CHAPTER
1
NUMBER SYSTEM
(i) Natural numbers: The counting numbers 1, 2, 3, 4, ...
are called Natural Numbers. The set of natural numbers is
denoted by N.
Thus N = {1, 2, 3, 4, ....}.
(ii) Whole numbers: Natural numbers including zero are
called whole numbers. The set of whole numbers is denoted
by W.
Thus W = {0, 1, 2, .........}
(iii) Integers: The numbers ... – 3, – 2, – 1, 0, 1, 2, 3 .... are
called integers and the set is denoted by Ι or Z. Thus Ι
(or Z) = {.. – 3, – 2, – 1, 0, 1, 2, 3...}
Note: (a) Positive integers Ι+ = {1, 2, 3 ....} = N
(b) Negative integers Ι– = {....., –3, –2, –1}.
(c) Non-negative integers (whole numbers) = {0, 1, 2, ...}.
(d) Non-positive integers = {......, –3, –2, –1, 0}.
(iv) Even integers: Integers which are divisible by 2 are called
even integers.
e.g. 0, ± 2, ± 4,.......
(v) Odd integers: Integers which are not divisible by 2 are
called odd integers.
e.g. ± 1, ± 3, ± 5, ± 7......
(vi) Prime numbers: Natural numbers which are divisible by 1
and itself only are called prime numbers.
e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ........
(vii) Composite number: Let ‘a’ be a natural number, ‘a’ is said
to be composite if, it has atleast three distinct factors.
e.g. 4, 6, 8, 9, 10, 12, 14, 15 .........
Note: (a) 1 is neither a prime number nor a composite number.
Basic Mathematics
(viii) Co-prime numbers: Two natural numbers (not necessarily
prime) are called coprime, if their H.C.F (Highest common
factor) is one.
e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8)
(15, 16) etc.
These numbers are also called as relatively prime numbers.
Note:
(a) Two prime number(s) are always co-prime but
converse need not be true.
(b) Consecutive natural numbers are always co-prime
numbers.
(ix) Twin prime numbers : If the difference between two
prime numbers is two, then the numbers are called twin
prime numbers.
e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
Note: Number between twin prime numbers is divisible by
6 (except (3, 5)).
(x) Rational numbers: All the numbers that can be
represented in the form p/q, where p and q are integers
and q ≠ 0, are called rational numbers and their set is
denoted by Q. Thus Q = {p/q : p, q ∈ Ι and q ≠ 0}. It
may be noted that every integer is a rational number
since it can be written as p/q. It may be noted that all
recurring decimals are rational numbers.
Note: Maximum number of different decimal digits in
is equal to q, i.e.
decimal digits.
p
q
11
will have maximum of 9 different
9
(xi) Irrational numbers: The numbers which can not be
expressed in p/q form where p, q ∈ Ι and q ≠ 0 i.e. the
numbers which are not rational are called irrational numbers
and their set is denoted by Qc. (i.e. complementary set of Q)
(b) Numbers which are not prime are composite numbers
(except 1).
e.g. 2 , 1 + 3 etc. Irrational numbers can not be
expressed as recurring decimals.
(c) ‘4’ is the smallest composite number.
Note: e ≈ 2.71 is called Napier’s constant and
π ≈ 3.14 are irrational numbers.
(d) ‘2’ is the only even prime number.
(xii) Real numbers: Numbers which can be expressed on
number line are called real numbers. The complete set of
rational and irrational numbers is the set of real numbers
and is denoted by R. Thus R = Q ∪ QC.
Negative side
–3
–2
–1
9 2
÷
7 7
9 7
=
×
7 12
3
=
4
=
Positive side
0
1
2
2
3 Real line
All real numbers follow the order property i.e. if there are two
distinct real numbers a and b then either a < b or a > b.
Note:
(a) Integers are rational numbers, but converse need not
be true.
(b) Negative of an irrational number is an irrational
number.
(c) Sum of a rational number and an irrational number is
always an irrational number
e.g. 2 + 3
(d) The product of a non zero rational number and an
irrational number will always be an irrational number.
(e) If a ∈ Q and b ∉ Q, then ab = rational number, only if
a = 0.
(f) Sum, difference, product and quotient of two irrational
numbers need not be a irrational number or we can say,
result may be a rational number also.
(xiii) Complex number: A number of the form a + ib is called
a complex number, where a, b ∈ R and i = −1 . Complex
number is usually denoted by Z and the set of complex
number is represented by C. Thus C = {a + ib : a, b ∈ R
and i = −1 }
Note: It may be noted that N ⊂ W ⊂ Ι ⊂ Q ⊂ R ⊂ C.
Train Your Brain
Example 1: The value of 1.285714 ÷ 1.714285 =
______.
3
7
(a)
(b)
4
8
3
7
(c)
(d)
7
12
Sol. (a)
1.285714
= 1 + 0.285714
2 9
=1 + =
7 7
1.714285
5 12
=1 + =
7 7
∴1.285714 ÷ 1.714285
2
Example 2: Prove that the difference 1025 – 7 is divisible
by 3.
Sol.Write the given difference in the form 1025 – 7
= (1025 – 1) – 6. The number 1025 – 1 = 99..9
 is
25 digits
divisible by 3 (and 9). Since the numbers (1025 – 1)
and 6 are divisible by 3, the number 1025 – 7, being
their difference, is also divisible by 3 without a
remainder.
Concept Application
1. If x = 12 − 9, y = 13 − 10, and=
z
then which of the following is true?
(a) z > x > y
(b) z > y > x
(c) y > x > z
11 − 8,
(d) y > z > x
SOME IMPORTANT IDENTITIES
(a + b)2 = a2 + 2ab + b2 = (a – b)2 + 4ab
(a – b)2 = a2 – 2ab + b2 = (a + b)2 – 4ab
a2 – b2 = (a + b) (a – b)
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 = (a + b)3 – 3ab (a + b) = (a + b) (a2 + b2 – ab)
a3 – b3 = (a – b)3 + 3ab (a – b) = (a – b) (a2 + b2 + ab)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca = a2 + b2 + c2
1 1 1
+ 2abc  + + 
a b c
1
9. a2 + b2 + c2 – ab – bc – ca = [(a – b)2 + (b – c)2 + (c – a)2]
2
10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
1
= (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]
2
If a + b + c = 0 , then a3 + b3 + c3 = 3abc
11. a4 – b4 = (a + b) (a – b) (a2 + b2)
12. a4 + a2 + 1 = (a2 + 1)2 – a2 = (1 + a + a2) (1 – a + a2)
1.
2.
3.
4.
5.
6.
7.
8.
Dropper JEE Module-1 Mathematics PW
Train Your Brain
Example 3: Show that the expression, (x2 – y z)3 + (y2 – z x)3 +
(z2 – x y)3 – 3 (x2 – y z) . (y2 – z x).(z2 – x y) is a perfect square
and find its square root.e
Sol.(x2 – yz)3 + (y2 – zx)3 + (z2 – xy)3 – 3(x2 – yz)
(y2 – zx) (z2 – xy) = a3 + b3 + c3 – 3abc
where a = x2 – yz, b = y2 – zx, c = z2 – xy
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
=
1
(a + b + c) ((a – b)2 + (b – c)2 + (c – a)2)
2
=
1 2 2 2
(x + y + z – xy – yz – zx) [(x2 – yz – y2 + zx)2
2
+
=
(y2 –
zx –
z2 +
xy)2 +
(z2 –
xy –
x2 +
1
Example 5. If x= = a , then what is the value of
x
1 1
x3 + x 2 + 3 + 2 ?
x x
(a) a3 + a2
(b) a3 + a2 – 5a
(c) a3 + a2 – 3a –2
(d) a3 + a2 – 4a –2
Sol. (c)
Given, x +
Now, x 3 + x 2 + 13 + 12 =  x 3 + 13  +  x 2 + 12 
x
x
x  
x 

3
yz)2]
= a3 – 3a + a2 – 2 = a3 + a2 – 3a –2.
Concept Application
+{y2 – z2 + x (y – z)}2 + {z2 – x2 + y (z – x)}2]
1 2 2 2
=
(x + y + z – xy – yz – zx) (x + y + z)2
2
[(x –
= (x + y +
=
z)2
(x2 + y2 + z2 –
(x3 + y3 + z3 –
y)2 +
(y –
z)2 +
xy – yz –
zx)2
(z –
x)2]
3xyz)2
(which is a perfect square) its square roots are
3
3
3
± ( x + y + z − 3 xyz )
Example 4. If x2 – 4x + 1 = 0, then what is the value of
1
x3 + 3 ?
x
Sol.
x2 – 4x + 1 = 0
4 ± 16 − 4 × 1× 1 4 ± 2 3
=
=2± 3
2 ×1
2
3
3
1
1
∴ x 3 + 3 =2 + 3 +
=2 + 3 +
3
x
2+ 3
⇒x=
(
(
)
)

 = 2+ 3



2 − 3 ×1

 2+ 3 2− 3

(
= 2
3
(
(
)
3
(
) + (2 − 3)
)(
)
+ ( 3 ) + 3× 2 × 3 (2 + 3 ) + 2
− ( 3 ) − 3× 2 × 3 (2 − 3 )
3
3
3
= 8 + 18 + 8 + 18 = – 50. Similarly for
3
x =2 − 3, x +
2. If x1/3 + y1/3 + z1/3 = 0, then what is (x+y+z)3 equal to?
(a) 1
(b) 3
(c) 3xy
(d) 27xyz
3. If a + b + c = 0, then what is the value of
a 2 + b2 + c2
( a − b ) + (b − c ) + (c − a )
2
2
2
1
(a) 1
(b) 3
(c)
(d) 0
3
4. Prove that (1 + x)(1 + x2)(1 + x4)(1 + x8)(1 + x16)
(1 − x 32 )
=
(1 − x)
5. If x, y, z are all different real numbers, then prove that
2
 1
1
1
1
1
1 
+
+
= 
+
+
 .
( x − y ) 2 ( y − z ) 2 ( z − x) 2  x − y y − z z − x 
6. The product (x + y)(x – y)(x2 + xy + y2)(x2 – xy + y2)
simplifies to
7. Find the real values of p, q, r satisfying (2p – 3)8
+ (1 – q)6 + (4 – 3r)4 = 0.
3
3
)
2
1
1 
1


=  x +  − 3 x +  +  x +  − 2
x
x 
x


1 2 2 2
(x + y + z – xy – yz – zx) [{x2 – y2 + z(x – y)}2
2
1
=
a
x
1
=52
x3
INDICES
If ‘a’ is any non zero real or imaginary number and ‘m’ is the
positive integer, then am = a · a · a. ... a (m times). Here a is called
the base and m is called the index, power or exponent.
Law of indices:
1. a0 = 1, (a ≠ 0)
P
W Basic Mathematics and Logarithm
3
1
, (a ≠ 0)
am
3. am + n = am · an, where m and n are rational numbers
2. a–m =
am
, where m and n are rational numbers, a ≠ 0
an
5. (am)n = amn
4. am – n =
6. ap/q =
7.
q
ap
Example 8: Find the square root of 10 + 24 + 60 + 40

a

a
a , where m, n ∈ N and
(m, n ≥ 2) and a is positive rational number
8. a  b 
ab , a, b ∈ R and atleast one of a and b should
be positive
m n
N
=
ow, 4 6 12=
216, 3 7 12 =
2401, 5 12 15625
Hence, their ascending order is
12 216, 12 2401, 12 15625, i.e., 4 6, 3 7, 5
\ The descending order of magnitude of the
given radical is 5, 3 7, 4 6 .
m
nn
n
m
Sol.
= 10 + 24 + 60 + 40
= 10 + 2 6 + 2 15 + 2 10
3 is a surd and
3
5 is a surd and
3 is an irrational number.
3
5 is an irrational number.
3. p is an irrational number, but it is not a surd.
4.
3
3 + 2 is an irrational number. It is not a surd, because
3 + 2 is not a rational number.
Train Your Brain
4
4
Example 6: Simplify  3 6 a 9   6 3 a 9 
Sol.
a9(1/6)(1/3)4 · a9(1/3)(1/6)4 = a2 · a2 = a4.
Example 7: Arrange the following in ascending or
descending order of magnitude:
4
6, 7, 5
3
1/ 4 3
1/3
4
Sol. =
6 6=
, 7 7=
, 5 51/ 2
LCM of the denominators of the exponents of
these three terms, 4, 3 and 2 is12.
Now express the exponent of each term, as a
fraction in which then denominator is 12.
1
3
( 6 )=
1
4
( 7 )=
1
6
12
( 56 )=
4
12
6=
6=
3
12
7=
7=
2
12
5=
5=
4
1
3 12
12
216
1
4 12
12
2401
12
15625
1
2+ 3+ 5
)
2
2 + 3 + 5.
=
If a is a positive rational number, which is not the nth power (n is any
natural number) of any rational number, then the irrational number
± n a are called simple surds or monomial surds.
Every surd is an irrational number (but every irrational number is
not a surd). So, the representation of monomial surd on a number
line is same as that of irrational numbers.
Examples:
2.
(
=
SURDS
1.
( 2 + 3 + 5 ) + 2 2 ( 3) + 2 3 ( 5 ) + 2 2 ( 5 )
=
Concept Application
8. If the surds 4 4, 6 5, 8 6 and 12 8 are arranged in
ascending order from left to right, then the third surd
from the left is
(a) 12 8
(b) 4 4
(c)
9.
6
8
(d)
6
6
5
15 − 2 56 3 7 + 2 2 = ______.
(a) 0
(b)
(c) 1
(d)
2
6
2
10. If x = 21/3 – 2, then x3 + 6x2 + 12x = _________.
(a) 6
(b) –6
(c) 8
(d) –8
−2
−3
−3
1 1 1
11. If a = 4 2 , b = (125) 3 , c = 81 4 then + + =
a b c
−1
 a+ b
 a+ b
12. Simplify a 
 + b 

 2b a 
 2a b 
−1
RATIO
(i) If A and B be two quantities of the same kind, then their
A
ratio is A : B; which may be denoted by the fraction
B
(This may be an integer or fraction)
a ma
(ii) A ratio may represented in a number of ways e.g. =
b mb
na
=
=...... where m, n,...... are non-zero numbers.
nb
Dropper JEE Module-1 Mathematics PW
(iii) To compare two or more ratio, reduce them to common
denominator.
( x + y + z) − ( y + z) ( x + y + z) − ( x + z)
=
=
(iv) Ratio between two ratios may be represented as the
ratio of
9
9
−3
−4
two integers
2
2
e.g. a : c : a b = ad or ad : bc
b d c d bc
(v) Ratios are compounded by multiplying them together i.e.
a c e
ace
⋅ ⋅ ..... = .....
b d f
bdf
(vi) If a : b is any ratio then its duplicate ratio is a2 : b2; triplicate
ratio is a3 : b3..... etc.
(vii) If a : b is any ratio, then its sub-duplicate ratio is a1/2 : b1/2;
sub-triplicate ratio is a1/3 : b1/3 etc.
PROPORTION
( x + y + z) − ( x + y)
9
−2
2
x
y
z
= = =
⇒x:y:z=3:1:5
3 / 2 1/ 2 5 / 2
a 3b 2c 6d
3a b 6c 2d
Example 10: If
, then
a 3b 2c 6d 3a b 6c 2d
the correct statement is
(a) ad = bc
(b) ac = bd
ab
(d) a + d = b + c
(c) c =
d
Sol. (a) Apply C and D the the given equation
=
4c 12d 4d 12c
c 3d d 3c
⇒
2a 6d
2b 6a
a 3b b 3a
⇒ bc + 3ac + 3bd + 9ad = ad + 3ac + 3bd + 9bc
⇒ 8ad = 8bc
⇒ ad = bc
When two ratios are equal, then the four quantities compositing
a c
them are said to be proportional. If = , then it is written as
b d
a : b = c : d or a : b : : c : d
(i) ‘a’ and ‘d’ are known as extremes and ‘b and c’ are known
as means.
(ii) An important property of proportion Product of extremes
= product of means.
(iii) If a : b = c : d, then b : a = d : c (Invertando)
(iv) If a : b = c : d, then a : c = b : d (Alternando)
a+b c+d
(v) If a : b = c : d, then
(Componendo)
=
b
d
a −b c−d
(vi) If a : b = c : d, then
=
(Dividendo)
b
d
a+b c+d
(vii) If a : b = c : d, then
=
(Componendo and
a −b c−d
dividendo)
a c e
a + c + e + .....
(viii) If = = = ..... , then each
b d f
b + d + f + ......
Sum of the numerators
=
Sum of the denominators
a c e
xa + yc + ze + ......
(ix) If = = = ..... , then each =
b d f
xb + yd + zf + ......
1/ n
x+ y y+z z+x
Example 9: If = =
, then find x : y : z.
2
3
4
Sum of the numerators
Sol. Each =
Sum of the denominators
2( x + y + z ) x + y + z
=
=
and therefore each
9
9/2
P
W Basic Mathematics and Logarithm
Concept Application
13. Solve the equations
x2 1 x2 1
x2 1 x2 1
2x2 1
2
INTERVALS
Intervals are basically subsets of R and are commonly used in
solving inequalities or in finding domains. If there are two
numbers a, b ∈ R such that a < b, we can define four types of
intervals as follows :
 xa n + yc n + ze n 
a c e
(x) If = = = ..... , then each =  n
n
n 
b d f
 xb + yd + zf 
Train Your Brain
⇒
Open Interval (a, b)
{x : a < x < b} i.e. extreme points are not includes
Closed Interval [a, b]
{x : a ≤ x ≤ b} i.e. extreme points are includes
It can possible when a and b are finite
Semi-Open Interval(a, b]
{x : a < x ≤ b} i.e. a is not include and b is include
Semi-Closed Interval [a, b)
{x : a ≤ x < b} i.e. a is include and b is not include
Note:
1. The infinite intervals are defined as follows :
(i) (a, ∞) = {x : x > a}
(ii) [a, ∞) = {x : x ≥ a}
5
(iii) (– ∞, b) = {x : x < b}
(iv) (∞, b] = {x : x ≤ b}
(v) (– ∞, ∞) = {x : x ∈ R}
2. x ∈ {1, 2} denotes some particular values of x, i.e. x = 1, 2
3. If there is no value of x, then we say x ∈ φ (null set)
GENERAL METHOD TO
SOLVE INEQUALITIES
Method of Intervals (Wavy Curve Method)
 ( x – b1 ) k1 ( x – b2 ) k2 − − − ( x – bn ) kn 
Let g(x) = 
...(i)
r1
r
rn 
 ( x – a1 ) ( x – a2 ) 2 − − − ( x – an ) 
Where k1, k2 ..... kn and r1, r2 ........ rn ∈ N and b1, b2..... bn and a1,
a2 ... an are real numbers.
Then to solve the inequality following steps are taken.
Steps: Points where numerator becomes zero are called zeros or
roots of the function and where denominator becomes zero are
called poles of the function.
(i) First we find the zeros and poles of the function.
(ii) Then we mark all the zeros and poles on the real line and
put a vertical bar to divide the real line in many intervals.
(iii) Determine sign of the function in any of the interval and
then alternates the sign in the neghbouring interval if the
poles or zeros dividing the two interval has appeared odd
number of times otherwise retain the sign.
(iv) Thus we consider all the intervals. The solution of the g(x)
> 0 is the union of the intervals in which we have put the
plus sign and the solution of g(x) < 0 is the union of all
intervals in which we have put the minus sign.
Train Your Brain
( x –1) 2 ( x + 4)
Example 11: Solution
< 0 is
(2 – x)
( x –1) 2 ( x + 4)
( x + 4)
Sol.
<0⇒
>0
(2 – x)
( x – 2)
+
–4
+
2
–
⇒ (– ∞, – 4) ∪ (2, ∞).
3
Example 12: The solution of
is
Sol.
6
( x + 4) 4 ( x –1)3
> 0
( x – 2)
( x + 4) 4/3 ( x –1)3
x –1
>0⇒
>0
( x – 2)
x–2
+
1
+
2
–
⇒ (– ∞, 1) ∪ (2, ∞) Excluding – 4.
Example 13. Find the range of x, so that following
expressions are defined.
(a) (x – 1) (x – 2) (x – 3)2 (x – 4)5 (x – 5) > 0
( x 1( x 2)
0
(b)
( x 3)
Sol. (a)
x (,1) (2, 3) (3, 4) (5, )
+
+
–
1
2
+
3
–
4
+
5
(b) x [1, 2] (3, )
–
+
1
–
2
+
3
Concept Application
14. Find the number of integer values of variable x
satisfying the following pair of inequalities.
( x 1)( x 4)
0 & x 2 6 x 27 0
x 3
LOGARITHM FUNCTION
Introduction
The kinetics of some chemical reactions (which rely on the
concentration of one or more of the reactants), radioactive decay
(used in carbon dating), are examples of exponential decay.
A nautilus's shell develops in such a way that each new
chamber is roughly a replica of the previous one but is enlarged
by a fixed amount. This results in the creation of the elegant
shape known as a logarithmic spiral. The pace of growth is
related to the existing size of many objects in nature. So, it
is extremely important to comprehend the mathematics that
explains the behaviour of entities like birth rates if we are
to forecast how they will behave in the future. Logarithmic
scales (the Richter scale in the case of an earthquake) are useful
in offering a means to handle the numerical values of these
measures.
Dropper JEE Module-1 Mathematics PW
Definition
The logarithm of the number N to the base ‘a’ is the exponent
indicating the power to which the base ‘a’ must be raised to obtain
the number N.
This number is designated as logaN.
Hence logaN = x ⇔ ax = N, a > 0, a ≠ 1 and N > 0
If a = 10, then we write log b rather than log10b
a = e, we write ln b rather than logeb
The existence and uniqueness of the number logaN follows
from the properties of an exponential functions.
Domain
The existence and uniqueness of the number loga N can be
determined with the help of set of conditions, a > 0 & a ≠ 1 &
N > 0.
FUNDAMENTAL IDENTITY
(i) loga1 = 0
(ii) logaa = 1
(iii) log1/a a = –1
(a > 0, a ≠ 1)
(a > 0, a ≠ 1)
(a > 0, a ≠ 1)
It states that ratio of logarithm of two numbers is independent of
their common base
Symbolically
log a M
= logb M (a > 0, M > 0, b > 0)
log a b
Proof:
Let logbM = x
⇒ M = bx
⇒ logaM = logabx
⇒ logaM = x · logab
log a M
= x = logb M
log a b
Important Results
1
(i) Base power formula: log ak M = loga M
k
Proof:
log a M log a M 1
=
log=
(M ) =
log a M
ak
log a a k k log a a k
(ii) a logb c = c logb a
FUNDAMENTAL
LOGARITHMIC IDENTITY
log a c .logb a
a c log b a
Proof:
=
a logb c a=
(a log
=
)
(c)logb a
... (i)
... (ii)
GRAPH OF LOGARITHMIC
FUNCTIONS
If a > 0, a ≠ 1, then the function y = loga x, x ∈ R+ (set of positive real
numbers) is called the logarithmic Function with base a.
Y
PRINCIPAL PROPERTIES
Let M & N are arbitrary positive numbers, a > 0, a ≠ 1,
and x, y are any real numbers, then:
(i) loga (M · N) = loga M + loga N;
Proof:
Let logaM = x and logaN = y
⇒ M = ax and N = ay
Now, MN = axay = ax+y ⇒ logaMN = x + y
In general
loga(x1 x2 ... xn) = logax1 + loga x2 + ... + loga xn
(ii) loga(M/N) = loga M − loga N
Proof:
Let logaM = x and logaN = y
P
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logaM/N = x – y
BASE CHANGING THEOREM
⇒
Remember
log102 ≈ 0.3010; log103 ≈ 0.4771
ln 2 ≈ 0.693; ln 10 ≈ 2.303
log N
a a = N, a > 0, a ≠ 1 & N > 0
Proof :
logaN = x
N = (a)x
by equation (i) & (ii)
N = (a)logaN
⇒ M = ax and N = ay
Now, M/N = ax/ay = ax–y ⇒
(iii) logaMα = α·loga M
O
y = log2x
y = log4x
y = log10x
+
Domain : R
Range : R
X
Nature : one-one
y = log1/10x
y = log1/4x
y = log1/2x
Note: (i) If the number and the base are on the same side of the
unity, then the logarithm is positive.
(ii) If the number and the base are on the opposite sides of
unity, then the logarithm is negative.
EXPONENTIAL FUNCTION
If a > 0, a ≠ 1 then the function defined by f (x) = ax, x ∈ R is called
an Exponential Function with base a.
7
Y
–x
–x
y = 4 y = 10
–x
y=2
y =10
x
x
y = 4 y = 2x
Domain : R
Range : R
+
a>1
Nature : one-one
0<a<1
COMMON AND
NATURAL LOGARITHM
log10N is referred as a common logarithm and logeN is called
as natural logarithm of N to the base Napierian and is popularly
written as ln N. Note that e is an irrational quantity lying between
2.7 to 2.8 Note that eln x = x.
X
O
Train Your Brain
LOGARITHMIC EQUATION
The equality loga x = loga y is possible if and only if x = y
i.e. loga x = loga y ⇔ x = y
Always check validity of given equation, (x > 0, y > 0, a > 0,
a ≠ 1)
Example 14: How many solutions are there for equation
log4 (x – 1) = log2 (x – 3)?
Sol. log4 (x – 1) = log2 (x – 3)
⇒ log22 (x – 1) = log2 (x – 3)
1
log2 (x – 1) = log2 (x – 3)
2
⇒ log2 (x – 1)1/2 = log2 (x – 3)
⇒ (x – 1)1/2 = (x – 3)
⇒ x – 1 = x2 – 6x + 9
⇒ (x – 2) (x – 5) = 0
⇒ x = 2, 5
But x – 1 > 0 and x – 3 > 0
x > 1 and x > 3
So only one solution x = 5
Example 15: Solve the logarithmic inequality log1/5
4x + 6
≥0.
x
Sol. Since log1/5 1 = 0, the given inequality can be written as.
⇒
LOGARITHMIC INEQUALITY
Let ‘a’ is a real number such that
(i) If a > 1, then logax > loga y ⇒ x > y
(ii) If a > 1, then logax < α ⇒ 0 < x < aα
(iii) If a > 1, then logax > α ⇒ x > aα
(iv) If 0 < a < 1, then logax > loga y ⇒ 0 < x < y
(v) If 0 < a < 1, then logax < α ⇒ x > aα
Form - I : f (x) > 0, g(x) > 0, g(x) ≠ 1
FormCollection of system
g ( x) > 1
 f ( x) ≥ 1,
(a) logg(x) f (x) ≥ 0 ⇔ 
0 < f ( x) ≤ 1, 0 < g ( x) < 1
, 0 < g ( x) < 1
 f ( x) ≥ 1
(b) logg(x) f (x) ≤ 0 ⇔ 
g ( x) > 1
0 < f ( x ) ≤ 1 ,
 f ( x) ≥ g ( x) a
(c) logg(x) f (x) ≥ a ⇔ 
a
0 < f ( x ) ≤ g ( x )
,
g ( x) > 1
, 0 < g ( x) < 1
0 < f ( x ) ≤ g ( x ) a
(d) logg(x) f (x) ≤ a ⇔ 
a
 f ( x) ≥ g ( x)
,
g ( x) > 1
, 0 < g ( x) < 1
From - II : f (x) > 0, g(x) > 0, f(x) > 0, f(x) ≠ 1
Form Collection of system
f ( x) ≥ g ( x), φ( x) > 1,

(a) logφ(x) f (x) ≥ logφ(x) g(x) ⇔ 
0
<
f ( x) ≤ g ( x);0 < φ( x) < 1

 0 < f ( x) ≤ g ( x), φ( x) > 1,
(b) logφ(x) f (x) ≤ logφ(x) g(x) ⇔ 
 f ( x) ≥ g ( x) > 0, 0 < φ( x) < 1
8
4x + 6
≥ log1/5 1
x
When the domain of the function is taken into account
the inequality is equivalent to the system of inequalities.
log1/5
 4x + 6
 x > 0,

 4x + 6 ≤ 1
 x
Solving the inequalities by using method of intervals
–3 
x ∈  –2,

2 

Example 16: For x ≥ 0, what is the smallest possible value
of the expression log(x3 – 4x2 + x + 26) – log(x + 2) ?
Sol. log
( x 3 − 4 x 2 + x + 26)
( x + 2)
= log
( x 2 − 6 x 2 + 13)( x + 2)
( x + 2)
Dropper JEE Module-1 Mathematics PW
= log
(x2
– 6x + 13)
= log{(x –
3)2
[⸪ x ≠ – 2]
+ 4}
∴ Minimum value is log 4 when x = 3
2
Example 17: Given log2a = s, log4b = s2 and log c2 (8) = s 3 + 1 .
a 2 b5
as a function of ‘s’ (a, b, c > 0, c ≠ 1).
c4
Sol. Given log2a = s ...(i)
Write log2
log2b = 2s2 ...(iii)
...(iv)
to find 2 log2a + 5 log2b – 4 log2c
⇒ 2s +
10s2
log6 x < 0
log1/7 x > 0
log1/9 x < 0
log2 (x – 1) > 1
(iv)
(vi)
(viii)
(x)
19. Solve for x:
(i) log4 (2x – 3) < 2
(iii) log16 (log4 (x) > 1
20. Prove that : 2
log 2 3
=3
log2 x ≤ 0
log1/8 x ≥ 0
log1/e x ≤ 0
log1/2 (x – 2) ≤ 1
(ii) log1/2 (3x – 2) ≥ 3
(iv) log1/2 (log1/4 (x)) < 1
log3 2
21. If log1227 = a find the value of log616 in term of a.
...(ii)
3
s +1
2
2 log c
s3 + 1
⇒
=
⇒ 4 log2c = 3(s3 + 1)
2
3log 2
log8c2 =
(iii)
(v)
(vii)
(ix)
–
3(s3
+ 1)
1
Example 18. If log 4 M + 4 log 4 N = 1+ log.008 5 then the
4
value of MN16 = k.21/3, where k is equal to
(a) 8
(b) 32
(c) 36
(d) 40
log 2 5
1
Sol. (b)
log 2 M 2 log 2 N 1 8
log 2 (.008)
ABSOLUTE VALUE FUNCTION /
MODULUS FUNCTION
This is also known as absolute value function and denoted by
 x, x≥0
f(x) = |x| i.e. f(x) = 
− x , x < 0
Domain of this function is set of all real numbers because
f(x) exists for all x ∈ R but |x| ≥ 0 so range is all non-negative real
numbers.
f(x)
log 2 10 1
3 3 log 2 10
1 2
1 3 3
y
log 2 M 1/ 8 log 2 N 2 1 log 2 MN
16 1/ 8
MN 16 1/ 8
MN
16
2
22 / 3
16 / 3
32(2 )
1/ 3
x
O
x
x
y
Domain = R; Range = [0, ∞) or R+ ∪ {0}
Properties of modulus : For any a, b R
(i) | a | ≥ 0
(ii) | a | = | –a |
Concept Application
15. Solve the inequality log1/3 (5x – 1) > 0.
16. If m1 = log8 16, m2 = log81 27, m3 = log1/3 1/9, m4
= log1/3 9 3 then m1 . m2 . m3 ∙ m4
17. If log7 (log3 (log2 x)) = 0, then find log0.125 x.
18. Solve for x:
(i) log3 x > 0
(ii) log5 x ≥ 0
P
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(a) | a |n = | an |
(b) | an | = an , where n is even and n ∈ z
(iii) a ≥ a, a ≥ −a
(iv) ab = a b
(v)
a
a
=
b
b
Note: | f (x) | + | g(x) | = | f (x) + g(x) |
⇒ f (x) · g(x) ≥ 0
9
Concept Application
Train Your Brain
Example 19 The absolute value of sum of real solutions of
log2 |x2 + 5x + 4| = log2 3 + log2 |x + 1| is
(a) 8
(b) 6
(c) 7
(d) 5
| ( x + 1)( x + 4) |
= log 2 3
Sol. log 2
| x + 1|
|x + 4| = 3
x + 4 = –3, +3
x = –7, –1 (rejected);
⇒ x = –7
22. The number of integers which does NOT satisfy
log|2x| (| x + 2| + | x – 2|) = 1 is
(a) 0
(b) 1
(c) 2
(d) 3
23. Solve for x:
(i) ||x – 2| –1| = 2
(ii) ||x – 3| –5| = 1
(iii) |||x – 5| –4| –3| = 2
Example 20: Solve the following linear equation
(i) x | x | = 4
INEQUALITIES INVOLVING
ABSOLUTE VALUE
(ii) | x – 3 | + 2 | x + 1 | = 4
Sol. (i) x | x | = 4
If x > 0
∴ x2 = 4
⇒
If x < 0 ⇒
possible
x = ±2
∴ x = 2 (⸪ x ≥ 0)
–x2 = 4 ⇒ x2 = –4 which is not
(i) | x | ≤ a (where a > 0)
It implies those values of x on real number line which are
at distance a or less than a from zero.
(ii) | x – 3 | + 2 | x + 1 | = 4
e.g. | x | ≤ 2 ⇒ –2 ≤ x ≤ 2
∴ –(x – 3) –2 (x + 1) = 4
⇒ –x + 3 –2 x – 2 = 4 ⇒
⇒ –3x = 3
⇒
–3x + 1 = 4
x = –1
–1 < x ≤ 3
⇒ –x + 3 +2x + 2 = 4
⇒ x = – 1 which is not possible
Case-III If x > 3
x – 3 + 2(x + 1) = 4
3x – 1 = 4
⇒
x = 5/3 which is not possible
∴ x = –1
Example 21: Number of real solutions of |x – 1| = |x – 2|
+ |x – 3| is
(a) 0
(b) 1
(c) 2
(d) more than 2
Sol.
Case-I: x ≤ 1, 1 – x = 2 – x + 3 – x
x = 4 (rejected)
Case-II: 1 < x ≤ 2, x – 1 = 2 – x + 3 – x, x = 2
Case-III: 2 < x < 3, x – 1 = x – 2 + 3 – x, x = 2
Case-IV: x ≥ 3, x – 1 = x – 2 + x – 3
x = 4
10
|x| < 3 ⇒ –3 < x < 3
In general, |f(x)| ≤ a (where a > 0) ⇒ –a ≤ f (x) ≤ a.
(ii) | x | ≥ a (where a > 0)
∴ –(x – 3) + 2 (x + 1) = 4
a
0
⇒ –a ≤ x ≤ a
Case-I If x ≤ – 1
Case-II If
–a
⇒ x = 2, 4
It implies those values of x on real number line which are
at distance a or more than a from zero
–a
0
a
⇒ x ≤ –a or x ≥ a
e.g. | x | ≥ 3 ⇒ x ≤ –3 or x ≥ 3
| x | > 2 ⇒ x < –2 or x > 2
In general, |f(x)| ≥ a ⇒ f(x) ≤ – a or f (x) ≥ a
(iii) a ≤ | x | ≤ b (where a, b > 0)
It implies those value of x on real number line whose
distance from zero is equal to a or b or lies between a and b
–b
–a
0
a
b
⇒ [–b, –a] ∪ [a, b]
e.g. 2 ≤ x ≤ 4 ⇒ x ∈ [–4, –2] ∪ [2, 4]
(iv)
If | x + y | = | x | + | y |, xy ≥ 0
If | x – y | = | x | + | y |, xy ≤ 0
If | x + y | = || x | – | y ||, xy ≤ 0
If | x – y | = || x | – | y ||, xy ≥ 0
Dropper JEE Module-1 Mathematics PW
Train Your Brain
Example 22: Solve x2 – 4| x | + 3 < 0.
Sol.
x2 – 4| x | + 3 < 0
⇒ (| x | – 1) (| x | – 3) < 0
⇒1<|x|<3
⇒ – 3 < x < –1 or 1 < x < 3
⇒ x ∈ (−3, −1) ∪ (1,3)
Example 23: Solve 1 ≤ | x – 2 | ≤ 3
Sol.
1≤|x–2|≤3
⇒ –3 ≤ x – 2 ≤ – 1 or 1 ≤ x – 2 ≤ 3
⇒ –1 ≤ x ≤ 1 or 3 ≤ x ≤ 5
⇒ x ∈ [–1, 1] ∪ [3, 5]
Example 24: Solve | x –1 | + | x –2| + | x –3| ≥ 6,
Sol.
For x ≤ 1, the given inequation becomes
1 – x + 2 – x + 3 – x ≥ 6
⇒ –3x ≥ 0
⇒ x ≤ 0 and for x ≥ 3, the given equation becomes
x –1 + x –2 + x – 3 ≥ 6
⇒ 3x ≥ 12
⇒ x ≥ 4
For 1 < x ≤ 2
we get x – 1 + 2 – x + 3 – x ≥ 6
⇒ –x+4≥6
i.e. – x ≥ 2 ⇒ x ≤ –2 Not possible
For 2 < x < 3,
We get x – 1 + x – 2 + 3 – x ≥ 6
⇒ x ≥ 6 not possible
Hence solution set is (–∞, 0] ∪ [4, ∞)
i.e. x ≤ 0 or x ≥ 4
Concept Application
24. Solve x − 1 − 2 < 5
25. Solve |x2 – 2x| + |x – 4| > |x2 – 3x + 4|.
x3
1.
26. Solve
x 1
27. Solve for x:
(i) |x – 1| ≥ 1
(iii) 1 < |2x + 1| < 3
(v) –1 ≤ |3x – 1| ≤ 5
(ii) |x – 2| < 1
(iv) 1 ≤ |1 – 2x| ≤ 3
(vi) –6 ≤ |1 – 3x| ≤ –1
Short Notes
Some Important Identities
1.
2.
3.
4.
5.
6.
7.
(a + b)2 = a2 + 2ab + b2 = (a – b)2 + 4ab
(a – b)2 = a2 – 2ab + b2 = (a + b)2 – 4ab
a2 – b2 = (a + b) (a – b)
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 = (a + b)3 – 3ab (a + b) = (a + b) (a2 + b2 – ab)
a3 – b3 = (a e– b)3 + 3ab (a – b) = (a – b) (a2 + b2 + ab)
8. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca = a2 + b2 + c2 +
1 1 1
2abc  + + 
a b c
1
9. a2 + b2 + c2 – ab – bc – ca = [(a – b)2 + (b – c)2 + (c – a)2]
2
10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
1
= (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]
2
If a + b + c = 0 , then a3 + b3 + c3 = 3abc
11. a4 – b4 = (a + b) (a – b) (a2 + b2)
12. a4 + a2 + 1 = (a2 + 1)2 – a2 = (1 + a + a2) (1 – a + a2)
P
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Laws of Indices
1. am × an = am+n
2. am ÷ an = am–n
3. (am)n = (an)m = amn
m
−
m
4.  a  n =  b  n
 
 
b
a
m
–n
5. a ÷ b = am × bn
6. ( n a ) n = a , where, n ∈ N, n ≥ 2 and a is positive rational
number
7. n a n b = n ab , where n ∈ N, n ≥ 2 and a, b are rational
number
8. a ×=
b
ab , a, b ∈ R and atleast one of a or b should be
positive.
Ratio and Proportion
a, b, c, d are in proportion. Then,
a c
=
b d
a b
(ii)
= .........(alternendo)
c d
(i)
11
a+b c+d
....(componendo)
=
b
d
a−b c−d
(iv)
......(dividendo)
=
b
d
a+b c+d
(v)
......(componendo and dividendo)
=
a−b c−d
b d
(vi)
= ........(invertendo)
a c
a c e
(vii) If = =
= ......, then each
b d
f
a + c + e + ......
Sum of the numerators
=
b + d + f + ...... Sum of the denominators
a c e
xa + yc + ze + ...
(viii) If = =
=......, then each =
xb + yd + zf + ...
b d
f
(iii)
Solving of Inequalities by Wavy Curve Method
Step 1: Obtain critical points by equating all factors to zero.
Step 2: Plot the critical points on the number line in the increasing
order.
Step 3: Put plus sign in the right most interval.
Step 4: Now, if a root is repeated even times, the sign of the
function will remain the same in the two adjacent sub-intervals of
the root. (when we are moving from right to left)
Step 5: If a root is repeated for odd times the sign of the function
will be different in the two adjacent sub intervals of the root.
(when we are moving from right to left)
Properties of Logarithm
1. loge(ab) = logea + logeb ; (a, b > 0)
a
2. log e=
  log e a − log e b ; (a, b > 0)
b
3. logeam = mlogea ; (a > 0, m ∈ R)
4. logaa = 1 ; (a > 0, a ≠ 1)
5. log b a =
m
1
log b a ; (a, b > 0, b ≠ 1 and m ∈ R – {0})
m
1
; (a, b, > 0 and a, b ≠ 1)
log a b
log m a
7. log b a =
; (a, b, m > 0 and m, b ≠ 1)
log m b
6. log b a =
8. a loga m = m ; (a > 0, a ≠ 1 m > 0)
9. a logc b = b logc a ; (a, b, c > 0 and c ≠ 1)
 x > y, if m > 1
10. If logmx > logmy ⇒ 
 x < y, if 0 < m < 1
(m,x,y > 0, m ≠ 1)
11. logma = b ⇒ a = mb ; (m, a > 0, m ≠ 1 ∈ real number)
a > mb ; if
12. logma > b ⇒ 
b
a < m ; if
a < mb ; if
13. logma < b ⇒ 

b
a > m ; if
m >1
0 < m <1
m >1
0 < m <1
Logarithmic Equations
(i) An equation of the form loga f(x) = b, (a > 0), a ≠ 1 is
equivalent to the equation. f(x) = ab, (f(x) > 0)
 g ( x) > 0

(ii) If loga f(x) > logag(x) and a > 1, then ⇒  f ( x) > 0
 f ( x) > g ( x)

 f ( x) > 0

(iii) If loga f(x) > logag(x) and a < 1, then ⇒  g ( x) > 0
 f ( x) < g ( x)

Definition of Modulus
 x, if x ≥ 0
|x| = 
− x, if x < 0
Properties of Modulus
Let 'a', 'b' are positive real number then,
(i) |x|= a ⇒ x = ± a
(ii) |x| ≤ a ⇒ –a ≤ x ≤ a ⇒ x ∈ [–a, a]
(iii) |x| < a ⇒ –a < x < a ⇒ x ∈ (–a, a)
(iv) |x| ≥ a ⇒ x ≤ –a or x ≥ a ⇒ x ∈ (–∞, –a] ∪ [a, ∞)
(v) |x| > a ⇒ x < –a or x > a ⇒ x ∈ (–∞, –a) ∪ (a, ∞)
(vi) a ≤ |x| ≤ b ⇒ x ∈ [–b, –a] ∪ [a, b]
(vii) a < |x| < b ⇒ x ∈ (–b, –a) ∪ (a, b)
(viii) |x + y| ≤ |x| + |y|; equality holds when xy ≥ 0
(ix) |x – y| ≥ ||x| – |y||; equality holds when xy ≥ 0
(x) |x + y| ≥ ||x| – |y||; equality holds when xy ≤ 0
(xi) |x – y| ≤ |x| + |y|; equality holds when xy ≥ 0
Solved Examples
1. The value of 81(1/log53) + 27 log936 + 34/log79 is equal to
(a) 49
(b) 625
(c) 216
(d) 890
Sol. (d)
12
81(
1/ log5 3)
+ 27 log9 36 + 34/ log7 9
1
3. log3 36
2
=
34 log3 5 + 3
4
3/2
+ 34 log9 7
=3log3 5 + 3log3 36 + 3log3 7
= 54 + 363/ 2 + 7 2 = 890
4/2
Dropper JEE Module-1 Mathematics PW
2. Values of x satisfying the equation
5
log52 x + log5x   = 1 are
x
(a) 1
1
(c)
25
Sol. (a,b,c)
(b) 5
(d) 3
1
≤ log0.1 x ≤ 2, then
2
(a) Maximum value of x is
4. If
1
(b) x lies between
and
100
1
10
1
10
1
(c) Minimum value of x is
10
1
(d) Minimum value of x is
100
5
=1
x
⇒ (log5x)2 + log5x5 – log5xx = 1
Sol. (a,b,d)
log 5 5
log 5 x
2
⇒ ( log 5 x ) +
1
−
=
1/ 2
2
log 5 5 + log 5 x log 5 5 + log 5 x
1
1
1
≤ log0.1x ≤ 2 ⇒   ≥ x ≥  
2
log 5 x
1
 10 
 10 
⇒ (log5x)2 +
1
−
=
1 + log 5 x 1 + log 5 x
log 3 135 log 3 5
5. Let N =
. Then N is
−
Let log5x = t
log15 3 log 405 3
1
t
(a) a natural number
(b) a prime number
1
∴ t2 +
−
=
1
+
t
1
+
t
(c) a rational number
(d) an integer
(log5x)2 + log5x
t 2 (1 + t ) + 1 − t
1
⇒
=
1+ t
⇒ t3 + t2 + 1 – t = 1 + t
t3 + t2 – 2t = 0
t (t2 + t – 2) = 0
t (t – 1) (t + 2) = 0
t = 0, 1, – 2
∴ log5x = 0, 1, –2
1
∴ x = 1, 5,
25
3. The equation log x2 16 + log2x 64 = 3 has
(a) One irrational solution
(b) No prime solution
(c) Two real solutions
(d) One integral solution
Sol. (a,b,c,d)
log x 64
4
log x 2 +
2
log x 2 x
2 x log x 2 6 log x 2
⇒
+
=
3
1 + log x 2
x
Let α = logx2
6α
2α +
=3
1+ α
2α + 2α2 + 6α – 3 – 3α = 0
⇒ 2α2 + 5α – 3 = 0
⇒ (α + 3) (2α – 1) = 0 ⇒ a = – 3, 1/2
∴ logx2 = – 3 ⇒ x = 2–1/3 (Irrational)
1
or logx2 = ⇒ x = 4 (Integer)
2
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Sol. (a,b,c,d)
= log3135 log315 – log35 log3405
= log3(5 × 33) . log(5 × 3) – log3 5. log3 (5 × 34)
= (log35 + log333) (log35 + log33) – log35 (log35 +
log334)
= (x + 3) (x + 1) – x (x + 4)
{Let log35 = x}
= x2 + 4x + 3 – x2 – 4x = 3
which is Prime, rational Integer and natural number
6. Let y =
log 2 3·log 2 12·log 2 48·log 2 192 + 16
Find y ∈ N.
– log212·log248 + 10.
Sol. [0006]
y=
log 2 3·log 2 12·log 2 48·log 2 192 + 16
– log212 · log248 + 10
=
log 2 3 · ( 2 + log 2 3) ( 4 + log 2 3) ( 6 + log 2 3) + 16 −
Let us put log23 = x
=
( 2 + log 2 3) ( 4 + log 2 3) + 10
x ( 2 + x ) ( 4 + x ) ( 6 + x ) + 16 − ( 2 + x ) ( 4 + x ) + 10
2
2
2
= ( x + 6 x ) ( x + 6 x + 8 ) + 16 − ( x + 6 x + 8 ) + 10
Put again x2 + 6x = α
= α ( α + 8 ) + 16 − ( α + 8 ) + 10
= α 2 + 8α + 16 − ( α + 8 ) + 10
= ( α + 4 ) − ( α + 8 ) + 10
2
= ( α + 4 ) − ( α + 8 ) + 10 = y = 6.
13
7. Sum of all the solutions of the equation
log6(x2 – 1) – log 6 ( x − 6) = log6(x + 1)2 is a + b ,(a, b εN).
Then a + b is equal to
2
Sol. log6(x2 – 1) – log 6 ( x − 6) 2= log 6 ( x + 1) 2
⇒ log 6
( x − 1)( x + 1)
= log 6 | x − 6 |
( x + 1) 2
 ( x − 1) 
⇒ log 6  =
 log 6 | x − 6 |
 ( x + 1) 
x −1
= | x−6|
⇒
x +1
Case-I: x ≥ 6
⇒ x – 1 = x2 – 5x – 6
⇒ x2 – 6x – 5 = 0
⇒ (x – 3)2 = 14
⇒ x = 3 ± 14
x = 3 – 14 < 1 rejected
x = 3 + 14 accepted
Case-II: x < 6
x – 1 = –(x2 – 5x – 6)
⇒ x2 – 4x – 7 = 0
(x – 2)2 = 11
x= 2 ± 11
x= 2 + 11 (accepted)
x= 2 − 11 (accepted)
Sum of roots = 7 + 14
⇒ a = 7, b = 14 ⇒ a + b = 21
8. Match the column
Column-I
Column-II
A. The roots of log2(x +e) = p. Positive Number
log2x + log2e is a
B. The solution of log1/5 (2x2 q. Rational Number
+ 5x +1) < 0 contains
C. log π
r. Irrational Number
π
is
sin
6
D. log 5.log 20 + log 2 2
10
10
10
simplifies to
14
s. Negative Number
Sol. A→(p, r) ; B→(p, q, r, s) ; C→(r, s) ; D→(p, q)
(A) x + e = xe
x(e – 1) = e
e
x=
e −1
(B) 2x2 + 5x + 1 > 1 and 2x2 + 5x + 1 > 0
⇒ 2x2 + 5x + 1 > 1
⇒ (x)(2x + 5) > 0
−5 

⇒ x ∈  −∞,  ∪ (0, ∞)

2 
2
(D) (1 − log10 2)(1 + log10 2) + log10
2
2
2
1
⇒ 1 − log10 2 + log 10 2 =
9. The largest integral value of x satisfying
18 x − 5 ≤ 2(18 x + 12) − 18 x + 5 is
(a)
(b)
(c)
(d)
Sol. (d)
0
1
2
no integral value of x possible
Let 18x = p
p − 5 +
p + 5 ≤ 2( p + 12)
⇒ p − 5 + p + 5 + 2 p 2 − 25 ≤ 2 p + 24
⇒ p 2 − 25 ≤ 12 ⇒ p 2 ≤ 169 ⇒ p ≤ 13
Also p ≥ 5
Thus 5 ≤ p ≤ 13 log185 ≤ x ≤ log1813
10. Solve
(e − sin x)( x − 2)
≥ 0.
( x + 4)
Sol. Zeroes x = 2, Pole x ≠ –4
e – sin x > 0 always positive
(e − sin x)( x − 2)
≥0
( x + 4)
Final solution x ∈ (–∞, –4) ∪ [2, ∞)
Dropper JEE Module-1 Mathematics PW
Exercise-1 (Topicwise)
BASIC CONCEPTS AND NUMBER SYSTEM
1. If x – a is a factor of x3 – a2x + x + 2, then ‘a’ is equal to
(a) 0
(b) 2
(c) –2
(d) 1
2. If x, y are integral solutions of 2x2 – 3xy – 2y2 = 7, then value
of |x + y| is
(a) 2
(b) 4
(c) 6
(d) 2 or 4 or 6
3. If a, b, c are real, then a(a – b) + b(b – c) + c(c – a) = 0, only
if
(a) a + b + = 0
(b) a = b = c
(c) a = b or b = c or c = a
(d) a – b – c = 0
4. If 2x3 – 5x2 + x + 2 = (x – 2) (ax2 – bx – 1), then a & b are
respectively
(a) 2, 1
(b) 2, – 1
(c) 1, 2
(d) –1, 1/2
5. The value of [e] – [– π] is, where [.] denotes greatest integer
function.
(a) 5
(b) 6
(c) 7
(d) 8
6. If
L
1
7 8
1
7 6
1
3 8
1
52
1
5 6
1 2 a 2 b , then a. b is equal to
(a) 30
(b) 45
(c) 8
(d) 16
7. If a, b, c are real and distinct numbers, then the value of
(a − b)3 + (b − c)3 + (c − a )3
is
(a − b)(b − c)(c − a )
(a) 1
(b) a b c
(c) 2
(d) 3
8. The remainder obtained when the polynomial
1 + x + x 3 + x 9 + x 27 + x81 + x 243 is divided by x – 1 is
(a) 3
(b) 5
(c) 7
(d) 11
9.
1
1
1
+
+
1 + log b a + log b c 1 + log c a + log c b 1 + log a b + log a c
has the value equal to
(a) abc
(c) 0
1
(b)
abc
(d) 1
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W Basic Mathematics and Logarithm
10. log 7 log 7 7( 7 7 ) =
(a) 3 log2 7
(b) 1 – 3 log37
(c) 1 – 3log72
(d) 1 – 10 log27
LOGARITHM AND ITS PRINCIPLE
PROPERTIES
11.
1
log
bc
abc
+
1
log
ca
abc
+
(a) 1/2
(c) 2
1
log
ab
abc
has the value equal to
(b) 1
(d) 4
1
. Then the value of 1000 x is equal to
12. If logx log18 ( 2 + 8) =
3
(a) 8
(b) 1/8
(c) 1/125
(d) 125
13. Number of real solutions of the equation log10 ( − x ) = log10
x 2 is :
(a) none
(c) exactly 2
(b) exactly 1
(d) 4
14. Greatest integer less than or equal to the number log2 15.log1/6
2.log3 1/6 is
(a) 4
(b) 3
(c) 2
(d) 1
log
a
2
3
− 3log27 ( a +1) − 2a
simplifies to
15. The ratio
7 4 log49 a − a − 1
(b) a2 + a – 1
(a) a2 – a – 1
2
(d) a2 + a + 1
(c) a – a + 1
2
21/4
16. If 32 log3 x – 2x – 3 = 0, then the number of values of ‘x’
satisfying the equation is
(a) zero
(b) 1
(c) 2
(d) More than 2
17. Anti logarithm of 0.75 to the base 16 has the value equal to
(a) 4
(b) 6
(c) 8
(d) 12
18. The sum of all the solutions to the equation 2logx – log
(2x – 75) = 2:
(a) 30
(b) 350
(c) 75
(d) 200
INEQUALITIES
2
19. If the solution set of the inequality log 0.9 log 5 ( x + 5 + x )
> 0 contains ‘n’ integral values, then n equals to
(a) 7
(b) 8
(c) 6
(d) 10
15
20. If log0.5 log5 (x2 – 4) > log0.51, then ‘x’ lies in the interval
(a)
(c)
( −3, − 5 ) ∪ (
( 5, 3 5 )
5, 3)
(b)
( −3, − 5 ) ∪ (
5, 3 5 )
(d) φ
(a) exactly four
(x2
21. Solution set of the inequality 2 − log2
+ 3x) ≥ 0 is:
(a) [− 4, 1]
(b) [− 4, − 3) ∪ (0, 1]
(c) (− ∞, − 3) ∪ (1, ∞)
(d) (− ∞, − 4) ∪ [1, ∞)
MODULUS FUNCTION
(c) x = −
5 2
(b) x = − ,
2 5
11 13
,
7 7
(c) exactly two
27. Simplify: 7
log3 5
(a) 0
3 7
(d) x = − ,
7 5
+3
23. If |x2 – 2x – 8| + |x2 + x – 2| = 3 |x + 2|, then the set of all real
values of x is
(a) [1, 4] ∪ {–2}
(b) [1, 4]
(c) [–2, 1] ∪ [4,∞)
(d) (–∞, –2] ∪ [1, 4]
24. The complete set of real ‘x’ satisfying ||x – 1| – 1| ≤ 1 is:
(a) [0, 2]
(b) [− 1, 3]
(c) [− 1, 1]
(d) [1, 3]
25. The number of real roots of the equation |x|2 – 3|x| + 2 = 0 is
(a) 1
(b) 2
(c) 3
(d) 4
log3 7
−5
(b) 1
− 7 log5 3
(c) 3
(d) 4
3x 4 + x 2 − 2 x − 3 5 x 4 + 2 x 2 − 7 x + 3
=
3x 4 − x 2 + 2 x + 3 5 x 4 − 2 x 2 + 7 x − 3
(b) x = 4, 1
(c) x = 3, 8
(d) x = 1, 5
(a)
=
b,by
1
2
−10 x + 3
(d) exactly one
log5 7
(a) x = 5, 2
x
29. =
If a
2
(b) exactly three
28. Solve the equation
22. Solutions of |4x + 3| + |3x – 4| = 12 are
7 3
(a) x = − ,
3 7
26. Number of real solution (x) of the equation | x − 3 |3 x
= 1 is
3
(b)
c and
=
cz
1
3
a then the value of xyz is
(c)
1
6
(d)
1
12
log a log b log c
, then aa . bb . cc =
30. If = =
b−c c−a a −b
(a) 3
(b) 1
(c) 4
(d) 2
31. The number of prime numbers satisfying the inequality
x2 − 1
< 3 is
2x + 5
(a) 1
(b) 2
(c) 3
(d) 4
Exercise-2 (Learning Plus)
3
(
x x
3
1. If x = x. x
(a) 1
(c) 0
) , then x =
x
1
.2
..
9998
6. The value of log 2 .log 3 ....log100 100
(a) 0
(b) 1
(c) 2
(d) 100!
(b) –1
(d) 2
is
2. Logarithm of 32 5 4 to the base 2 2 is
(a) 3.6
(b) 5
(c) 5.6
(d) 10
7. The number of solution of log2(x + 5) = 6 – x is
(a) 2
(b) 0
(c) 3
(d) 1
3. If x = loga(bc), y = logb(ca), z = logc(ab), then which of the
following is equal to 1
(a) x + y + z
(b) (1 + x)–1 + (1 + y)–1 + (1 + z)–1
(c) xyz
(d) x + y – z
8. If log102 = 0.30103, log103 = 0.47712, the number of digits
in 312 × 28 is
(a) 7
(b) 8
(c) 9
(d) 10
4. The solution of the equation log7 log5
(a) x = 2
(c) x = 4
5. The value of ( 0.05 )log
1
(a) 81
(b)
81
16
(
)
x2 + 5 + x =
0.
(b) x = 3
(d) x = – 2
20
( 0.1+ 0.01+ 0.001+........)
(c) 20
9. Exhaustive set of values of x satisfying log|x| (x2 + x + 1)
≥ 0 is
(a) (–1, 0)
(b) (–∞, –1) ∪ (1, ∞)
(c) (–∞, ∞) – {–1, 0, 1} (d) (–∞, –1) ∪ (–1, 0) ∪ (1, ∞)
10. The set of real values of x satisfying
log1/ 2 ( x 2 − 6 x + 12 ) ≥ −2 is
is
(d)
1
20
(a)
(c)
( −∞, 2]
[ 4, +∞ )
(b) [2, 4]
(d) [3, 8]
Dropper JEE Module-1 Mathematics PW
11. If log0.04(x – 1) ≥ log0.2(x – 1) then x belongs to the interval
(a) (1, 2]
(b) (–∞, 2]
(c) [2, + ∞)
(d) [2, 2]
12. The set of real value(s) of p for which the equation
| 2x + 3 | + | 2x – 3 | = px + 6 has more than two solutions is :
(a) [0, 4)
(b) (– 4, 4)
(c) R – {4, – 4, 0}
(d) {0}
13. Let 3a = 4, 4b = 5, 5c = 6, 6d = 7, 7e = 8 and 8f = 9. The value
of the product (abcdef) is :
(a) 1
(b) 2
6
(c)
(d) 3
14. There are two positive solutions to the equation log2x2 +
3
log4 2x = − . The product of these two solution is:
2
(a)
1
32
(b)
1
8
(c)
1
2
(d)
1
21
15. Number of real value of x satisfying the equation
log 2 (2 x 2 2 ) (a) 0
x2 1
x 2
2
(b) 1
(d) 3
16. Number of values of x satisfying the equation
x 4 | x |log 2 ( x
2
12 )
175
12
(c)
353
24
349
24
112
(d)
3
(b)
22. Which is the correct order for a given number a, a > 1
(a) log2 a < log3 a < loge a > log10 a
(b) log10 a < log3 a < loge a > log2 a
(c) log10 a < loge a < log2 a > log3 a
(d) log3 a < loge a < log2 a > log10 a
23. The smallest integral value of x such that
is
(a) 400
(c) 401
24. The ratio
2
log
(a) 2
(b) 3
(c) 4
(d) 5
(a) a2 – a – 1
(c) a2 – a + 1
25. 10 p q
(a) rq/r
(c) 1
(a) 15
(b) 16
(c) 17
(d) 18
a
2
3
(b) a2 + a – 1
(d) a2 + a + 1
= 1 and logq (logr (logp x)) = 0 then ‘p’ equals
(b) rq
(d) rr/q
26. Which one of the following is the smallest?
 1 
(c) 

 log10 π 
17. The number of zeros after decimal before the start of any
significant digit in the number N = (0.15)20 are :
1
10
− 3log27 ( a +1) − 2a
simplifies to
7 4 log49 a − a − 1
21/4
(a) log10p
is:
x2 x2 (b) 20
(d) 399
log (log (log r x ))
is :
(c) 2
(a)
(b)
3
log10 π2


1

(d) 
 log10 π 
27. The set of all solutions of the inequality (1/ 2) x
contains the set
(a) (–∞, 0)
(b) (–∞, 1)
(c) (1, ∞)
(d) (3, ∞)
2
−2 x
< 1/ 4
18. If n, k ∈ N, then the smallest value of k such that
k + 2k + 3k ... 24k = n3 is
(a) 100
(b) 90
(c) 120
(d) 60
28. The value of b satisfying thee quation, loge2 ∙ logb625
= log1016 ∙ loge10 is
(a) 5
(b) 7
(c) 9
(d) 10
a b c
6 and
b c a
29. Consider the quadratic equation, (log108)x2 - (log105)x
= 2(log210)−1 − x. Which of the following quantities are
irrational.
(a) Sum of the roots
(b) Product of the roots
(c) Sum of the coefficients (d) Discriminant
19. If a, b, c ∈ R and a, b, c ≠ 0 such that
b c a
a 3 b3 c 3
8 then 3 3 3 3 is equal to
a b c
b
c
a
(a) 81
(b) 48
(c) 72
(d) 84
20.
If 3x = 5y, then how many ordered pairs (x, y), x, y ∈ R satisfy
the given equation?
(a) 0
(b) 1
(c) 2
(d) Infinite
21. The value of x + y + z satisfying the system of equations
log2 x + log4 y + log4 z = 2 is
log3 y + log9 z + log9 x = 2
log4 z + log16 x + log16 y = 2
P
W Basic Mathematics and Logarithm
30. Which of the following statements are true?
(a) log23 < log1210
(b) log65 < log78
(c) log326 > log29
(d) log1615 > log1011 > log76
31. If x + y = a and x2 + y2 = b, then the value of (x3 + y3) is
(a) ab
(b) a2 + b
(c) a + b2
(d)
3ab − a 3
2
17
32. If x + y + z = 0, then a factor of the expression (x + y)3 +
(y + z)3 + (z + x)3 is
(a) 3(x + y)(y + z)(z + x) (b) 3xyz
(c) (x + y – z)
(d) (x – y + z)
39. Match the columns:
Column-I
33. The number of real solution/s of the equation 9log3(logex)
= loge x – (loge x)2 + 1 is:
(a) 0
(b) 1
(c) 2
(d) 3
34. The set of all the solutions of the inequality log1–x (x – 2)
≥ –1 is
(a) (–∞, 0)
(b) (2, ∞)
(c) (–∞, 1)
(d) f
x2 − 1
≥ 0 & x 2 − 5 x + 2 ≤ 0 is:
35. The complete solution of
x+3
 5 − 17 5 + 17 
 5 − 17 
,
(a) x ∈ 
 (b) x ∈ 1,

2 
2 
 2

(c) x ∈ (−3, −1]
Column-I
A.
25
5 x − 24 =
5x
(2x+1)(5x)
B.
= 200
2/x
1/x
C. 4 – 5(4 ) + 4 = 0
D. 22x+1 – 33(2x–1) + 4 = 0
E. 2 x −1 ⋅ 4 x +1
= 16
8 x −1
Column-II
p. –3
–2
–1
0
1
u. 2
F. 32x+1 + 10 (3x) + 3 = 0
G. 64(9x) – 84(12x) + 27(16x) = 0 v. 3
H. 52x – 7x – 52x(35) + 7x(35) = 0 w. None
(a) A → u, B → u, C → t, D → q, v, C → p,q,r,s,t,u,v
F → w, G → t,u, H → s
(b) A → q, B → t, C → u, D → v, C →t,u,v F → w,
G → t,H → q
(c) A → p, B → u, C → s, D → q, C → p,q, F → s, G → u,
H→q
(d) A → q, B → s, C → r, D → s, C →q,r, F → q, G → u,
H→q
18
3
B.
log 4/3 (1.3) + 3
q.
5
C.
log 2 − 3 (2 + 3) + 6
r.
4
D. logtan 20° tan 70° + 4
s.
2
E. log
cot 40° tan 50°
t.
0
F. log0.125 (8) + 8
u.
–1
G.
log1.5 (0.6) + 9
v.
8
H.
log 2.25 (0.4)
w. 7
I.
log10 (0.9)
x.
1
(c) A → s, B → v, C → t, D → p, E → t, F → u, G → w,
H → x, I → w
(d) A → q, B → s, C → r, D → v, E → u, F → v, G → v, H
→ w, I → x
40. Match the following columns.
Column-I
A.
If a = 3
(
Column-II
)
8+ 2 7 − 8−2 7 ,
p. −1
=
b
(42)(30) + 36 then the value of
logab is equal to
B.
q.
r.
s.
t.
p.
(b) A → s, B → r, C → q, D → p, E → t, F → u, G → v,
H → w, I → x
36. The number of the integral solutions of x2 + 9 < (x + 3)2
< 8x + 25 is:
(a) 1
(b) 3
(c) 4
(d) 5
38. Match the values of x given in Column-II satisfying the
exponential equation given in Column-I (Do not verify).
Remember that for a > 0, the terms ax is always greater than
zero ∀ x ∈ R.
A. logsin 30° (cos 60°) + 1
(a) A → q, B → p, C → s, D → v, E → u, F → u, G → q,
H → w, I → x
(d) x ∈ (−3, −1] [1, ∞)
37. Number of non-negative integral values of x satisfying the
2
1
2x −1
−
−
≥ 0 is
inequality 2
x − x + 1 x + 1 x3 + 1
(a) 0
(b) 1
(c) 2
(d) 3
Column-II
If a =
=
b
4+2 3 − 4−2 3,
q. 1
(42)(30) + 36 then the value of
logab is equal to
C.
D.
3 2 2 , b =−
3 2 2 then the
If a =+
value of logab is equal to
If a = 7 + 7 2 − 1 , b = 7 − 7 2 − 1 ,
then the value of logab is equal to
(a)
(b)
(c)
(d)
r.
2
s.
2 + 2log23
A → s, B → p, C → q, D → p
A → r, B → p, C → r, D → p
A → r, B → s, C → p, D → p
A → p, B → q, C → p, D → r
41. If log2 (log3 (log4 (x))) = 0 and log3 (log4 (log2 (y))) = 0 and
log4 (log2 (log3 (z))) = 0 then find the sum of x, y and z is
Dropper JEE Module-1 Mathematics PW
42. Let log2 x + log4 x + log8 x = logk x for all x ∈ R+. If k = b a
where a, b ∈ N then find the smallest positive value of
(a + b).
43. Solve the following inequalities:
(i) logx (4x – 3) ≥ 2
1
(ii) log (3 x2 +1) 2 <
2
(iii) log x2 (2 + x) < 1
1 

44. Solve for x: log 4 + 1 + =
log 3 log ( (3)1/ x + 27 )
2
x


45. Simplify:
1
81log5 9 + 3
(i)
409
(ii) 5
(iii)
1
log1/5
2
10
3
log 6 3
+ log
1
2 + log(16)
2
2
2


log 25 7
(
7)
?
− (125)log25 6  =





(b + c)(c + a)
logb (logb N )
logb a
63. ( x 2 − 4) x 2 − 1 < 0
=?
64. Solve for x:
D = (log5 49)(log7 125)
=
Find P log  N  | N + A + D + 6 | − log 5 2,
 A− 
 10 
48. Solve for x:
(i) log2 (log1/2 (x)) < 2
(ii) log1/2 (log3 (x)) > 3
(iii) (log2 (x) – 1)(log3 (x) – 2) ≤ 0
(iv) (log2 (x) – 1)(log1/2 (x) – 2) ≤ 0
log10 ( x − 3)
1
=
2
log10 ( x − 21) 2
(ii) log2 (x + 1) > log3 (x + 1)
(iii) log1/2 (x – 1) > log1/3 (x – 1)
50. If a + b + c = 1, a2 + b2 + c2 = 9, a3 + b3 + c3 + 1, then find
1 1 1
value of + + .
a b c
51. a + b + c = 10 and ab + bc + ac = 20 then find the value of
a3 + b3 + c3 – 3abc
b)3
+ (b –
c)3
+ (c –
(i)
a)3
= 3(a – b)(b – c)
1
3
≤ | 2 x − 1| ≤
2
5
(iii) 2 ≤ | 4 − 5 x | ≤
10
3
1
(ii) − ≤ | 3 x − 4 | ≤ 2
3
(iv) 3 ≤ |x2 – 1| ≤ 8
65. Solve for x:
(i) |||x + 5| –3| –1| = 2
(ii) ||||x – 5| –7| –3| –2| = 1
66. Solve for x:
(i)
49. Solve for x:
53. The value of
(a 2 − b 2 )3 + (b 2 − c 2 )3 + (c 2 − a 2 )3
= (a + b)
(a − b)3 + (b − c)3 + (c − a )3
61. a


4
1


?
=

 + log1/ 2 

62. x > 1 − x
 7+ 3
 (10 + 2 21) 
47. If N = 7 log49 900 , A = 2log2 4 + 3log2 4 + 4log2 2 − 4 log 2 3,
52. Prove that (a –
(c – a)
−100
5
starts in   is equal to [Use: log102 = 0.3010]
4
57. Number of real solution of log5 [2 + log3 (x + 3)] = 0 is
58. Prove that a2 + b2 + c2 – ab – bc – ac is non-negative for
all real values of a, b and c. Also prove that equality holds
when a = b = c.
59. If 4A + 9B = 10C, where A = log164, B = log39 & C = logx83,
then find x.
60. Prove that
46. log3 |3 – 4x| > 2
(i)
55. If logx–3(2x – 3) is a meaningful quantity then find the interval
in which x must lie.
56. Number of cyphers after decimal before a significant figure
(iii)
(| x | − 1)
≤0
(| x | − 2)
(ii)
(| x | − 1)(| x | − 3)
≥0
| x |2 − 2 | x |
(| x |2 − 5 | x | + 6)
≤0
(4 − | x |2 )
(iv) (||x – 1| –2| – 3)(|x – 2| –3) ≥ 0
(v) (|||x – 1| –2| –1| –2)(|x – 2|) ≥ 0
67. Solve for x:
(i)
x−3
≤1
x +1
(iii) 1 +
3
>2
x
(ii)
x +1
( x + 1) 2
+ | x + 1| =
|x|
x
(iv) |2x – 1| + |4 – 2x| < 3
| x + 2|
5 5 5 5......... - is
54. If x =3 7 + 5 2 −
1
3
7+5 2
, then the value of x3 + 3x – 14
is equal to
P
W Basic Mathematics and Logarithm
 1  2 − | x|
>9
(v)  
3
68. Solve: ||x2 – 2x + 6| – |x + 6|| = |x2 – 3x|
69. Solve
5 x (sin x − π)
≥ 0.
( x 2 − 4)
19
Exercise-3 (JEE Advanced Level)
1. log10(log2 3) + log10(log3 4) + log10(log4 5) + ... + log10(log1023
1024) simplifies to
(a) a composite
(b) a prime number
(c) rational which is not an integer
(d) an integer
2. The number of positive integers not satisfying the inequality
log2(4x – 2.2x + 17) > 5.
(a) 2
(b) 3
(c) 4
(d) 1
2
3. If log 4 {log 3 {log 2 ( x − 2 x + a )}} is defined ∀ x ∈ R, then
the set of values of ‘a’ is
(a) [9, ∞)
(b) [10, ∞)
(c) [15, ∞)
(d) [2, ∞)
4. Find the
 x1/ x
log  1/( x+1)
x
(a) 1
value of x that satisfies the equation

1
=
 5050
(b) 10
(c) 100
(d) 1000
5. Number of positive solution which satisfy the equation log2x⋅
log4x log6x = log2x ⋅ log4x + log2x ⋅ log6x + log4x ⋅ log6x?
(a) 0
(b) 1
(c) 2
(d) infinite
log5 5log5 ( x )
6. The value of x satisfying the equation ( 5)log5 5
is
(a) 1
(b) 2
(c) 4
(d) 8
=2
7. If |x –
– 6 |x – 3| – 23 < 4, then
(a) x ∈ (–12, 6)
(b) 17 integers satisfy the inequality
(c) 11 non-negative integers satisfy the inequality
(d) 6 negative integers satisfy the inequality
8. The roots of the equations |x| = 49
(a) One positive number greater than 1 only
(b) Two real number
(c) Two irrational number
(d) One negative rational number
9. If x1, x2 are the solution of the inequality
(x – 4)2 + (x – 6)4 = 34, then
(a) x1 + x2 = 10
(b) x1x2 = 21
(c) |x1 – x2| = 2 2
(d) Both x1, and x2 are irrational
20
11. Indicate all correct alternatives, where base of the log
is 2.
The equation x(3/4) (log2x)² + log2x−(5/4) = 2 has:
(a) At least one real solution
(b) Exactly three real solutions
(c) Exactly one irrational solution
(d) Imaginary roots
12. Solution set of the inequality
2

x3 
32
4
( log 2 x ) −  log1 2  + 9 log 2  2  < 4 log1 2 x
8 
x 

(a, b) ∪ (c, d) then the correct statement is
(a) a = 2b and d = 2c
(b) b = 2a and d = 2c
(c) logcd = logba
(d) there are 4 integers in (c, d)
(
)
2
is
COMPREHENSION BASED QUESTIONS
Comprehension (Q. No. 13 to 14): Let α and β are the solutions
of the equation ( x )
= 5 where α ∈ I and β ∈ Q Then
[Use: log102 = 0.3010, log103 = 0.4771]
log5 x −1
13. The number of significant digits before decimal in (α)10 is
(a) 13
(b) 14
(c) 15
(d) None of these
3|2
1

+ log 1 27 + log343 81
2



7
10. Which of the following is true?
(a) (log102)2 + 1 > log104
(b) log1090 > log550
(c) log4 log3 log2 16 > log16 4
(d) 2(log10 3)2 –3(log10 2)2 > (log10 2) × (log10 3)
include
14. The value of (β)log25 9 is
1
(a)
3
1
(c)
5
(b) 5
(d) 9
MATCH THE COLUMN TYPE QUESTIONS
15. Match the Column:
Column-I
Column-II
A. The value(s) of x, which does not p.
2
satisfy the equation log22 (x2 – x) – 4
log2(x – 1) log2x = 1, is (are)
B. The value of x satisfying the equation q.
3
2log2 e
ln 5log5 7
log7 10log10 (8 x −3)
= 13, is
Dropper JEE Module-1 Mathematics PW
r.
 1
1 
+
 is
The number N= 
 log 2 π log 6 π 
less than
D. Let l = (log34 + log29)2 – (log34 – s.
4
C.
log29)2 and m = (0.8) (1 + 9log3 8 )
then (l + m) is divisible by
5
log 5
65
t.
(a)
(b)
(c)
(d)
6
A → r, t, s; B → q; C → r, s, q; D → q, r
A → q, r, s, t; B → p; C → q, r; D → r, s
A → q, r, s, t; B → p; C → q, r, s, t; D → q, r, s
A → t; B → s; C → q, t; D → r, s
16. Match the columns:
Column-I
A.
B.
3 2+2 3
If p =
If r =
3 5+ 3
5− 3
If t =
6)
p is
(1/ r ) is
15
then log 9 + 2
C.
D.
then log (5+ 2
3 2 −2 3
3+ 6
5 3 − 2 12 − 32 + 50
3 2
3+ 6
−
4 3
6+ 2
+
q.
2
r.
–1
s.
1
then
log 3 t 2 is
If k =
Column-II
p.
0
6
2+ 3
then loge (k + 1) is
(a)
(b)
(c)
(d)
A → t, B → s, C → r, D → q
A → s, B → r, C → q, D → p
A → r, B → p; C → r, D → s
A → t, B → q, C → s, D → r
NUMERICAL BASED QUESTIONS
17. Find the number of integral solution of the equation
log x ( x + | x − 2 |) = logx(5x – 6 + 5|x – 2|).
18. Positive numbers x, y and z satisfy xyz = 1081 and (log10x)
(log10yz) + (log10y)(log10z) = 468.
Find the value of
( log10 x ) + ( log10 y ) + ( log10 z )
2
2
2
19. If (x1, y1) and (x2, y2) are the solution of the system of
equation
log225(x) + log64(y) = 4
logx(225) – logy(64) = 1,
then find the value of log30(x1y1x2y2).
20.
Number of ordered pairs (a, b), a,b ∈N such that a2 = 4002
+ b2 is
P
W Basic Mathematics and Logarithm
21. Suppose n be an integer greater than 1. Let an =
1
.
log n 2002
Suppose b = a2 + a3 + a4 + a5 and c = a10 + a11 + a12 +
a13 + a14. Then find the value of (b – c).
22. If the product of alll solutions of the equation
(2009) x
= (2009)log x ( 2010) can be expressed in the lowest
2010
m
form as
then the value of (m − n) is
n
23. If complete solution set of inequality log1/2 (x + 5)2 > log1/2
p2 + q2 + r 2
(3x − 1)2 is (−∞, p) ∪ (q, r) ∪ (s, ∞) then find
s2
24. Find the real solutions to the system of equations
log10(2000xy) − log10x ⋅ log10y = 4
log10(2yz) − log10y ⋅ log10z = 1
and log10(zx) − log10z ⋅ log10x = 0
25. Solve the equation x 0.5 log
26. Solve for x
2
x ( x −x )
= 3log9 4.
(a) log2x+3(x2) < log2x+3(2x + 3)
(b) logx+3(x2 – x) < 1
27. If x and y satisfying both the equations
log3 x + log2 y = 2;
3x – 2y = 23
then sum of all the values of x and y is
28. Solve for x:
(i) logx (2) . log2x 2 = log4x 2
(ii) 5loga x + 5 x loga 5 =
3(a > 0)
(log
x)
+
4
(iii) x
= 32
(iv) logx+1 (x2 + x – 6)2 = 4
(v) | x − 1|(log x )
2
− log x 2
=
( x − 1)3
(vi) x + log10 (1 + 2x) = x ∙ log10 5 + log10 6
29. Find the product of the positive roots of the equation
2005( x)log2005 x = x 2 .
30. If a = log12 18 & b = log24 54 then find the value of ab
+ 5(a – b).
31. Find the number of real solution of (x2 + x – 2)3 + (2x2 – x – 1)3
= 27(x2 – 1)3
32. Solve: log 3 ( x + | =
x − 1|) log 9 (4 x − 3 + 4 | x − 1|)
33. Prove that: 2

 log a 4 ab + logb 4 ab − log a


4
b
+ logb
a
4
a 
⋅ log a b
b

 2
=  loga b
2
if b ≥ a > 1
if 1 < b < a
34. Find all the solutions of the equation | x − 1|(log x )
= |x – 1|3, where base of logarithm is 10.
2
− log x 2
21
1
35. Solve the following equations for x & y: log100 | x + y | =
2
log10 y – log10 |x| = log100 4.
36. Solve for x −1 ≤
1 + x2
≤1
2x
x − 5 − 9 − x > 1: x ∈ Z
39. Solve for x
40. If
( x − sin1)( x − sin 2)
≤0
37. Solve for x
( x − sin 3)( x − sin 4)
(| x | − 1)(| x | − 3)
≥0
(| x | −4)(| x | −7)
38. Solve for x
( x − 3)
−| x|
x
( x − 4) 2 (π − x)
− x (− x 2 + x − 1)(| x | −9)
satisfying the inequality is
< 0 then no. of integers x
Exercise-4 (Past Year Questions)
9. If a, b are the roots of the equation
JEE MAIN
1. The number of real roots of the equation 5 + |2x – 1| = 2x
(2x – 2) is
(2019)
x − (5 + 3
2
log3 5
−5
log5 3
) x + 3(3
1
2
(log3 5) 3
(log5 3) 3
−5
)
−1 =
0
(2019)
1
1
and ,
(2022)
(a) 3x2 – 20x – 12 = 0
(b) 3x2 – 10x – 4 = 0
2
(c) 3x – 10x + 2 = 0
(d) 3x2 – 20x + 16 = 0
3. Solution set of 3x(3x – 1) + 2 = |3x – 1| + |3x – 2| contains
(2020)
| x 3 | 1
0 and
10. Let S x [6, 3] {2, 2} :
| x | 2
2
T = {x ∈ Z : x – 7 |x| + 9 ≤ 0}. Then the number of elements
is S ∩ T is
(2022)
(a) 7
(b) 5
(c) 4
(d) 3
(a) 3
(b) 2
(c) 4
then the equation, whose roots are (d) 1
2. The sum of the solutions of the equation | x 2 | x ( x 4) 2 0, ( x 0) is equal to
(a) 9
(b) 12
(c) 4
(d) 10
(a) Singleton set
(b) Two elements
(c) At least four elements (d) Infinite elements
4. The product of the roots of the equation
9x2
(2020)
– 18|x| + 5 = 0, is :
25
25
5
5
(b)
(c)
(d)
9
81
9
27
5. The number of solutions of the equation log 4 (x – 1)
= log2(x – 3) is
(2021)
(a)
6. The number of the real roots of the equation (x + 1)2 + |x – 5|
27
=
is
(2021)
4
1
is equal to
(2021)
7. The value of 3 1
4
1
3
1
4
3 ...
(a) 1.5 + 3
(b) 2 + 3
(c) 3 + 2 3
(d) 4 + 3
8. The number of elements in the set
{x ∈ : (|x| – 3) |x + 4| = 6} is equal to
(a) 3
(b) 2
(c) 4
(d) 1
22
11. Let p and q be two real numbers such that p + q = 3 and p4
1 1
+ q4 = 369. Then p q
2
is equal to _________.
(2022)
JEE ADVANCED
12. Let (x0, y0) be the solution of the following equations
(2x)ln2 = (3y)ln3
3lnx = 2lny. Then x0 is
1
1
(a)
(b)
6
3
13. The value of
(2011)
1
(c)
2
(d) 6


1
1
1
 1

6
+
log
4
–
4
–
4
–
....
3

 is
3
2
3
2
3
2
3
2
2


(2012)
14. If 3x = 4x – 1 , then x =
(2021)
(a) 2 log 3 2
2 log 3 2 – 1
(b)
(2013)
2
1
(c)
(d) 2 log 2 3
2 – log 2 3
2 log 2 3 – 1
1 – log 4 3
1
1
15. The value of ( (log 2 9) 2 ) log2 (log2 9) × ( 7) log4 7 is ______
(2018)
Dropper JEE Module-1 Mathematics PW
ANSWER KEY
CONCEPT APPLICATION
1. (b)
2. (d)
11. [60]
12. [2ab]
6. x6 – y6
3. (c)
5
4
13. x = ± 4
7. p = 3/2, q = 1, r = 4/3 14. [6]
1 2
15.  ,  16. [–5]
5 5
8. (d)
9. (c)
17. [–1]
10. (b)
18. (i) (1, ∞) (ii) [1, ∞)
 2 17 
 3 19 
(iii) (0, 1) (iv) (0, 1] (v) (0, 1) (vi) (0, 1] (vii) (1, ∞) (viii) [1, ∞) (ix) (3, ∞) (x) [3/2, ∞) 19. (i) x ∈  ,  (ii) x ∈  , 
2 2 
 3 24 
12 − 4a
 1
16
(iii) x ∈ (4 , ∞) (iv) x ∈  0, 
21.
22. (d)
23. (i) {–1, 5} (ii) {6, –6, 7, –1} (iii) (14, –4, 0, 10, 2, 8)
3+ a
 2
24. (–6, 8) 25.(0, 2) ∪ (4, ∞) 26. x ≥ 1
27. (i) x ∈ (–∞, 0) ∪ (2, ∞) (ii) 1 < x < 3 (iii) x ∈ (–2, –1) ∪ (0, 1)
(iv) x ∈ (–∞, 0) ∪ (1, ∞) (v) –4/3 ≤ x ≤ 2 (vi) x ∈ φ
EXERCISE-1 (TOPICWISE)
1. (c)
2. (b)
3. (b)
4. (a)
5. (b)
6. (d)
7. (d)
8. (c)
9. (d)
10. (c)
11. (b)
12. (d)
13. (c)
14. (c)
15. (d)
16. (b)
17. (c)
18. (d)
19. (b)
20. (a)
21. (b)
22. (c)
23. (a)
24. (b)
25. (d)
26. (b)
27. (a)
28. (c)
29. (d)
30. (b)
31. (d)
EXERCISE-2 (LEARNING PLUS)
1. (a)
2. (a)
3. (b)
4. (c)
5. (a)
6. (b)
7. (d)
8. (c)
9. (d)
10. (b)
11. (c)
12. (d)
13. (b)
14. (a)
15. (a)
16. (d)
17. (b)
18. (b)
19. (c)
20. (d)
21. (c)
22. (b)
23. (c)
24. (d)
25. (a)
26. (c)
27. (c)
28. (a)
29. (d)
30. (b)
31. (d)
32. (a)
33. (b)
34. (d)
35. (b)
36. (d)
37. (d)
38. (a)
39. (b)
40. (c)
41. x = 64, y = 16, z = 9 42. 75
43. (i) x ≥ 3, (ii) |x| > 1 (iii) x ∈ (–2, 1) u (2, ∞)
45. (i) → 1 (ii) → 6 (iii) → 20
1

46.  −∞,  2

47. [2]
 1
48. (i) x ∈ 1, 38  ,




 1
(iv) x ∈  0,  ∪ [ 2, ∞ ) 49. (i) x ∈ {5}, (ii) x ∈ (0, ∞), (iii) x ∈ (1, 2) 50. [1]
 4
55. x ∈ (3, 4) ∪ (4, ∞)
56. [9]
 1 1  3 4
64. (i) x ∈  ,  ∪  , 
 5 4  4 5
57. [4]
2 
(ii) x ∈  , 2 
3 
44. (φ)
(ii) x ∈ (1, 31/8),
51. [4w]
53. [5]
(iii) x ∈ (2, 9),
54. [0]
 5 −1 
61. logbN 62. x ∈  2 ,1
63. x ∈ (–2, –1) ∪ (1, 2)


 2 2   6 22 
(iii) x ∈  ,  ∪  ,  (iv) x ∈ (–3, –2) ∪ (2, 3)
15 5   5 15 
59. [10]
65. (i) x ∈ {–11, –5, –1} (ii) x ∈ {–8, –6, –2, –4, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18}
66. (i) x ∈ (–2, –1] ∪ [1, 2) (ii) x ∈ (–∞, –3] ∪ (–2, –1] ∪ [3, ∞) ∪[1. 2) (iii) x ∈ (–∞, –3] ∪ [3, ∞)
(iv) x ∈ (–∞, –4] ∪ [–1, 5] ∪ [6, ∞) (v) x ∈ (–∞, –4] ∪ [6, ∞] ∪ [6, ∞) ∪ {2}
67. (i) x ∈ [1, ∞] (ii) x ∈ [0, ∞] ∪ {–1} (iii) x ∈ (–1, 0) ∪ (0, 3)
68. [–6, ∞)
(iv) x ∈ φ (v) x ∈ (2, 6)
69. x ∈ (–2, 2)
P
W Basic Mathematics and Logarithm
23
EXERCISE-3 (JEE ADVANCED LEVEL)
1. (d)
2. (a)
3. (b)
4. (c)
5. (c)
6. (c)
11. (a, b, c) 12. (b, c)
13. (b)
14. (a)
15. (c)
16. (b)
21. [–1]
23. [17/3] 24. [x = 1; y = 5; z = 1; or x = 100; y = 20; z = 100] 25. [2]

22. [1]
−3

26. (a)  x ∈  ,3  − {−1, 0}
 2 


27. [5]
7. (b, c)
17. [1]
8. (b, c)
9. (a,c,d) 10. (a,d)
18. [5625] 19. [12]
20. [0]
(b) [x ∈ (–3, –2) ∪ (–1, 0) ∪ (1, 3)]
28. (i) [2] (ii) [3/2] (iii) [2.22] (iv) [2.022] (v) [1]
29. [(2008)2] 30. [1]
32. [x ∈ [0, 1] ∪ {4}]
36. {1, –1} 37. [x ∈ (sin4, sin3) ∪ [sin1, sin2)]
39. x ∈ (8.323,9)
34. [1000 or 1/10]


 10 20 
35.  x ∈  ,  ; y ∈ (−10, 20) 
3
3




38. x ∈ (–∞, –7] ∪ (–4, –3] ∪ [–1, 1] ∪ [3, 4] ∪ [7, ∞]
40. (infinite)
EXERCISE-4 (PAST YEAR QUESTIONS)
JEE Main
1. (d)
2. (d)
3. (a)
4. (b)
5. [1]
6. [2]
7. (a)
8. (b)
9. (b)
10. (d)
11. [4]
JEE Advanced
12. (a)
24
13. (d)
14. (a,b,c) 15. [8]
Dropper JEE Module-1 Mathematics PW