3𝑞
Solve the system of equations using matrix method
1.
Three shopkeepers A, B and C are using polythene, handmade bags (prepared by persons) and newspapers
envelopes as carry bags. It is found that the shopkeepers A,B ,C are using(20,30,40) ,(30,40,20) ,(40,20,30)
polythene ,handmade bags and newspapers envelopes respectively. The shopkeepers A, B,C spent Rs 250,Rs
220 and Rs 200 on these carry bags respectively. Find the cost of each carry bags using matrices. Keeping in
mind the social and environmental conditions, which shopkeeper is better and why?
2.
In a survey of 25 richest person of three localities x, y ,z. It is found that in locality A 7 believes in honesty,12
in hard work and 6 in unfair means. While in Y 10 believes in honesty 11 in hard work , 4 in unfair means and
in Z, 8 believes in honesty ,12 in hard work 5 in unfair means. If the income of 25 richest persons of locality
x, y and z are respectively Rs.1,19,000,Rs.1,18,0001,18,000per day then find income of each type of person
by matrix method. In your view which type of person is better for society and for nation.
SOLUTIONS (USING MATRIX METHOD)
1. Given system of equations is x+3y+4z=8,2x+y+2z=5,5x+y+z=7
1 3 4 𝑥
8
[2 1 2] [𝑦] = [5]or AX=B, |𝐴| = 11 ≠ 0
5 1 1 𝑧
7
A is invertible
−1
𝐴
=
−1
X=𝐴
𝐴𝑑𝑗 𝐴
|𝐴|
−1
1
2
= [ 8 −19 6 ]
11
−3 14 −5
1
𝑥
−1
1
2 8
11
1
1
𝑦
B → [ ]= [ 8 −19 6 ] [5] = [11] → x= 1, y= 1, z = 1
11
11
𝑧
−3 14 −5 7
11
2. Given system of equations is
5 10 5 𝑥
32500
𝑦
[5 8 7] [ ] = [30500]or AX=B, |𝐴| = 40 ≠ 0
6 8 6 𝑧
31000
−1
𝐴
=
𝐴𝑑𝑗 𝐴
|𝐴|
−1
X=𝐴
−8
= [ 12
40
−8
A is invertible
−20 30
0
−10]
20 −10
1
𝑥
−8 −20 30 32500
1
B → [𝑦]= [ 12
0
−10] [30500]
40
𝑧
−8 20 −10 31000
x= 1500, y= 2000, z = 1000
Continuity and Differentiability
5, 𝑖𝑓 𝑥 ≤ 2
1. (x)= {𝑎𝑥 + 𝑏, 2 < 𝑥 < 10is continuous function.
21, 𝑖𝑓 𝑥 ≥ 10
1−𝑐𝑜𝑠𝑥
2. f(x)={
𝑘𝑥 2
,
3,
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0
𝑑2 𝑦
3. Find,𝑑𝑥 2 when a function is given by
𝑡
x= a(cost + log(tan2)), y = a sint
𝑑𝑦
−1
4. if 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥=0 then prove that𝑑𝑥 = (1+𝑥)2
5. if y = 𝑒 𝑎𝑐𝑜𝑠
−1 𝑥
Solution
𝑑2 𝑦
𝑑𝑦
then show that (1-x2)𝑑𝑥 2 - x𝑑𝑥 - 𝑎2 𝑦 = 0
5, 𝑖𝑓 𝑥 ≤ 2
1. asf(x)= {𝑎𝑥 + 𝑏, 2 < 𝑥 < 10is continuous function.
21, 𝑖𝑓 𝑥 ≥ 10
∴ lim+ 𝑓(𝑥)= f(2)
𝑥→2
∴ lim+ 𝑎𝑥 + 𝑏=5
𝑥→2
2a+b = 5…(1)
∴ lim− 𝑓(𝑥)= f(10)
𝑥→10
∴ lim− 𝑎𝑥 + 𝑏 = 21
𝑥→10
10a+b = 21… (2)
a=2, b=1
1−𝑐𝑜𝑠𝑥
2.As f(x)={
𝑘𝑥 2
,
3,
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0
lim 𝑓(𝑥)= f(0)
𝑥→0
.lim
𝑥
2
𝑥2
2𝑠𝑖𝑛2
𝑥→0 4𝑘
=3
4
1
.2𝑘 = 3
K =1/6
2.
𝑡
3.x= a(cost + log(tan2)), y = a sint
𝑑𝑥
𝑐𝑜𝑠2 𝑡
1
. 𝑑𝑡 = a(-sint +𝑠𝑖𝑛𝑡) =a 𝑠𝑖𝑛𝑡
𝑑𝑦
. 𝑑𝑡 = a cos t
𝑑𝑦
.𝑑𝑥 =tan t
𝑑2 𝑦
𝑠𝑖𝑛𝑡
.𝑑𝑥 2 = 𝑠𝑒𝑐 2 𝑡*𝑎 𝑐𝑜𝑠2 𝑡
4. 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥=0
.𝑥√1 + 𝑦= -𝑦√1 + 𝑥
−𝑥
. y=1+𝑥
𝑑𝑦
−1
.𝑑𝑥 =(1+𝑥)2
5. if y = 𝑒 𝑎𝑐𝑜𝑠
𝑑𝑦
−1 𝑥
.𝑑𝑥 = 𝑒 𝑎𝑐𝑜𝑠
𝑑2 𝑦
−𝑎
−1 𝑥
√1−𝑥 2
𝑑2 𝑦
𝑥
𝑑𝑦
.√1 − 𝑥 2 𝑑𝑥 2 - √1−𝑥 2 𝑑𝑥 = 𝑒 𝑎𝑐𝑜𝑠
𝑑2 𝑦
𝑑𝑦
then show that (1-x2)𝑑𝑥 2 - x𝑑𝑥 - 𝑎2 𝑦 = 0
−1 𝑥
−𝑎2
√1−𝑥 2
𝑑𝑦
(1-x2)𝑑𝑥 2 - x𝑑𝑥 - 𝑎2 𝑦 = 0
INCREASING AND DECREASING FUNCTIONS
1. Find the intervals in which the function f given by f(x) = 4𝑥 3 – 6 𝑥 2 - 72𝑥 + 30 is strictly
increasing .
2. Find the intervals in which the function f given by f(x) = -2𝑥 3 - 9 𝑥 2 - 12𝑥 +1 is strictly
decreasing .
SOLUTIONS: INCREASING AND DECREASING FUNCTIONS
1. by f(x) = 4𝑥 3 – 6 𝑥 2 - 72𝑥 + 30
f ’(x) = 12 𝑥 2 - 12𝑥 - 72 = 12(𝑥 2 –x – 6 ) = 12( x-3 ) ( x+2)
Critical points are -2,3.
Interval
x-3
x+2
12( x-3 ) ( x+2)
+
x -2
Nature
increasing
-2
x
+
3
x
3
+
+
2. f(x) = -2𝑥 3 - 9 𝑥 2 - 12𝑥 +1
f ’(x) = -6𝑥 2 - 18𝑥 - 12 = -6(𝑥 2 +3x + 2
Critical points are -1, -2.
Interval
x +1
x -2
-2 x -1
+
x -1
+
Decreasing
increasing
) = -6( x +1 ) ( x+2)
x+2
+
+
-6( x+1 ) (x+2)
+
-
Nature
Decreasing
increasing
Decreasing
TANGENTS AND NORMALS
1. Find the equation of tangent to the curve y = √3𝑥 − 2 , which is parallel to the line
4x -2y + 5 =0.
2. Find the equations of the tangent to the parabola 𝑦 2 = 4𝑎𝑥 at the point ( a𝑡 2 , 2𝑎𝑡 )
3. SOLUTION:TANGENTS AND NORMALS
1.curve y = √3𝑥 − 2 ,
𝑑𝑦
𝑑𝑥
=
3
2√3𝑥−2
Slope of the line
Given
3
2√3𝑥−2
4x -2y + 5 =0 is 2.
41
=2
=> x = 48and y =
3
4
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
3
41
y - 4 = 2 ( x - 48 )
=>
48x -24 y -23 = 0
2. parabola 𝑦 2 = 4𝑎𝑥
𝑑𝑦
2y𝑑𝑥 = 4a
𝑑𝑦
=>𝑑𝑥 =
2𝑎
𝑦
𝑑𝑦
=>𝑑𝑥 at the point ( a𝑡 2 , 2𝑎𝑡 ) =
1
𝑡
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
1
y – 2at = 𝑡 ( x - a𝑡 2 )
=>
x – ty + a𝑡 2 = 0
RATE OF CHANGE QUANTITIES
1. A particle move along the curve 6y=𝑥 3 +2. Find the points on the curve at which the
Y coordinate is changing 8 times as fast as the x –coordinate.
1
2. The radius of an air buble is increasing at the rate of 2cm/s. At what rate is the
volume of The buble increasing when the radius is 1cm?
SOLUTIONS: RATE OF CHANGE QUANTITIES
1.Here 6y =𝑥 3 + 2 𝑎𝑛𝑑
𝑑𝑦
𝑑𝑡
𝑑𝑥
=8 𝑑𝑡
Solving them we get x=±4
At x=4 y=11
Atx=-4, y=
−32
3
Thus requird points are (4,11) and (-4,
2.
−32
3
)
Let r be the radius of bubble and v be the volume of bubble at any time t.
Then
𝑑𝑟
𝑑𝑡
1 𝑐𝑚
=2
𝑠
𝑎𝑛𝑑 𝑟 = 1𝑐𝑚
By solving we get
𝑑𝑣
𝑑𝑡
=2π𝑐𝑚3 /s
MAXIMA AND MINIMA
1. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R
is 8/27 of the volume of the sphere.
Solution:
Volume of the cone = (1/3) πr2h = (1/3) π (R2 – x2 ) ( R + x) ------ (1)
3V/ π = R3 – x3 +R2 x – x2 R
R
Z(x) = R3 – x3 +R2 x – x2 R
Z’ (x) = -3x2 + R2 – 2xR
For maximum volume Z’ (x) = 0 ==> -3 x2 – 2x R + R2 = 0 x = R/3 or x = - R
x= -R is not possible .
Z’’ (x) = -6x -2R < 0
Therefore volume is maximum. In (1) substituting the value of x
x
r
Volume of the cone = (1/3) π (8R2)/9 (4R/3) = (8/27) ( 4/27) π R3 = (8/27) volume of
the sphere.
2. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a
square from each corner and folding up the flaps to form the box. What should be the
side of the square to be cut off so that the volume of the box is the maximumpossible.
Solution:
Volume of the box V(x) = x (18 – 2x) 2 = 324 x + 4x3 –72x2
V’ (x) = 324 + 12 x2 – 120 x
18-x
For maximum volume V’ (x) = 0  324 + 12 x2 - 144 x = 0.
=> x = 9 or x =3
X = 9 is not possible , therefore x = 3.
V’’ (x ) = 24 x – 144. V’’ (x) at x = 3 = -72 < 0.
Therefore volume of the box is maximum.
INDEFINITE INTEGRALS
(5𝑥+3)𝑑𝑥
1. Evaluate:∫ √𝑥 2
+4𝑥+10
Solution:
𝑑
5𝑥 + 3 = 𝜆(𝑑𝑥 (𝑥 2 + 4𝑥 + 10)) + µ
5𝑥 + 3 = 𝜆(2𝑥 + 4) + µ
𝜆 =
5
2
,
4𝜆 + µ = 3
µ = 3 − 10 = −7
5𝑥 + 3 =
5
(2𝑥 + 4) – 7 = 5𝑥 + 10 – 7 = 5𝑥 + 3
2
(5𝑥 + 3)𝑑𝑥
5
(2𝑥 + 4)𝑑𝑥
𝑑𝑥
= ∫
− 7∫
√𝑥 2 + 4𝑥 + 10 2 √𝑥 2 + 4𝑥 + 10
√𝑥 2 + 4𝑥 + 10
=
5 2𝑡𝑑𝑡
𝑑𝑥
∫
− 7∫
2 √𝑡 2
√𝑥 2 + 2𝑥(2) + 4 + 6
5
= ∫ 𝑑𝑡 − ∫
1
𝑑𝑥
√(𝑥 + 2)2 + (√6)2
18-x
= 5𝑡 – 7 𝑙𝑜𝑔 |(𝑥 + 2) + √𝑥 2 + 4𝑥 + 10| + 𝑐
= 5√𝑥 2 + 4𝑥 + 10 − 7 𝑙𝑜𝑔 |𝑥 + 2 + √𝑥 2 + 4𝑥 + 10| + 𝑐
4
2. Evaluate:∫1 (𝑥 2 − 𝑥)𝑑𝑥 as the limit of a sum.
Solution: Here 𝑎 = 1, 𝑏 = 4 𝑎𝑛𝑑 𝑛ℎ = 𝑏 − 𝑎 = 3
𝑓(𝑥) = (𝑥 2 − 𝑥)
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = lim ℎ[𝑓(𝑎) + 𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ) + … … . . + 𝑓(𝑎 + ̅̅̅̅̅̅̅
𝑛 − 1ℎ)]
ℎ→0
𝑎
4
∫ (𝑥 2 − 𝑥)𝑑𝑥 = lim ℎ[𝑓(1) + 𝑓(1 + ℎ) + 𝑓(1 + 2ℎ) + … … . . + 𝑓(1 + ̅̅̅̅̅̅̅
𝑛 − 1ℎ)]
ℎ→0
1
̅̅̅̅̅̅̅
= lim ℎ[{(ℎ)2 + ℎ} + {(2ℎ)2 + 2ℎ} + {(3ℎ)2 + 3ℎ} + … … . . + {(𝑛
− 1ℎ)2 + ̅̅̅̅̅̅̅
𝑛 − 1ℎ}]
ℎ→0
= lim ℎ[{12 + 22 + 32 + ⋯ … + (𝑛 − 1)2 }(ℎ)2 + {1 + 2 + 3 + ⋯ … (𝑛 − 1)}ℎ]
ℎ→0
= lim ℎ[
ℎ→0
𝑛(𝑛 − 1)(2𝑛 − 1) 2 𝑛(𝑛 − 1)
ℎ +
ℎ]
6
2
𝑛ℎ(𝑛ℎ − ℎ)(2𝑛ℎ − ℎ) 𝑛ℎ(𝑛ℎ − ℎ)
= lim [
+
]
ℎ→0
6
2
=
3.3.6
6
+
3.3
2
=
27
2
.
𝑑𝑥
3. Evaluate:∫ √𝑥 2
+4𝑥+10
Solution:
𝐼=∫
=∫
𝑑𝑥
√𝑥 2 + 2𝑥(2) + 4 + 6
𝑑𝑥
√(𝑥 + 2)2 + (√6)2
= 𝑙𝑜𝑔 |(𝑥 + 2) + √𝑥 2 + 4𝑥 + 10| + 𝐶
𝑑𝑥
4. Evaluate:∫ √𝑥 2
+8𝑥+10
Solution:
𝐼=∫
=∫
=∫
𝑑𝑥
√𝑥 2 + 8𝑥 + 10
𝑑𝑥
√𝑥 2 + 2𝑥(4) + 16 − 6
𝑑𝑥
√(𝑥 + 4)2 − (√6)2
= 𝑙𝑜𝑔 |(𝑥 + 4) − √𝑥 2 + 8𝑥 + 10| + 𝐶
𝑑𝑥
5. Evaluate:∫ √𝑥 2
+8𝑥+10
Solution:
𝐼=∫
=∫
=∫
𝑑𝑥
√6 − 4𝑥 − 𝑥 2
𝑑𝑥
√6 − (4𝑥 + 𝑥 2 )
𝑑𝑥
√10 − (𝑥 2 + 2.2. 𝑥 + 4)
𝑑𝑥
=∫
= 𝑠𝑖𝑛−1
2
√(√10) −(𝑥+4)2
6. Evaluate:∫ 2𝑥 2
𝑥+2
+6𝑥 +5
𝑥+4
√10
+𝐶
𝑑𝑥
Solution:
x+2 = A (4x +6) + B
Equating the coefficients of x and the constant terms from both sides we get
4A = 1 and 6A + B = 2 => a = ¼ and
B = ½∫ 2𝑥 2
I1 =∫
𝑑𝑡
𝑡
𝑥+2
4𝑥+6
𝑑𝑥 = ¼ ∫ 2𝑥 2
+6𝑥 +5
= log(2x2 + 6x 5)
𝑑𝑥 + ½ ∫ 2𝑥 2
+6𝑥 +5
𝑑𝑥
+6𝑥 +5
= I1 + I2
2 x2 + 6x + 5 = ( x + 3/ 2)2+ ( ½)2
I2 = tan-1 2( x + 3/2)
Required answer = I1 + I2
𝑑𝑥
7. Evaluate: ∫ 𝑥 2 −6𝑥+13
Solution:
x2 – 6x +13 = ( x-3)2 + 22
𝑑𝑥
∫ 𝑥 2 −6𝑥+13 = ½ tan-1 t/2 + c
𝑥−3
= ½ tan-1
2
+c
𝑑𝑥
8. Evaluate: ∫ 9𝑥 2 +6𝑥+5
Solution:
9x2 + 6x - 5 = 9( x +1/3)2 - (√6/3)2
𝑑𝑥
3𝑥+1−√6
∫ 9𝑥 2 +6𝑥+5 = 27/4log3𝑥+1+√6 + c
𝜋/4
9. Evaluate: ∫0
log( 1 + tan 𝑥)𝑑𝑥
Solution: 𝜋/4
I = ∫0
log( 1 + tan 𝑥)𝑑𝑥
𝜋
𝑎
𝑎
=∫04 log( 1 + tan(𝜋/4 − 𝑥))𝑑𝑥; using property∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥
𝜋/4
=∫0
𝜋/4
= ∫0
1−𝑡𝑎𝑛 𝑥
log (1 + 1+𝑡𝑎𝑛 𝑥) 𝑑𝑥
𝜋/4
= ∫0
2𝐼 =log 2.𝜋/4
𝐼 =
𝜋 log 2
8
.
2
log (1+𝑡𝑎𝑛𝑥 ) 𝑑𝑥
𝜋/4
log 2 − 𝐼 2𝐼 = ∫0
log 2𝑑𝑥
AREA UNDER BOUNDED CURVE
1. Find area of region bounded by triangle whose vertices are (-1,2), (1,5) and (3,4).
Solution: let A be (-1,2), B be (1,5), C be (3,4).
Equation of AB : y = 3/2x +7/2
Equation of BC : y = 11/2 – x/2
Equation of AC : y = x/2 + 5/2
1
3
3
Required area = ∫−1 𝐴𝐵 𝑑𝑥 + ∫1 𝐵𝐶 𝑑𝑥 − ∫−1 𝐴𝐶 𝑑𝑥
= 7 + 9 -12 = sq. units.
Questions based on Homogeneous Differential Equations
𝑑𝑦
1. Solve the differential equation: 𝑥 2 𝑑𝑥 = 2𝑥𝑦 + 𝑦 2 .
𝑑𝑦
𝑦
𝑦
2. Solve the differential equation: 𝑑𝑥 − 𝑥 + 𝑐𝑜𝑠𝑒𝑐 (𝑥 ) = 0; 𝑦 = 0 𝑤ℎ𝑒𝑛 𝑥 = 1.
𝑑𝑦
𝑦
3. Solve the differential equation: 𝑥 𝑑𝑥 = 𝑦 − 𝑥 tan (𝑥 ).
Solutions:
𝑑𝑦
1. 𝑥 2 𝑑𝑥 = 2𝑥𝑦 + 𝑦 2
𝑑𝑦
𝑑𝑥
=
2𝑦
𝑥
𝑦 2
+ (𝑥 ) … (1)
𝑑𝑦
𝑑𝑣
Put 𝑦 = 𝑣𝑥 𝑔𝑖𝑣𝑒𝑠 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
Substitute in equation (1), we get
𝑣+𝑥
𝑥
𝑑𝑣
= 2𝑣 + 𝑣2
𝑑𝑥
𝑑𝑣
= 𝑣2 + 𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
=
+𝑣
𝑥
𝑣2
Integrating both the sides, we get
∫
𝑑𝑣
𝑑𝑥
=∫
+𝑣
𝑥
𝑣2
∫
𝑑𝑣
𝑑𝑥
=∫
𝑣(𝑣 + 1)
𝑥
∫
((1 + 𝑣) − 𝑣)𝑑𝑣
= log|𝑥| + 𝑙𝑜𝑔𝑐
𝑣(𝑣 + 1)
1
1
∫ 𝑑𝑣 − ∫
𝑑𝑣 = log|𝑥| + 𝑙𝑜𝑔𝑐
𝑣
𝑣+1
log|𝑣| − log|𝑣 + 1| = log|𝑐𝑥|
log |
𝑣
| = log |𝑐𝑥|
𝑣+1
𝑣
= 𝑐𝑥
𝑣+1
𝑦
𝑥 = 𝑐𝑥
𝑦
+1
𝑥
𝑦
𝑦+𝑥
2.
𝑑𝑦
𝑑𝑥
= 𝑐𝑥is the required solution.
𝑦
𝑦
− 𝑥 + 𝑐𝑜𝑠𝑒𝑐 (𝑥 ) = 0 . . . (1)
𝑑𝑦
𝑑𝑣
Put 𝑦 = 𝑣𝑥 𝑔𝑖𝑣𝑒𝑠 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
Substitute in equation (1), we get
𝑣+𝑥
𝑑𝑣
− 𝑣 + 𝑐𝑜𝑠𝑐𝑒𝑐 𝑣 = 0
𝑑𝑥
𝑑𝑣
= −𝑐𝑜𝑠𝑒𝑐 𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
=−
𝑐𝑜𝑠𝑒𝑐 𝑣
𝑥
𝑑𝑥
sin 𝑣 𝑑𝑣 = −
𝑥
Integrating both the sides, we get,
𝑑𝑥
∫ sin 𝑣 𝑑𝑣 = − ∫
𝑥
− cos 𝑣 = − log|𝑥| − 𝑐
𝑦
cos = log|𝑥| + 𝑐
𝑥
Now put 𝑦 = 0 𝑤ℎ𝑒𝑛 𝑥 = 1
cos 0 = log 1 + 𝑐 𝑜𝑟 𝑐 = 1
𝑦
So solution becomes cos 𝑥 = log|𝑥| + 1.
𝑥
𝑑𝑦
𝑦
3. 𝑥 𝑑𝑥 = 𝑦 − 𝑥 tan (𝑥 )
𝑑𝑦 𝑦
𝑦
= − tan ( ) . . . . (1)
𝑑𝑥 𝑥
𝑥
𝑑𝑦
𝑑𝑣
Put 𝑦 = 𝑣𝑥 𝑔𝑖𝑣𝑒𝑠 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
Substitute in equation (1), we get
𝑣+𝑥
𝑑𝑣
= 𝑣 − tan 𝑣
𝑑𝑥
𝑑𝑣
= − tan 𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
=−
tan 𝑣
𝑥
𝑑𝑥
cot 𝑣 = −
𝑥
Integrating both the sides
𝑑𝑥
∫ cot 𝑣 𝑑𝑣 − ∫
𝑥
log|𝑠𝑖𝑛𝑣| = − log|𝑥| + log 𝑐
𝑐
log|𝑠𝑖𝑛 𝑣| = log | |
𝑥
𝑐
𝑆𝑖𝑛 𝑣 =
𝑥
𝑦
𝑐
𝑦
𝑆𝑖𝑛 𝑥 = 𝑥 𝑜𝑟 𝑥 𝑆𝑖𝑛 𝑥 = 𝑐 is the required solution.
𝑥
VECTORS
1. If 𝑎⃗ = 2𝑖 + 2𝑗 + 3𝑘 , 𝑏⃗⃗ = −𝑖 + 2𝑗 + 𝑘𝑎𝑛𝑑𝑐⃗ = 3i+j are such that 𝑎⃗ + 𝜇𝑏⃗⃗ is perpendicular to 𝑐⃗, then
find the value of 𝜇 .
⃗⃗ 𝑐⃗ + 𝑐⃗. 𝑎⃗.
2. If 𝑎⃗, 𝑏⃗⃗, 𝑐⃗are unit vectors such that 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = ⃗⃗0⃗⃗ .Find the value of 𝑎⃗. 𝑏⃗⃗ + ⃗𝑏.
SOLUTION
1.𝑎⃗ + 𝜇𝑏⃗⃗ = (2𝑖 + 2𝑗 + 3𝑘) + 𝜇(−𝑖 + 2𝑗 + 𝑘) = (2- 𝜇)𝑖 + (2+2 𝜇)j+ (3+ 𝜇)𝑘𝑎⃗ + 𝜇𝑏⃗⃗ is
perpendicular to 𝑐⃗, ∴ ( 𝑎⃗ + 𝜇𝑏⃗⃗ ) 𝑐⃗= 0 , 3(2- 𝜇) + (2+2 𝜇) = 0 ,∴ 𝜇 = 8.
2.𝑎⃗, 𝑏⃗⃗, 𝑐⃗are unit vectors ∴ |𝑎⃗|= |𝑏⃗⃗| =|𝑐⃗|= 1 , 𝑎⃗. (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) = 𝑎⃗. ⃗⃗0⃗⃗ = 0 ,
|𝑎⃗|2 + 𝑎⃗. 𝑏⃗⃗ + 𝑎.
⃗⃗⃗⃗ 𝑐⃗= 0, ∴ 𝑎⃗. 𝑏⃗⃗ + 𝑎.
⃗⃗⃗⃗ 𝑐⃗ = −1,
similarily, 𝑏⃗⃗. 𝑎⃗ + ⃗⃗⃗⃗
𝑏. 𝑐⃗ = −1 and 𝑐⃗. 𝑏⃗⃗ + 𝑐⃗. 𝑎⃗ = -1 , Add all above three equation , we get 2(𝑎⃗. 𝑏⃗⃗ +
⃗⃗⃗⃗
⃗⃗ 𝑐⃗ + 𝑐⃗. 𝑎⃗ = −3
𝑏. 𝑐⃗ + 𝑐⃗. 𝑎⃗ ) = -3, 𝑎⃗. 𝑏⃗⃗ + ⃗𝑏.
2