Example 5: Trick:

The domain of the function 1  x2 1

y  2
 1
x x2
 
f x  x  x 2  4  x  4  x is
1
(A) [-4, ¥) (B) [-4, 4] Now, since, is always > 0
x2
(C) [0, 4] (D) [0, 1] Þ y>1
Solution: (D) Þ y Î (1, ¥)
Therefore, option (B) is correct.
We know that, f x   x  x2  4  x  4  x
Example 7:
Clearly, f(x) is defined if
Find the range of the function
4 + x ≥ 0 Þ x ≥ –4 x 2  11x  31
4–x≥0Þx£4  
f x  2
x  11x  30
, where x Î R.

x(1 – x) ≥ 0 Þ x ≥ 0 or x £ 1
Solution: [1, 7/3)
Therefore, domain of
f = (–¥, 4] Ç [–4, ¥] Ç [0, 1] = [0, 1] x 2  11x  31
Here, f x   x 2  11x  30
Hence, 4th option is correct.

Co-Domain and Range of a Function  

f x  1
1
x  11x  31
2
 1
1
2
 11  3
If a function f is defined from a set of A to x   
set B, then for f: A ® B. Set A is called the  2 4

domain of function f and set B is called the 7 −11

Now, the maximum value of f(x) is at x =
co-domain of function f. The set of all f- 3 2
images of the elements of A is called the And the minimum value of f(x) is 1 at x = ¥
range of function f. In other words, we can Therefore, the range of f(x) is [1, 7/3)
say, domain = all possible values of x for
Note: For two sets A and B, the number of
which f(x) exists.
functions from A to B are |B| |A|. For exam-
Range = for all values of x, all possible values ple, let A = {1, 2, 3, 4} and B = {p, q, r, s, t}.
of f(x) or the set of all the outputs is known Therefore, the number of functions = BA = 54
as range. = 625 functions from A to B.
Example 6:
Types of Functions
1 + x2
Find the range of . One–one function (injective)
x2
(A) (0, 1) (B) (1, ¥) A function f: A ® B is said to be a one–one
function if different elements of A have dif-
(C) [0, 1] (D) [1, ¥)
ferent images in B.
Solution: (B)
Number of one–one functions from A to
1 + x2 B having m and n elements respectively
Let y =
x2
1 =
Þ x2y = 1 + x2 Þ x2(y – 1) = 1 Þ x2 =
y −1 To check the injectivity of a function:
Now, x2 ≥ 0
1 1. Take two arbitrary elements x1 and x2 in
Þ ≥ 0 Þ (y – 1) ≥ 0 Þ y > 1 as y cannot the domain of f.
y −1
be 1 2. Check whether f(x1) = f(x2).

Functions and Graphs 31

 Example 18: Put x = 4, y = 1
If f(x + y) = f(x) + f(y) + f(xy) and f(1) = 1, then f(5) = 2f(4) + f(1)
find the value of: = 2[15 f(1)] + f(1)
f(2) + f(3) + ∙∙∙ + f(11). = 31 f(1)
Solution: 4,082 = 31 × 1 = 31
It is given that, f(x + y) = f(x) + f(y) + f(xy) Now we can find the pattern:
Put x = y = 1 3, 7, 15, 31,…,
f(1 + 1) = f(1) + f(1) + f(1) 2
2 – 1, 3
2 – 1, 4
2 – 1, 211 – 1,
f(2) = 3f(1) = 3 × 1 = 3 Now required sum:
Now, put x = 2, y = 1 f(2) + f(3) + ∙∙∙ +, f(11)
f(3) = f(2) + f(1) + f(2) (22 – 1) + (23 – 1) + (24 – 1) + ∙∙∙ + (211 – 1)
= 2 f(2) + f(1) = (22 + 23 + ∙∙∙ + 211) – 10
= 2[3f(1)] + f(1) = (20 + 21 + 22 + 23 + ∙∙∙ + 211) – (20 + 21) – 10
= 7 f(1) = 7 × 1 = 7 = 212 – 1 – 3 – 10
Putting, x = 3, y = 1 = 4,096 – 14
f(4) = f(3) + f(1) + f(3) = 4,082.
= 2(f3) + f(1)
= 2(7 f(1)) + f(1)
= 15 f(1)
= 15 × 1 = 15.

44 Functions and Graphs

 Practice Exercise – 1

Level of Difficulty – 1 Level of Difficulty – 2

1 n 
1. If ( ) =
2
, ¹ 2, then find 6. Let f n      , where [ ] denotes
2  2 100 

     
f f f f f 3 an integral part of . Then, the value of
100

15   
f n is:
n 1
( )
11 (A) 50
−37 (B) 51
(B) (C) 1
7
(D) None of these
(C) 3
(D) 1 7. ( + ) = ( ) ( ) for all , , (4) = +5.
What is (-12)?
2. The domain of the function ( ) = log (A) 125
(5 – 6 – 2) is: 1
(B)
( ) (5, 6) 125
(B) (2, 3) (C) 1728
(C) (2, ¥) 1
(D)
(D) None of these 1728
3. If ( ) = 3 + 1, and the value of the e 8. Let ( ) be a function satisf ing ( ) ( ) =
pression (2 ) – 2 ( ) = 4373, then value ( ) for real , . If (2) = 4, then what is
of x is:  1
value of f   ?
( ) 8 2
(B) 3 9. Let ( ) be the greatest integer func-
(C) 2 tion and ( ) be the modulus func-
(D) 9 tion, then what is the value of
 5   5 
4. The domain of the function 
gof   
  fog  ?
 3   3 
 
f x  4 x  4  2 x  10 is:
(A)
3
5
(A) [1, 5]
11
(B) [-1, 5] (B)
(C) [5, ¥] 3
(D) [–1, ¥] 2
(C)
7
5. If 3 2 – 2 – 2
+ 4, then [ (2, 3), (−1, 1)]
(D) 3
is equal to:
(A) -68 10. Let ( ) = 3 – 4 + , and (0) and (1) are
(B) 95 of opposite signs, then which of the fol-
(C) 251 lowing is necessarily true?
(D) 232 (A) -1 < < 2
(B) 0 < < 3
(C) -2 < < 1
(D) -3 < < 0

Functions and Graphs 45

 Level of Difficulty – 3 14. Find (13) where ( + 2) = 2 ( ) – ( + 1)
and (10) = 20, (15) = 108.
2
11. For a function ( ), ( ) + ( – 1) = and (A) 24
(20) = 40. Find (40). (B) 28
(C) 44
12. The operation ( ) is defned by, (i) (1) =
(D) Cannot be determined
2, (ii) ( + ) = ( ) ( ), for all positive
n
   2
 ; if
1
integers and , if  x 1
f x  1022 then, 15. Let ( ) =
 2

1
¹ 1, –1
is:
and 1 if = 1, –1.
(A) 8
(B) 9 x1
Let ( ) = ; if ¹ 1
(C) 10 x1
(D) 11 and 3, if = 1. What are the minimum

13. The domain of the function ( ) = log7 possible values of

f x
?
 
(log3 (log5 (20 – 2 – 91))) is: 1 g x  
(A)
(A) (7, 13) 2
(B) (8, 12) (B) –1
(C) (7, 12)
1
(D) (12, 13) (C)
3
1
(D)
4

46 Functions and Graphs

  1 n
 
f  f 2  f 1  Also, a sum of terms, S =
2
[ + ]=
2 10
f 
 1 f 1   1 1
f 
2
[43 + 79] = 5 × 122 = 610

2 f 2   2 4 Taking equation (i)

9. (D) (40) = 610 + (20)
If it is given that ( ) = [ ] and ( ) = | | (40) = 610 + 40
(40) = 650
 5 
Then,   = [-1.66] = (-2) = 2
 3  12. (B)
 5  Since ( + ) = ( ) × ( ) and (1) = 2
fog   
  fog  1.66  f 1.66  1 Putting = = 1
 3 
Now, (1 + 1) = (1) × (1)
 5   5  Þ (2) = 2 × 2 = 4
Then, gof    fog    2  1  3.
 3   3  If = 2, = 1
Hence, (D) option is correct. Þ (3) = (2) × (1) = 4 × 2 = 8
Similarly, (4) = (3) (1) = 8 × 2 = 16
10. (B) The pattern followed: ( ) = 2
(0) = n
Now,  x 1
 
f x  1022 = (1) + (2) + (3)
(1) = 1 – 4 + = –3
+ ∙∙∙ + ( ) = 1022
Since they are of opposite sign, ( – 3) < 0
Þ 21 + 22 + 23 + ∙∙∙ + 2 = 1022
Þ 0< <3 The series is a GP with frst term, = 2
Hence, (B) option is correct. and a common ratio = 2

11. 650 Þ Sum of GP =


a rn  1 
Here, ( ) = 2 – ( – 1)  r  1
(40) = 402 – (39)
= 402 – [392 – (38)] Therefore,

2 2n  1   1022
= 402 – 392 + (38) 2  1
= 402 – 392 + 382 – (37) 1022
= 402 – 392 + 382 – 372 + (36) 2 –1= = 511
2
2 = 512
= 402 – 392 + 382 – 372 + 362 + 2 = 29
352 + ∙∙∙ + (20) On comparing both the sides, we get
= (40 + 39) (40 – 39) + (38 + or, = 9
37) + (38 – 37) + (36 + 35) Hence, (B) option is correct.
(36 – 35) + ∙∙∙ + (22 + 21) (22 –
13. (B)
21) + (20)
A logarithm function of the form log is
= [79 + 75 + 71 + ∙∙∙ + 43] + (20)
true,
= [43 + 47 + ∙∙∙ + 71 + 75 + 79) +
if > 0 and = is written as:
(20)
=
(40) = s + (20) ...(i)
For s: = 79; = 43; =4 Since ( ) = log7 (log3 (log5 (20 – 2 – 91)))
Since we know that Þ log3 (log5 (20 – 2 – 91)) > 0
= + ( – 1) Þ log5 (20 – 2 – 91) > 30
79 = 43 + ( – 1) 4 Þ log5 (20 – 2 – 91) > 1
36 Þ 20 – 2 – 91 > 51
= –1 or = 10
4

48 Functions and Graphs

 Þ 2 – 20 + 96 < 0 This function is greater than 0.
Þ ( – 8) ( – 12) < 0
Þ 8 < < 12 Let =
x 2
1
2
Therefore, domain of ( ) = (8, 12). ( x  1)
Hence, (B) option is correct. Þ 2
( – 1) + 2 + ( – 1) = 0 which is quad-
14. (C) ratic in .
( + 2) = 2 ( ) – ( + 1) Discriminant should be greater than 0
Let (11) = 4 2
– 4( – 1)2 ≥ 0
Þ (12) = 2 (10) – (11) = 40 – 1
Also, (13) = 2 (11) – (12) = 2 – 40 + Þ ≥
2
Þ (13) = 3 – 40
(14) = 2 (12) – (13) When = 1, then
 1
f x
= 80 – 2 – 3 + 40 = 120 – 5 g x 3
Now, (15) = 2 (13) – (14) = 6 – 80 – 120 + 5 Therefore, either the value should be greater
Þ 108 = 11 – 200 1 1
Þ 11 = 308 than or equal to .
2 3
Þ = 28
Hence, (D) option is correct.
Therefore, (13) = 3 – 40 = 84 – 40 = 44.

15. (D)

   x
f x
2
1    x 1   x  1
2

g x x 2
 1  x  1 ( x  1)
2

Functions and Graphs 49

 Practice Exercise – 2

Level of Difficulty – 1 6. A function ( ) is an even function, if

(- ) = ( ), where ∈ R.
5x  2 A function ( ) is an odd function, If
1.  
If f x 
3x  5
and ( ) = 16 – 15, then
( ) = – ( ), where ∈ R.
the value of ( ( (5))) is: If ( ) = 4 2 – 5 , then ( ) + (- ) is:
(A) 33 (A) Always odd
(B) 49 (B) Always even
(C) 65 (C) Always equal to 1
(D) 81 (D) Neither odd nor even
2. If ( ) = + and [ [ ( )]] = 64 + 147, 7. A unction is defned as ( ) = sin
the value of 2 + 3 is: (sin – 4) + 3 for all the real values of .
(A) 29 Find the ma imum value of ( ).
(B) 35
8. ( ) ( ) = ( ) + ( ) + ( ) – 2 where and
(C) 24
are positive real numbers. If (3) = 10
(D) 31
and (2) = 5, fnd (12).
3. For a positive number , [ ] = the largest
9. Given that ( ) = 1, (2 ) = 4 ( ) + 5 and
integer less than and { } = the smallest
( + 2) = ( ) + 3 + 5 for all real values
integer greater than , what is the value
of . What is the value of (20)?
of [7.2] + {8.3} – [4] – [6] + {8.7} – {9.5}
– [8.1]? 10. Find the range of value of for which [ ]
(A) –3 + [2 ] + [3 ] = 8, where is a real number
(B) –7 and [ ] is the greatest integer less than
(C) –5 or equal to .
(D) –1 4
(A) 1 ≤ ≤
3
4. Let ( ) be a function satisf ing ( ) ( ) =
5
( ) for all real values of and . If (3) = (B) <
3
 1 3 5
6, then what is the value of f   ? (C)  
(A) 1 3 2 3
1 (D) None of these
(B)
6 Level of Difficulty – 2
1
(C) 1 1
2  
11. If f x 
x
 
and x  x  , then which
x
(D) 6
of the following is true?
5. For a positive integer , ( ) is such that
( + ) = ( × ), where is an integer
     g f  x    1
(A) f g x

and (1) = 6. If the value of (1,007) = , (B) f  g  x    g  f  x    1

the value of will be.
f  g  x 
(A) 6 (C) 1
(B) 5 g f  x  
(C) 4
f  g  x 
(D) 3 (D)  1
g f  x  

50 Functions and Graphs

12. For how many integral values of the 18. Which of the following function is an
16  x 2 even function?
function f x     x 
is defned? æ1+ ö 5
(A) log ç ÷
log x 2   ç 5 ÷
 5  è1- ø

13. ( ) is function such that ( × ) =

 ,
f a
 1 3 
2

f  b (B)
3
( < ) for all positive real values of and 12

. Find (0.125), given that (8) = 0.5. (C) e

(D) All of these
(A) 0.125
(B) 0.25 19. Let ( ) be a fourth-degree polynomial,
(C) 0.50 the coecient o whose highest power
(D) None of these is 3. If (2) = 4, (3) = 9, (4) = 16 and (5)
1  x2 = 25, fnd ( ).
 
14. If f x 
1  x2
, then fnd the value o
(A) 3 4 + 14 3 + 72 2 + 154 + 120
 1  x2  (B) 3 4 + 42 3 + 214 2 + 462 + 360
f
 1  x 2 
. (C) 3 4 – 14 3 + 72 2 – 154 + 120
  (D) 3 4 – 42 3 + 214 2 – 462 + 360
(A) 2
20. The function ( ) denotes a parabola
x 2 + 2x
(B) which cuts the -axis at the points =
3+ x –15 and = –8. If ( ) = ( ), then fnd the
2 value of + , if > .
(C)
1 (A) 0
x2 + 2
x (B) –7
2 (C) 23
2
(D) 2
(D) –23
+1
(E) –11.5
15. The range of the function ( ) = | | – is:
(A) (-¥, ¥)
Level of Difficulty – 3
(B) [–2, 0]
(C) [0, 1] 21. ( ) is a function such that (2) = (5) =
(D) [0, ¥) (8) = 1 and (11) = 163. Find (3).
16. Find , when ( ) = 3 – 1 and ( ( (2 ))) 22. If ( 2
– )= 4
–2 3
– 4 – 5, then +5 2
= 365.
(A) 5 fnd the approximate value o
f f 10
.
  
(B) 7 f 10  
(C) 9 (A) 139
(D) None of these (B) 143
(C) 150
17. Let ( ) be a function such that ( + 1) (D) 172
+ ( – 1) = ( ) for every real . Then, for
what value of is the relation ( + ) 23. Find (13) where ( + 2) = 2 ( ) – ( + 1)
= ( ) necessarily true for every real ? where (10) = 20, (15) = 108.
(A) 3 (A) 24
(B) 4 (B) 28
(C) 2 (C) 44
(D) 6 (D) Cannot be determined

Functions and Graphs 51

 24. Find the sum  3 1    3 2    3 3    3 4      3 150  æ 1 ö
          27. If ( ) + (1 – ) = 8, fnd the value o f ç ÷
è 100 ø
Note: [ ] ® Represents the greatest in-
teger less than or equal to . æ 2 ö æ 3 ö æ 98 ö æ 99 ö
(A) 430 +fç ÷ +fç ÷ +... f ç ÷ +fç ÷
è 100 ø è 100 ø è 100 ø è 100 ø
(B) 130
(C) 510
.
(D) 530
(A) 400
1 x  3 x  x 3
 
25. If f x  log   and g x   
, (B) 392
1 x  1 3 x 2 (C) 396
then fnd the value o ( ) (D) 404
(A) 3f x   28. For two sets A and B, (A) = 6 and (B) =
(B) f  x  5. Find the number of functions that are
not onto that can be defned rom A to B.
2
(C)  f  x  29. A function ( ) exists for all the real val-
ues of such that (3 – ) = (3 + ) and
(D)  3f x   (10 – ) = (10 + ). If (3) = 0, then what
is the minimum number of roots for ( )
26. Consider the function ( ) = ( + 5) ( + 7)
= 0, if lies between –300 and 300 (both
( + 9) … ( + 97) ( + 99). The number of
inclusive)?
integers for which ( ) < 0 is:
(A) 24
(B) 23
30. log 1  2

 2  24  log 1 11. The range of
4 4
(C) 46 is:
(D) 48
(A) (-5, 7)
(B) (-4, 6)
(C) (-5, -4) È (6, 7)
(D) (-¥, -4) È (6, ¥)

52 Functions and Graphs

 Solutions

1. (C)  1
5x  2      
f 3  f 3 f 3 f  
We have, f x   
3x  5
3
1  1

5 5  2 27 f 

\ f 5  

3 5  5 10
 
f 3 3
 1 1
 27  Therefore, f   
5 2 3 6
  
f f 5   10 
 27 

155
31
5 Hence option (B) is the correct answer.
3 5
 10  5. (A)
\ ( ( (5))) = 16(5) – 15 = 65 Let = 1 and = 0
Thus, option (C) is correct. Then ( + 0) = (1 × 0)
(1) = (0)
2. (A) Since (1) = 6 is given in the question.
( )= + \ (1) = (0) = 6
[ ( )] = [ + ] If we consider = 1,007 and = 0
= ( + )+ Then, (1,007 + 0) = (1,007 × 0)
= 2 + + (1,007) = (0) = 6
[ ( ( ))] = ( 2 + + ) (1,007) = 6 =
2
= [ + + ]+ Thereore, it is confned that this is a
3 2
= + + + constant function for all values of .
Given, this is equal to 64 + 147 Hence option (A) is the correct answer.
3
Þ = 64 and 2 + + = 147
= 4 and ( 2 + + 1) = 147 6. (B)
147 Since, the given function is ( ) = 4 2
Þ = =7 –5
21
If we check for ( ) + (– ) = 4 2 – 5 +
\ 2 + 3 = 2(4) + 3(7) 4(– )2 – 5(– ) = 8 2
= 8 + 21 = 29. where 8 2 is always even.
Hence option (A) is the correct answer. Hence, (B) is the correct answer.
3. (D) 7. 8
[7.2] + {8.3} – [4] – [6] + {8.7} – {9.5} – [8.1]? The e pression ( ) can be written as:
= 7 + 9 – 3 – 5 + 9 – 10 –8 ( ) = sin² – 4 sin + 3 = sin² – 4 sin +
= –1 4–4+3
Hence, option (D) is the correct answer. = (sin – 2)2 – 1
4. (B) To ma imize the value of ( ), the value
We have ( ) = ( ) ( ) of sin will be –1.
Also, (3) = 6 Hence, the ma imum value of ( ) =
The above expression can be written as: (–1 – 2)² – 1 = 8.

 1  1 8. 148
 
f 9    f 9  f   ( )( )= ( )+ ( )+ ( )–2
 3 3
(3) (2) = (3) + (2) + (6) – 2
 1
  
f 3  f 33 f   10(5) = 10 + 5 + (6) – 2
3

Functions and Graphs 53

 Þ (6) = 50 – 15 + 2 = 37 Thus, the required option (B) is the correct
Put = 6, = 2 answer.
(6) (2) = (6) + (2) + (12) – 2
(12) = 37(5) – 37 – 5 + 2 = 145. 12. 4
16  x 2
9. 393 We have, f x     x 
(1) = 1 log x 2  
Put = 1 in ( + 2) = ( ) + 3 + 5  5
Þ (1 + 2) = (1) + 3 + 5 or (3) = 1 + 3 + 5 = 9 We know that the base of a log cannot
Put = 3 in ( + 2) = ( ) + 3 + 5 be 1. The value inside the square should
Þ (3 + 2) = (3) + 3 × 3 + 5 be positive. Also, the argument of the log
or (5) = 9 + 9 + 5 = 23 should be positive.
Put = 5 in (2 ) = 4 ( ) + 5
The argument is always positive.
(2 ) = 4 ( ) + 5 5
or (10) = 4 (5) + 5 = 97 So, ¹0
Put = 10 in (2 ) = 4 ( ) + 5 x
(20) = 4 (10) + 5 = 4 × 97 + 5 = 393. Also, it cannot be one as the denom-
5
10. (C) inator would become 0.
By observing we can fnd that > 1 and < 2. Hence, ¹ 5 and ¹ –5
Else the RHS ≠ + 8. Also, 16 – 2 ≥ 0 Þ –4 £ £ 4
So, the combinations are [ ] = 1, [2 ] = 2 Now, + 2 > 0 Þ > –2
or 3, and [3 ] = 4 or 5. Also, + 2 ¹ 1 Þ ¹ –1
The combinations that give RHS = 8 are 1 Hence, Î (–2, –1) È (–1, 0) È (0, 4]
+ 2 + 5 or 1 + 3 + 4. The integral values of are 1, 2, 3, 4.
For any value of , the case of ‘1 + 2 + 5’ is Hence, there are four integral values of
not possible. Hence, it has to be the case of for which ( ) is defned.
3
‘1 + 3 + 4’. Which will occur when, ≥ 13. (A)
2
5   ;a  b
f a
and < .
3

f ab   f  b
3 5
Hencethesolutionis  £ < . Now (8) = 0.5 (given)
2 3
11. (B)   ;(1  8)
f 1
1 
 f 1 8  
f 8
We have, f x   x
and
æ 1ö
( ) 1
( ( ))
g x = x - f g x = f ç x - ÷  =
x è xø
1
1  f 8 
f  1
x- f 8
x
= 2
x  f  1  f 8 .f 8
x - 1 (1) = 0.5 × 0.5 = 0.25
 1
  
And g f x  g  
x
Now, for (1/8):
æ 1ö
fç ÷
1 x 2  x 2  1 æ1 ö æ1 ö
Also, f ç ´ 8 ÷ = è ø ; ç < 8 ÷
1 1 1 8
  x  
 x 
=
x 1 x x   è8 ø f (8) è 8 ø
x
 1
 x 2  1 f 
     
\f g x g f x  2
x
 
 x 
 1 f 1   

8
x  1   f 8  
54 Functions and Graphs
  1 For ( ( (2 ))): ( ( (2 ))) = (18 – 4) = 3(18
  
 f    f 1  f 8 = 0.25 × 0.5 = 0.125 – 4) – 1 = 54 – 12 – 1 = 54 – 13
8
14. (C) Þ 54 – 13 = 365 Þ 54 = 378 Þ = 7
Hence, option (B) is the correct answer.
1  x2
Since f x    1  x2 2 17. (D)
 1  x2  Given that ( + 1) + ( − 1) = ( )
1
 1  x2   1  x 2  Putting ( + 1) at the place of gives us
Then, f   
 1  x 2 
=
1 x 
2 the following:
  2
1 
 1  x 2  ( + 2) + ( ) = ( + 1)
  Adding these two equations, we get
1 4 2 2 ( + 2) + ( − 1) = 0
1
1 4  2 2 Þ ( + 3) + ( ) = 0
=
1 4 2 2 Þ ( + 4) + ( + 1) = 0
1 Þ ( + 5) + ( + 2) = 0
1 4  2 2
Þ ( + 6) + ( + 3) = 0
1  x 4  2x 2  1  x 4  2x 2
Þ ( + 6) − ( ) = 0.
=
1  x 4
 2x 2  Hence, ( + 6) = ( ).
4 2
1  x  2x  1  x  2x 2 4 So, = 6
Hence, option (D) is the correct answer.
1  x 4
 2x 2 
18. (C)
4x2
=    
If f  x  f x , then it is an even
2x 4 + 2 function.
2
2 After verif ing the options, we can see
= 4
+1 option (C) is an even function.
2 æ 1+ x 5 ö
= 4
1 ( )
(A) f x = log ç
ç 1- x 5 ÷÷ ⇒
2
+ 2
è ø
æ 1 + ( - x )5 ö æ 1 - x5 ö
 1  x2 
So, f  
2 ( )
f - x = log ç
ç 1 - ( - x )5 ÷÷
= log çç 5 ÷ ( )
÷¹f x
 1  x 2  è ø è 1 + x ø
  x2  1
2
x2
Hence option (C) is the correct answer. (B) f x 
 1 3 
x

⇒
3x
15. (D) 2

When < 0 f x  

 1 3 
x

f x 
Then, ( ) = – – = –2 3 x
So, ( ) will be a positive number. 12
12
  x   ex 12  f x ,
Therefore, all the positive numbers come
 
(C) f x  e x ⇒ f  x  e    
in this range. is an even function.
Also, if ≥ 0 then ( ) = 0
19. (D)
\ The range of the function is [0, ¥).
Given (2) = 4, (3) = 9, (4) = 16 and (5) = 25
16. (B) Thus ( ) = 2
( )=3 –1 But given ( ) is a fourth-degree polyno-
Þ (2 ) = 3(2 ) – 1 = 6 – 1 mial with the coecient o highest pow-
Now, ( (2 )) = (6 – 1) = 3(6 – 1) – 1 = er as 3.
18 – 3 – 1 = 18 – 4 So, ( ) = 3( – 2) ( – 3) ( – 4) ( – 5) + 2

Functions and Graphs 55

 So total valid cases will be = 24.
From  3 8  to  3 26  , value of each term
   
is 2, total sum = 2 × 19 = 38 27. (C)
From 3
27  to 3
63  , value of each ( ) + (1 - ) = 8
   
term is 3, total sum = 3 × 37 = 111  1   1 
f   f 1  8
From  3 64  to  3 124  , value of each  100   100 
   
term is 4, total sum = 4 × 61 = 244  1   99 
f f 8
From  3 125  to  3 150  , value of each  100   100 
   
term is 5, total sum = 5 × 26 = 130  2   98 
Similarly, f  f   9 and so on.
Therefore, required sum = 7 + 38 + 111 +  100   100 
244 + 130 = 530.
There are 99 terms, out of which we can
25. (A) make 49 pairs of value 8 each.
1 x   50 
 
We have: f x  log   and One middle term f   will be left.
1 x   100 
3 x  x 3  50   1
g x   1  3x 2
f f 
 100  2
 3 x  x 3 
   
\ fog x  f  g x   f 
 1  3 x 2 
 1
So total value will be = 49 × 8 + f  
  2
 Now ( ) + (1 – ) = 8.
3 x  x 3
1   3  1
  log  1  3 x  3 x  x 
2
= log  1  3 x 2
Put x =
 3 x  x 3   1  3x 2  3x  x 3  2
 
1   1  1
 1  3x 2  f    f 1   8
æ 1+ x 3 ö 2  2
= log ç ( ) ÷ æ1+ x ö
= log ç
3

 1
ç 3 ÷ ÷
2f    8
(
ç 1- x ÷
è ) ø
è1- x ø
2
1 x 
 3 log    1
1 x  f 4
2
= 3f x ( )  1
Hence, total value = 49 × 8 + f  
26. (A) 2
= 49 × 8 + 4
Total terms in ( ) = 48
= 392 + 4 = 396.
Product ® ( + 5) ( + 7) ( + 9) … ( + 97)
( + 99) < 0. 28. 13,825
When When ( ) = 6, and ( ) = 5
Positive Terms Negative Terms The total number of functions from to
is given by ( ) ( ) = 56
47 1 when = -6
The number of onto functions from to
45 3 when = -10 is given by
. . – 1 ( – 1) + 2 ( – 2) – 3 ( – 3) + ∙∙∙
. . Here, = 5, =6
. . = 5 – 1 (4)6 + 5 2 (3)6 – 5 3 (2)6 + 5 4 (1)6
6 5

= 56 – 5 (4,096) + 10 (729) – 10 (64) + 5

1 47 when = -98 = 56 – 13,825

Functions and Graphs 57

 \ The required number of functions = 56 = < 22.6
– {56 – 13,825} = 13,825. So, = 22
Since 3 has been counted twice, hence
29. 43 minimum number of roots
We have, (3 – ) = (3 + ) ....(i) = 22 + 22 – 1 = 43.
So, the function is s mmetric about 3.
(10 – ) = (10 + ) ....(ii) 30. (C)
Hence, it is also s mmetric about 10. x
 1
Now, if (3) = 0, then from 2, (10 – 7) =   decrease as increases …(i)
(10 + 7) Þ (3) = (17) = 0 4
a b
Using 1 again, (17) = (3 – (14)) = (3 – 14)  1  1
2
= (- 11) = 0 Let – 2 – 24 =   and 11 =  
4 4
All the roots with such as (–11), (3), (17)
are 0. Given > . \ (1) Þ 2 – 2 – 24 < 11
2
(–25), (–11), (3), (17), (31) are equal to 0. \ – 2 – 35 < 0 Þ ( – 7) ( + 5) < 0
So the roots are a series of arithmetic \ –5 < < 7 …(ii)
progression with common dierence 
For log 1 x 2  2 x  24  to be defned,
= 14 4
th term = + ( – 1) 2
– 2 – 24 must be positive
Þ 300 > 3 + ( – 1) × 14 i.e. ( – 6) ( + 4) must be positive
297 \ < –4 or > 6 …(iii)
–1<
14 From equations (ii) and (iii), the range of
< 22.22 is (-5, -4) È (6, 7).
= 22
Þ –300 < 3 + ( – 1) (–14)

58 Functions and Graphs