RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 1

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

Relation and its Types (b) reflexive but neither symmetric nor transitive
(c) an equivalence relation
1. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9),
(3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. The (d) symmetric but neither reflexive nor transitive
relation is Ans. (b)
(a) reflexive and symmetric only 3 2 2 3
Sol. x - 3x y - xy + 3y = 0
(b) an equivalence relation
(c) reflexive only Þ x x - y x + y - 3y x - y x + y = 0
(d) reflexive and transitive only
Þ x - 3y x - y x + y = 0
Ans. (d)

Sol. (i) (a, a) Î R"a Î A Þ (3,3) (6, 6) (9, 9) (12,12) Now, x = y" x, y Î N ´ N so reflexive

Þ R is reflexive But not symmetric & transitive

See, (3,1) satisfies but (1,3) does not.
(ii) (a, b) Î R Þ (b, a) Î R
Also (9,3) and (3,1) satisfies but (9,1) does not
Here, (3, 6) is present in R,
Functions and its classifications
but (6, 3) is not there in R.
Þ R is not symmetric
ax 2
4. Let f (x) = , x ¹ - 1 . The value of a for which
(iii) (3, 6) Î R and (6, 12) Î R Þ (3, 12) Î R x +1
f (a) = a, (a ¹ 0) is
Þ R is transitive
2. Let W denotes the words in the English dictionary. Define 1 1
(a) 1 - (b)
the relation R by R = {(x, y) Î W × W : the words x and y a a
have at least one letter in common}. Then, R is
(a) reflexive, symmetric and not transitive 1 1
(c) 1 + (d) -1
a a
(b) reflexive, symmetric and transitive
(c) reflexive, not symmetric and transitive Ans. (c)

(d) not reflexive, symmetric and transitive Sol. f(a) = a

Ans. (a) aa 2
Þ =a
Sol. R = {(x, y) Î w ´ w ; the words x and y have at least one a +1
letter in common} clearly R is reflexive and symmetric. For
transitive let x = abcd, y = defg, z = ghi Þ aa 2 = a 2 + a

(x, y) Î R and (y, z) Î R Þ (x, z) Î R Þ a a = a + 1 (Q a ¹ 0)

Þ R is not transitive 1
Þ a =1+
3. Let N be the set of natural numbers and a relation R on N a
be defined by
log 2 x + 3
R= x, y Î N ´ N : x 3 - 3x 2 y - xy 2 + 3y 3 = 0 . 5. The domain of f (x) = is
x 2 + 3x + 2
Then the relation R is:
(a) R – {–1, –2} (b) (–2, + ¥)
(a) reflexive and symmetric, but not transitive
 2 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS
(c) R – {–1, –2, –3} (d) (–3, + ¥) – {–1, –2} Ans. (b)
Ans. (d)
Sol. e x + ef (x) = e
2
Sol. x + 3x + 2 ¹ 0
ex = e - ef (x)
Þ (x + 1) (x + 2) ¹ 0 Þ x ¹ - 1, - 2 .........(i)
Q ex > 0 Þ e - ef (x) > 0 ef (x) < e
and x + 3 > 0 Þ x > – 3 ........ (ii)

(i) Ç (ii) Þ x Î (-3, ¥) - {-1, - 2} Þ f (x) < 1

6. The range of the function y = log3 (5 + 4x – x2) is Þ f (x) Î ( -¥, 1)

(a) (0, 2] (b) (–¥, 2]
x2 + x + 2
(c) (0, 9] (d) none of these 8. Range of the function f (x) = ; x Î R is
x2 + x +1
Ans. (b)
æ 11 ö
Sol. y = log 3 ( - (x 2 - 4x - 5)) (a) (1, ¥) (b) ç 1, ÷
è 7ø
= log3 (- ( (x - 2) 2 - 4 - 5))
æ 7ù æ 7ö
(c) ç 1, ú (d) ç 1, ÷
è 3û è 5ø
= log 3 - (x - 2) 2 - 9
Ans. (c)
Q for log a x Þ x > 0

\- x - 2
2
-9 > 0 x2 + x + 1 + 1
Sol. y=
x2 + x + 1
2
9- x -2 >0 1
y = 1+ 2
x + x +1
2
9- x -2 = 0+
min 2
æ 1ö 3
x2 + x + 1 = ç x + ÷ +
è 2 ø 4
5 + 4x - x 2
max
3
(x 2 + x + 1) min =
-b 4
x=
2a
(x 2 + x + 1) max ® ¥
-4
x= =2 Þ y min ® 1 + 0
2 -1

1 7
5 + 4 2 - 22 = 9 y max = 1 + =
3 3
4
y max = log 3 9 = 2

Þ range = -¥, log3 9 = -¥, 2 æ 7ù

Range = ç1, ú
è 3û
7. If ex + ef(x) = e, then range of the function of f is
(a) (–¥, 1] (b) (–¥, 1) x 1 p
9. The equation 2 sin2 . cos2 x = x + , 0 < x < has
(c) (1, ¥) (d) [1, ¥) 2 x 2
(a) one real solution
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 3
(b) no real solution
ìx + 4 – 4 £ x <1
(c) infinitely many real solutions = í –( x + 4) x < –4
î
(d) none of these
Ans. (b)
Sol. Q x>0

Þ By AM ³ GM

1
x+ ³2
x

2 x
LHS : 2 sin cos 2 x 2
2 12. The function f (x) = cos log x + x + 1 is :

= (1 - cos x) cos 2 x (a) even (b) odd

Clearly LHS < 1 (c) constant (d) None of these

Ans. (a)
Þ No solution

x - [ x] Sol. f (x) = cos [log (x + x 2 + 1) ]

10. Let f(x) = , x Î R, then the range of f is :
1 + x - [ x]
f (-x) = cos [log ( x 2 + 1 - x) ]
(a) [0, 1] (b) [0, 1/2]
(c) [0, 1/2) (d) (0, 1) é ìï ( x 2 + 1 - x) ( x 2 + 1 + x) üï ù
Ans. (c) = cos ê log í ýú
ê ï x 2
+ 1 + x ú
ë î þï û
Sol. For x Î I, f(x) = 0
For x Î R ~ I
é 1 ù
(x – [x]) < 1 = cos ê log ú
ëê x 2 + 1 + x ûú
2(x – [x]) < 1 + x – [x]
Thus f(x) < 1/2
= cos é - log ( x 2 + 1 + x) ù
11. The range of k for which ||x–1|–5| = k have four distinct êë úû
solutions -
(a) [0, 5] (b) (–¥, 5) = cos é log ( x 2 + 1 + x) ù
êë úû
(c) [0, ¥) (d) (0, 5)
Ans. (d) Q cos -q = cos q
Sol. f(x) = ||x–1| –5 | Þ f (x) is even function
Case I x ³ 1
13. Let f : R ® R be a function such that f (x) = x3 – 6x2 + 11x – 6. Then
f(x) = |x – 1 – 5|
(a) f is one-one and into (b) f is one-one and onto
ìx – 6 x³6 (c) f is many-one and into (d) f is many-one and onto
= |x – 6| = í – x + 6 1 £ x < 6
î Ans. (d)

Case II x < 1 Sol. f (x) = (x - 1) (x - 2) (x - 3)

f(x) = |x + 4|
 4 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

\ f (1) = f (2) = f (3) = 0 Also, f ¢ (x) = 2 ± sin x

So, the function is many-one. Since f ¢ (x) > 0

Thus it is one-one.

x2 – 4
16. Let f be a function from R to R given by f (x) = .
x 2 +1
Then f (x) is.
(a) one-one and into (b) one-one and onto
(c) many-one and into (d) many-one and onto
Ans. (c)
Sol. f (x) = f (–x). So, f is many-one.

5
Also f (x) = 1 - 2
> 1- 5 = - 4
x +1
Clearly, for x < 0, f (x) < 0 and goes on decreasing as x So, f is into.
decreases.
For x > 3, f (x) > 0 and goes on increasing as x increases. 17. f (x) = x + x 2 is a function from R ® R. Then f (x) is
\ f (x) can have call real values. So, f is onto. (a) injective (b) surjective
14. Let f : R ® R be a function such that (c) bijective (d) none of these
f (x) = x3 + x2 + 3x + sin x. Then Ans. (d)
(a) f is one-one and into (b) f is many-one and into
Sol. f (x) = x + | x | .
(c) f is one-one and onto (d) f is many-one and onto
Clearly f (–1) = f (–2) = ........... = 0.
Ans. (c)
So, f is many-one.
ìïæ 2
1 ö 11 üï Also, f (x) ³ 0 for all x.
Sol. f (x) = x íç x + ÷ + ý + sin x.
ïîè 2ø 4 ïþ
So, it is not surjective.
18. A function f : A ® B, where A {x : – 1 £ x £ 1} and
2 2
ìïæ 1 ö 11 üï B = {y : 1 £ y £ 2} is defined by the rule y = f (x) = 1 + x .
Clearly x íç x + 2 ÷ + 4 ý increases with x. Which of the following statement is true?
ïîè ø ïþ
(a) f is injective but not surjective
and its value change over R. (b) f is surjective but not injective
Also, -1 £ sin x £ 1. (c) f is both injective and surjective
So, the range of f = R (d) f is neither injective nor surjective
Hence, f is onto and one-one. Ans. (b)
15. Let f be a function from R to R given by f (x) = 2x + |cos x|. Sol. Since, A = {x : - 1 £ x £ 1}
Then f is
(a) one-one and into (b) one-one and onto and B = {y :1 £ y £ 2}

(c) many-one and into (d) many-one and onto For y = f (x) = 1 + x 2
Ans. (b)
Sol. Here, f (x) will give all real values for real values of x, Thus x = 1, y = 1 + (1) 2 = 2
it is onto.
and for x = -1, y = 1 + (-1)2 = 2
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 5

\ f is not injective. (one-one)

é1 3 ù
= 2 ê sin x - cos x ú + 1
-1 £ x £ 1 Þ 0 £ x 2 £ 1 ë 2 2 û
Þ 1 £ 1 + x2 £ 2 = 2 sin (x - 60º ) + 1
\ Here, " B there is a preimage.
Q - 1 £ sin q £ 1
Hence, f is surjective.
\ - 2 £ 2 sin (x - 60º ) £ 2 Þ - 1 £ 2sin (x - 60º ) + 1 £ 3.
ì x, if x is rational
19. f x =í and 21. If a function f : [2, ¥) ® B defined by f (x) = x2 – 4x + 5 is a
î0, if x is irrational bijection, then B is :
(a) R (b) [1, ¥)
ì0, if x is rational
g x =í . Then, f – g is (c) [4, ¥) (d) [5, ¥)
î x, if x is irrational
Ans. (b)
(a) one-one and into (b) neither one-one nor onto
(c) many one and onto (d) one-one and onto Sol. f (x) = x 2 - 4x + 5; x Î [2, ¥)
Ans. (d) For function to be bijective, it should be injective as well
as surjective
Sol. Let f (x) = f (x) - g (x)
f (x) = x 2 - 4x + 5; x Î [2, ¥)
ì x, x Î Q
=í From the graph it is injective surjective means
î - x, x Ï Q

for one-one
Take any straight line parallel to x-axis which will intersect
f (x) only at one point.

Þ f (x) is one-one
For onto

ì x, x Î Q
As, f (x) = í , which shows
î - x, x Ï Q B = Range of f (x)
y = x and y = – x for rational and irrational values Þ B Î [1, ¥).
Þ y Î real numbers. 22. Which of the following function has period p ?
\ Range = Codomain
æ 2px ö æ px ö
Þ f (x) is onto. (a) 2cos ç ÷ + 3sin ç ÷
è 3 ø è 3 ø
Thus, f – g is one-one and onto. (b) |tan x| + cos 2 x
20. If f : R ® S, define by f (x) = sin x – 3 cos x + 1, is onto,
æ pö æ pö
then the interval of S is (c) 4cos ç 2px + ÷ + 2sin ç px + ÷
è 2 ø è 4ø
(a) [0, 1] (b) [–1, 1]
(d) none of the above
(c) [0, 3] (d) [–1, 3]
Ans. (b)
Ans. (d)

Sol. f (x) = sin x - 3 cos x + 1

 6 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

2p 2p Sol. cos x is not periodic function

Sol. (a) T1 = = 3, T2 = =6
2p p
So clearly f (x) = cos x + sin 2 x is not periodic.
3 3
Composition of a function
LCM of T1 & T2 = 6
26. Let f (x) be a function defined on [0, 1] such that
2p
(b) T1 = p, T2 = =p
2 ì x x ÎQ
ï
f x =í
LCM of T1 & T2 = p ï1 - x, x Ï Q
î
2p 2p
(c) T1 = = 1, T2 = =2 Then for all x Î [0, 1], fof (x) is
2p p
(a) a constant (b) 1 + x
LCM of T1 & T2 = 2 (c) x (d) none of these

23. Let f (x) = cos 3x + sin 3 x. Then f (x) is Ans. (c)

(a) a periodic function of period 2p. ì x; x ÎQ

Sol. f (x) = í
î1 - x; xÏQ
(b) a periodic function of period 3 p.
(c) not a periodic function ì f ( x), f ( x) Î Q
f f ( x) = í
(d) none of these î1 - f ( x ), f ( x) Ï Q
Ans. (c)
ì x; x Î Q ì x; x Î Q
2p f (f (x)) = í =í
Sol. Period of cos 3x = T1 = î1 - (1 - x); x Ï Q î x; x Ï Q
3
Þ fof (x) = x.
2p
Period of sin 3x = T2 =
3 27. If f x = 2 - x and g x = 1 - 2 x , then the domain of
LCM of T 1 and T2 is not defined f [g (x)] is
Þ f (x) is not a periodic function. (a) (– ¥, 1/2] (b) [1/2, ¥)

24. The period of sin q is

2 (c) (– ¥, –3/2] (d) none of these

(a) p
2
(b) p Ans. (d)
3
(c) p (d) p/2 Sol. f (x) = 2 - x, g (x) = 1 - 2x
Ans. (b)
f (g (x)) = 2 - 1 - 2x
1 - cos 2q
Sol. sin 2 q =
2
1
Þ 1 - 2x ³ 0 Þ x £
2
æ 2p ö
Þ cos 2q period is ç ÷
è 2 ø 2 - 1 - 2x ³ 0
Þp
3
25. Which one is not periodic Þ 4 ³ 1 - 2x Þ x ³ -
2
2
(a) |sin 3x| + sin x (b) cos x + cos 2 x
2 2
(c) cos 4x + tan x (d) cos x + sin x
Ans. (b)
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 7

é 3 1ù p
Þ x Î ê- , ú . = cos–1(sin x) = –x
ë 2 2û 2

28. Let f (x) = sin x and g(x) = In |x|. If the ranges of the gofoh(x) = sin 2 (cos –1 x ) = 1 - x 2
composition functions fog and gof are R 1 and R 2
respectively, then fohog(x) = sin2(cos–1 x ) = 1 – x.
(a) R1 = {u : –1 < u < 1}, R2 = {v : –¥ < v < 0}
Thus no two composites are equal.
(b) R1 = {u : –¥ < u < 0}, R2 = {v : –1 < v < 0}
31. If f(g(x)) = | cos x |, g(f(x)) = cos2 x , then -
(c) R1 = {u : –1 < u < 1}, R2 = {v : –¥ < v < 0}
(d) R1 = {u : –1 < u < 1}, R2 = {v : –¥ < v < 0} (a) f(x) is a periodic function and g(x) is a non-periodic
function.
Ans. (d)
(b) f(x) is a non-periodic function and g(x) is a periodic
Sol. f (x) = sin x, g (x) = ln | x | function.
fog (x) = sin ln |x| (c) Both f(x) and g(x) are periodic functions
Þ range = [–1, 1] (d) Neither f(x) nor g(x) is a periodic function
Ans. (b)
sin ce - 1 £ sin x £ 1
Sol. Given, f(g(x)) = | cos x | = cos2 x ….(i)
(gof) (x) = ln |sin x|

Q - 1 £ sin x £ 1 Þ 0 £ | sin x | £ 1 g(f(x)) = cos2 x ….(ii)

Þ range of (gof) (x) = (-¥, 0] from (i) and (ii), f(x) = x . And g(x) = cos2x
Clearly f(x) is a non-periodic function and g(x) is a periodic
2
29. If g {f (x)} = |sin x| and f {g (x)} = sin x , then function.

32. Consider the functions f x = x and g(x) = 7x + b. If the

(a) f x = sin 2 x ,g x = x
function y = fog(x) passes through (4, 6) then the value of
(b) f x = sin x ,g x = x b is
(a) 8 (b) – 8
(c) f x = x 2 ,g x = sin x
(c) – 25 (d) 4 - 7 6
(d) f and g cannot be determined Ans. (a)
Ans. (a)
Sol. y= f g x = f 7x + b = 7x + b
Sol. g (f (x)) = | sin x |, f (g (x)) = (sin x )2
Given that y 4 = 6, so 28 + b = 6 Þ 28 + b = 36 Þ b = 8
Cleary f (x) = sin 2 x and g (x) = x
Inverse of a Function
30. If f(x) = sin2x, g x = x and h(x) = cos–1x, 0 £ x £ 1, then -

(a) hogof(x) = gofoh(x) a x - a -x

33. The inverse of the function f x = is
(b) gofoh(x) = fohog(x) a x + a -x
(c) fohog(x) = hogof(x) (where codomain of f(x) is (–1, 1))
(d) None of these
1 æ1- x ö 1 æ1+ x ö
Ans. (d) (a) log a ç ÷ (b) log a ç ÷
2 è1+ x ø 2 è1- x ø
Sol. hogof x = cos -1 sin 2 x
 8 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

æ 1+ x ö Sol. y = [1 - (x - 3) 4 ]1/7
(c) log a ç ÷ (d) none of these
è 1- x ø
4
y7 = 1 - x - 3
Ans. (b)
Interchange x « y
a x - a -x a 2x - 1
Sol. f (x) = =
a x + a -x a 2x + 1 Þ x 7 = 1 - (y - 3)4

a 2y - 1 Þ (1 - x 7 )1/4 = y - 3
x=
a 2y + 1
Þ y = 3 + (1 - x 7 )1/4 .
x +1
Þ a 2y = Functional Equations
1- x

1 æ 1+ x ö æ1ö 1
Þ y= log a ç ÷. 36. If 3 f (x) + 5 f ç ÷ = – 3, " x(¹ 0) Î R, then f (x) is
2 è1- x ø èxø x

34. Let f : [–1, ¥) ® R be given by f (x) = (x + 1)2 –1, equal to :

x > –1. Then f –1 (x), is :
1 æ3 ö 1 æ 3 ö
(a) ç + 5x – 6 ÷ (b) ç – + 5x – 6 ÷
(a) – 1 + x +1 16 è x ø 16 è x ø

(b) – 1 – x +1 1 æ3 ö
(c) ç - 5x - 6 ÷ (d) none of these
16 è x ø
(c) does not exist because f is not one-one
Ans. (b)
(d) does not exist because f is not onto
Ans. (d) æ1ö 1
Sol. 3f (x) + 5f ç ÷ = - 3 .........(1)
èxø x
Sol. f (x) = (x + 1)2 - 1
1
Replace x.by
x

æ1ö
Þ 3f ç ÷ + 5f (x) = x - 3 ..........(2)
èxø

Eq n (1) ´ 3 and eq n (2) ´ 5

æ1ö 3
9f (x) + 15 f ç ÷ = - 9
èxø x
Clearly range = [-1, ¥)
æ1ö
25 f (x) + 15f ç ÷ = 5x - 15
But co-domain = R èxø
Þ function is not onto Þ inverse of function does not Subtract
exist.
35. The inverse of the function y = [1 – (x – 3) ] is
4 1/7 3
16 f (x) = 5x - -6
7 1/4 7 1/4 x
(a) 3 + (1 – x ) (b) 3 – (1 – x )
7 1/4
(c) 3 – (1 + x ) (d) none of these 1 æ 3 ö
Þ f (x) = 5x - - 6 ÷
Ans. (a) 16 çè x ø
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 9
37. Let f : R ® R be a function given by f (x + y) = f (x) + f (y) for (a) 4a = 3b (b) 3a = 4b
all x, y Î R such that f (1) = a. Then, f (x) =
7p
(c) a – b = (d) none of these
(a) ax (b) ax 12
(c) xa (d) a + x
Ans. (a)
Ans. (b)
Sol. f (x + y) = f (x) + f (y) Sol. tan -1 tan x = x
Put x = y = 1
æ 5p ö
f (2) = f (1) + f (1) = 2a \a = tan -1 ç tan
è 4 ÷ø
Put x = 1, y = 2
f (3) = f (1) + f (2) = 3a ì æ p öü
= tan -1 í tan ç p + ÷ ý
In same manner f (x) = ax î è 4 øþ
38. Let f be a real valued function satisfying
f (x + y) = f (x) f (y) for all x, y Î R such that f (1) = a. Then, æ pö
= tan -1 ç tan ÷
f (x) = è 4ø
(a) ax (b) ax
(c) xa (d) none of these p
=
4
Ans. (a)
Sol. f (x + y) = f (x) f (y) and

Put x = y = 1 ì æ 2p ö ü
b = tan -1 í- tan ç ÷ý
f (2) = f (1) . f (1) = a2 î è 3 øþ
Put x = 1, y = 2
Þ f (3) = f (1) . f (2) = a3 ì æ p öü
= tan -1 í- tan ç p - ÷ ý
î è 3 øþ
In same manner f (x) = ax

æ 1 ö ì æ p öü
39. If a f (x + 1) + bf ç ÷ = x, x ¹ -1, a ¹ –b, then f (1) is equal = tan -1 í tan ç ÷ ý
è x +1 ø î è 3 øþ
to
p
(a) a + b (b) a2 – b2 =
3
1
(c) (d) f(1) = 0 \ 4a = p
a+b

Ans. (d) 3b = p

Sol. Put x = 0 \ 4a = 3b
Þ af(1) + bf(1) = 0 41. Which one of the following is correct?
Þ f(1) (a + b) = 0 (a) tan 1 > tan–1 1 (b) tan 1 < tan–1 1
Þ f(1) = 0 (c) tan 1 = tan–1 1 (d) None of the above
as a + b ¹ 0 Ans. (a)
Sol. tan 1 means (tan (57º)) approx
Simplification problems of ITF
and tan 57º > 1
æ 5p ö æ 2p ö
40. If a = tan–1 ç tan ÷ and b = tan–1 ç – tan ÷ , then -1 p
è 4 ø è 3 ø but tan (1) = <1
4
so tan 1 > tan–1 (1)
 10 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

ép -1 æ 3 öù x
42. The value of sin ê - sin çç - ÷÷ ú is : 1 - tan
êë 2 è 2 ø úû 2 = cos a
x
1 + tan
2
3 3
(a) (b) -
2 2
x
1 + tan
1 1 2 = 1
(c) (d) - Þ
2 2 x cos a
1 - tan
2
Ans. (c)
Apply C & D
ép æ 3 öù
Sol. = sin ê - sin -1 ç -
ç 2 ÷÷ ú 2 1 + cos a
ëê 2 è ø úû =
x 1 - cos a
2 tan
2
ép pù
= sin ê + ú
ë2 3û
x 1 - cos a
tan =
p 1 2 1 + cos a
= cos =
3 2
x
43. cot -1 cos a - tan -1 cos a = x, then sin x is equal 2 tan
sin x = 2
x
to 1 + tan 2
2
æaö 2 æaö
(a) tan 2 ç ÷ (b) cot ç ÷
è2ø è2ø æ 1 - cos a ö
2 çç ÷÷
æaö = è 1 + cos a ø
(c) tan a (d) cot ç ÷
è2ø (1 + cos a )2 + (1 - cos a )2
(1 + cos a )2
Ans. (a)

Sol. cot -1 ( cos a ) - tan -1 ( cos a ) = x 2 (1 - cos a )

=
2 (1 + cos a)
cot -1 ( cos a ) + tan -1 ( cos a ) - 2 tan -1 ( cos a ) = x

p a
- 2 tan -1 ( cos a ) = x 2sin 2
2 = 2
2 a
2 cos
2
p
2 tan -1 ( cos a ) = -x
2
æaö
= tan 2 ç ÷
p x è2ø
tan -1 ( cos a ) = -
4 2
1
44. If tan(cos–1 x) = sin æç cot -1 ö÷ , then x is equal to :
æp xö è 2ø
tan ç - ÷ = cos a
è4 2ø
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 11
Properties of ITF
5 5
(a) ± (b) ±
3 3
-1 p
46. If sin x = , for some x Î (–1, 1), then the value of
5 5
(c) ± (d) None of these
3 cos–1 x is :
Ans. (b) 3p 5p
(a) (b)
10 10
æ 1ö
Sol. tan (cos-1 x) = sin ç cot -1 ÷
è 2ø 7p 9p
(c) (d)
10 10
1- x2 2 Ans. (a)
=
x 5
p
Sol. If sin -1 x = ,
5
5 (1 - x 2 ) = 2x
We know
Squaring both sides
p
5 - 5x 2 = 4x 2 sin -1 x + cos -1 x =
2

9x 2 = 5 -1 p p 3p
so, cos x = - =
2 5 10
5
x=±
3 3 5
47. The value of cot -1 + sin -1 is :
4 13
45. cos [tan–1 {sin (cot–1 x)}] is equal to :
-1 63 12
x2 + 2 x2 + 2 (a) sin (b) sin -1
(a) (b) 65 13
x2 + 3 x2 +1
-1 65 5
(c) sin (d) sin -1
68 12
x2 +1
(c) (d) None of these Ans. (a)
x2 + 2

Ans. (c) æ3ö æ5ö

Sol. cot -1 ç ÷ + sin -1 ç ÷
è4ø è 13 ø
Sol. cos [tan -1 {sin (cot -1 x)}]
æ 4ö æ 5ö
é = tan -1 ç ÷ + tan -1 ç ÷
ïì æ 1 ö ïüù 3
è ø è 12 ø
= cos ê tan -1 ísin ç sin -1 ÷÷ ýú
ç
êë îï è x 2 + 1 ø þïúû
æ 4 5 ö
ç + ÷
é é 1 ùù = tan -1 ç 3 12 ÷
= cos ê tan -1 ê úú 4 5
çç 1 - ´ ÷÷
êë êë x 2 + 1 úû úû è 3 12 ø

é æ x 2 + 1 öù
= cos êcos -1 ç ÷ú æ 63 ö
ç x 2 + 2 ÷ú ç ÷ 63
ê
ë è øû = tan -1 ç 36 ÷ = tan -1
çç 16 ÷÷ 16
è 36 ø
x2 +1
=
x2 + 2
 12 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

æ ö x -1 x +1 p
63 Sol. tan -1 + tan -1 =
= sin -1 ç ÷ x+2 x+2 4
ç 632 + 16 2 ÷
è ø
æ a+b ö
Q tan -1 (a) + tan -1 (b) = tan -1 ç ÷
æ 63 ö è 1 - ab ø
= sin -1 ç ÷
è 65 ø
æ x -1 x + 1 ö
y ç + ÷
48. –1
If cos x – cos –1 = a, then 4x2 – 4xy cos a + y2 is p
2 \ tan -1 ç x + 2 x + 2 ÷ =
ç (x - 1) (x + 1) ÷ 4
equal to ç1- ÷
è (x + 2)2 ø
(a) –4 sin2 a (b) 4 sin2 a
(c) 4 (d) 2 sin 2a
Ans. (b) 2x
x +2 æ pö
Þ 2 2
= tan ç ÷
y (x + 4x + 4) - (x - 1) è 4ø
Sol. cos -1 x - cos -1 =a 2
2 (x + 2)

æ xy y2 ö
cos -1 ç + 1 - x2 1- ÷=a 2x (x + 2)
=1
ç 2 4 ÷ 4x + 5
è ø

2x 2 + 4x = 4x + 5
2
xy y
+ 1- x2 1 - = cos a
2 4 2x 2 = 5

5
y2 xy x=±
1- x2 1 - = cos a - 2
4 2
50. If 2 tan–1 (cos x) = tan–1 (2 cosec x), then the value of x is :
Square both sides.
3p p
y2 x 2 y2 x 2 y2 1 (a) (b)
1- - x2 + = cos 2 a + - 2. xy cos a 4 4
4 4 4 2
p
(c) (d) None of these
y 2 3
1- - x 2 = cos 2 a - xy cos a
4 Ans. (b)

4 - y 2 - 4x 2 = 4 cos 2 a - 4xy cos a Sol. 2 tan -1 (cos x) = tan -1 (2 cosec x)

4x 2 - 4xy cos a + y2 = 4sin 2 a æ 2 cos x ö

tan -1 ç 2 ÷
= tan -1 (2cosec x)
è 1 - cos x ø
x -1 x +1 p
49. If tan -1 + tan -1 = , then x is equal to :
x+2 x+2 4 2 cos x 2
2
=
1 1 sin x sin x
(a) (b) -
2 2
2 cos x - 2 sin x
=0
5 1 sin 2 x
(c) (d) ±
2 2

Ans. (c)
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 13

p p
Þ tan x = 1 so x = Sol. cos -1 x > - cos -1 x
4 2

11p p
51. The equation 2cos-1 x + sin -1 x = has : 2 cos -1 x >
6 2
(a) no solution (b) only one solution
p
(c) two solutions (d) three solutions cos -1 x >
4
Ans. (a)
p
11p x < cos
Sol. 2 sin -1 x + cos -1 x = 4
6
1
p 11p x<
+ sin -1 x = 2
2 6

4p é 1 ù
sin -1 x = so x Î ê -1, ú
3 ë 2û

Never possible, so no solution. 54. Set of values of x satisfying cos–1 x > sin–1 x

æaö æbö p
52. If tan -1 ç ÷ + tan -1 ç ÷ = , then x is equal to : æ 1ö é 1ö
x
è ø èxø 2 (a) ç 0, ÷ (b) ê0, ÷
è 2ø ë 2ø
(a) ab (b) 2ab
(c) 2ab (d) ab æ1 ö æ1 ù
(c) ç , 1÷ (d) ç , 1ú
è2 ø è2 û
Ans. (a)
Ans. (b)
-1 æaö -1 æ b ö p
Sol. tan ç x ÷ + tan ç x ÷ = 2
è ø è ø p
Sol. cos
-1
x> - cos -1 x Q x ³ 0
2

éa bù p
-1
êx+xú p cos -1 x >
tan ê ú= 4
ê1 - ab ú 2
ëê x 2 ûú p p
Þ < cos -1 x £
4 2
ab
Þ so 1 - =0
x2 0£ x <
1
2
Þ x 2 = ab
1
0£ x<
Þ x = ab 2
53. If cos–1 x > sin–1 x, then :
1
(a) x < 0 (b) –1 < x < 0 55. The value of cos (2 cos–1 x + sin–1 x) at x = is :
5
1 1 (a) 1 (b) 3
(c) 0 £ x < (d) -1 £ x <
2 2
2 6
Ans. (d) (c) 0 (d) -
5
Ans. (d)
 14 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

1 1
Sol. cos (2 cos -1 x + sin -1 x) = ? at x = +..... + tan -1 = tan -1 q , then q =
5 1+ n n +1

= cos (cos-1 x + sin -1 x + cos -1 x) n n +1

(a) (b)
n +1 n+2
æp æ 1 öö
= cos ç + cos -1 ç ÷ ÷
è 2 è 5 øø n n -1
(c) (d)
n+2 n+2
æ æ 1 öö
= - sin ç cos -1 ç ÷ ÷ Ans. (c)
è è 5 øø
1 n +1- n
Sol. tan -1 = tan -1
æ 2 6ö 1+ n n +1 1 + n n +1
= - sin ç sin -1 ÷
ç 5 ÷ø
è
= tan -1 n + 1 - tan -1 n
-2 6
= é -1 -1 -1 æ a - b ö ù
êQ tan (a) - tan (b) = tan ç 1 + ab ÷ ú
5
ë è øû
¥
æ 1 ö
56. The value of å tan -1 çè 1 + r + r 2 ÷ø is equal to : so that L.H.S. of the given equation is
r =0
= tan -1 2 - tan -1 1 + tan -1 3 - tan -1 2 + .... + tan -1 n + 1 - tan -1 n
p 3p
(a) (b) = tan -1 n + 1 - tan -1 1
2 4
p n +1-1
(c) (d) None of these = tan -1
4 1+ n +1
Ans. (a)
n
¥ = tan -1
æ 1 ö n+2
Sol. å tan -1 çè 1 + r + r 2 ÷ø
r =0
-1 n n
so that tan = tan -1 q Þ q = .
¥
æ 1 ö n+2 n+2
= å tan -1 ç 1 + r (r + 1) ÷
r =0 è ø ITF- Domain & Range
¥
æ r +1- r ö
= å tan -1 ç 1 + r (r + 1) ÷ –1 é æ x öù
r =0 è ø 58. The domain of sin êl og 3 ç 3 ÷ ú is
ë è øû
¥
(a) [1, 9] (b) [–1, 9]
= å [tan -1 (r + 1) - tan -1 r]
r =0 (c) [–9, 1] (d) [–9, –1]
Ans. (a)
= tan -1 1 - tan -1 0 + tan -1 2 - tan -1 1 + tan -1 3 - tan -1 2....

æxö
-1 p Sol. -1 £ log3 ç ÷ £ 1
....tan ¥ = è3ø
2

1 1 1 1 x
57. If tan -1 + tan -1 + tan -1 £ £3
1+ 2 1+ 2 3 1+ 3 4 3 3
 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 15

Þ 1£ x £ 9 æ p pö
Þ domain of log (cos x) = D3 = ç - , ÷
Domain = [1, 9] è 2 2ø

sin –1 x – 3 é pö
59. The domain of the function f (x) = is D1 Ç D2 Ç D3 Þ x Î ê 0, ÷
9– x 2 ë 2ø

61. If q = sin–1 x + cos–1 x – tan–1 x, x ³ 0, then the smallest

(a) [1, 2] (b) [2, 3]
interval in which q lies, is given by :
(c) [2, 3) (d) [1, 2)
Ans. (c) p 3p p
(a) £q£ (b) - £q£0
2 4 4
Sol. 9 - x 2 > 0 Þ (3 - x) (3 + x) > 0
p p p
(c) 0 £ q £ (d) £q£
4 4 2
Ans. (d)

Þ x Î ( -3, 3) ....... (1)

Sol. q = sin -1 x + cos -1 x - tan -1 x

Given x > 0 but x £ 1

And -1 £ x - 3 £ 1
[Domain of sin–1 and cos–1]
2£x£4 ....... (2)
p
(1) Ç (2) Þ x Î 2, 3 so q = + tan -1 x x Î (0, 1]
2

p p Þ 0 < x £1
60. The largest interval lying in æç - , ö÷ for which the
è 2 2ø
Þ 0 < tan -1 x £ p / 4
2 æx ö
function f (x) = 4- x + cos -1 ç - 1÷ + log (cos x) is Þ -p / 4 £ - tan -1 x £ 0
è2 ø
defined, is Þ +p / 4 £ +p / 2 - tan -1 x £ p / 2
æ p pö
(a) [0, p] (b) ç - , ÷ p / 4 £q £p / 2
è 2 2ø
62. Range of f(x) = sin–1 x + tan–1 x + sec–1 x is
é p pö é pö
(c) ê - , ÷ (d) ê 0, ÷ æ p 3p ö é p 3p ù
ë 4 2ø ë 2ø (a) ç , ÷ (b) ê , ú
è4 4 ø ë4 4 û
Ans. (d)
ì p 3p ü
(c) í , ý (d) None of these
- x2 æ p pö î4 4 þ
Sol. Domain of 4 = D1 = ç - , ÷
è 2 2ø
Ans. (c)
For Þ x = 0, - 1 Sol. f (x) = sin -1 x + tan -1 x + sec-1 x
x x -1
-1 £ -1 £ 1 Þ 0 £ £ 2 Domain sin x, x Î [-1, 1]
2 2

Þ 0 £ x £ 4 Þ D 2 = 0, 4 Domain sec -1 x, x Î [R - ( -1, 1)]

So, permissible value x = {–1, 1}.

æ p pö
Q for x Î ç - , ÷ , cos x > 0
è 2 2ø f (x) = sin -1 (1) + tan -1 (1) + sec-1 (1)
 16 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

p p 3p Sol. f (x) = sin 4 x + cos4 x

= + +0 =
2 4 4
4 4
-1 -1 -1 æ pö æ æp öö æ æp öö
or f (x) = sin (-1) + tan (-1) + sec (-1) f ç x + ÷ = ç sin ç + x ÷ ÷ + ç cos ç + x ÷ ÷
è 2 ø è è 2 øø è è 2 øø
p p p
=- - +p= = cos 4 x + sin 4 x = f (x)
2 4 4

p
ì p 3p ü Þ Period =
Range = í , ý 2
î4 4 þ
66. The period of the function f (x) = |sin 4x| + |cos 4x| is p/k.
63. Range of f (x) = sin–1x + cot–1x + tan–1x is
Then the value of k is
Ans. (8)
(a) [0, p] (b) é p , pù
êë 2 úû
p
Sol. We know period of f (x) = | sin x | + | cos x | is
ép ù 2
(c) ê , pú (d) [–p, p]
ë4 û Þ Period of f (x) = | sin 4x | + | cos 4x |
Ans. (a)
p
Sol. f (x) = sin -1 x + cot -1 x + tan -1 x p
= 2 =
4 8
p
= sin -1 x + 67. Let [x] denote the greatest integer £ x. If f (x) = [x] and
2
æ 8 ö æ 8 ö
p p g(x) = |x|, then the value of f ç g æç ö÷ ÷ - g ç f æç - ö÷ ÷ is
- £ sin -1 x £ è è øø è è 5 øø
5
2 2
Ans. (-1)
So 0 £ f (x) £ p
Sol. Given f (x) = [x] and g (x) = | x |
f (x) Î [0, p]
æ æ 8 öö æ 8 ö é8ù
Now, f ç g ç ÷ ÷ = f ç ÷ = ê ú = 1
Numerical Type Valued Questions è è 5 øø è 5 ø ë5û

64. Let n(A) = 4 and n(B) = 6. Then the number of one-one

æ æ 8 öö æ é 8ù ö
functions from A to B is and g ç f ç - ÷ ÷ = g ç ê - ú ÷ = g ( -2) = 2
è è 5 øø è ë 5û ø
Ans. (360)
Sol. n(A) = 4 and n(B) = 6 æ æ 8 öö æ æ 8 öö
\ f ç g ç ÷ ÷ - g ç f ç - ÷ ÷ = 1 - 2 = -1.
Let a, b, c, d be the elements of A 5
è è øø è è 5 øø
Number of way to select image of a from B = 6
68. If f : R ® R is given by
Number of way to select image of b from B = 5
Number of way to select image of c from B = 4 ì-1, when x is rational
f x =í ,
Number of way to select image of d from B = 3 î 1, when x is irrational

Total number of ways = 6 × 5 × 4 × 3 = 360

4 4
then ( fof ) 1 - 3 is equal to
65. The period of the function f (x) = sin x + cos x is p/k. Then
the value of k is Ans. (-1)
Ans. (2)
ì-1, when x is rational
Sol. Given, f (x) = í
î 1, when x is irrational

Now, (fof ) (1 - 3) = f [f (1 - 3] = f (1) = -1.

 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 17
69. The number of real solutions of the equation ex = x is Þ f(2) = 2f (1) = 10
Ans. (0) Also, f (3) = f (2) + f (1) = 15

Sol. Draw graph of y = ex and y = x clearly no solution Þ f (n) = 5n

Þ f (100) = 500.

æ 1 ö
73. If 2f (x + 1) + f ç ÷ = 2x and x ¹ –1, then f (2) is equal to
è x +1 ø
k/6. Then the value of k is.
Ans. (10)

1
Sol. Put x = 1, - is given function respectively, we get
2

æ1ö
2f (2) + f ç ÷ = 2 .........(i)
70. The number of real solutions of the equation log0.5 x = |x| is è2ø
Ans. (1)
æ1ö
Sol. draw graph of log 0.5 x and y = | x | and 2f ç ÷ + f (2) = -1 ..........(ii)
è2ø

5
On solving Equation (i) and (ii), we get f (2) = .
3
74. If f (x) = ax 2 + bx + c satisfies the identity
f (x + 1) – f (x) = 8x + 3 for all x Î R. Then, a + b =
Ans. (3)

Sol. é a x + 1 2 + b x + 1 + c ù - é ax 2 + bx + c ù = 8 x + 3
ë û ë û

Clearly one solution Þ éë 2ax + a + b ùû = 8 x + 3

71. The number of real solutions of the equation sin (ex) = 5x +
Þ 2a = 8 & a + b = 3
5–x is
Ans. (0) Þ a = 4 & b = -1

Sol. By AM ³ GM 75. sec2 (tan–1 2) + cosec2 (cot–1 3) is equal to

Ans. (15)
5x + 5- x ³ 2
Sol. sec 2 (tan -1 2) + cosec 2 (cot -1 3)
But sin e x Î [ -1, 1]
æ æ 5 öö
Þ No solution. = sec2 ç sec-1 ç ÷ ÷ + cosec2 (cosec-1 10)
ç ç 1 ÷÷
è è øø
72. If f is a real valued function such that f (x + y) = f (x) + f (y) and
f (1) = 5, then the value of f (100) is = 5 + 10 = 15
Ans. (500) 20 20

Sol. Given, f (x + y) = f (x) + f (y) 76. If å sin -1 x i = 10p , then å

i =1
x i is equal to :
i =1
For x = 1, y = 1 we get
Ans. (20)
f (2) = f (1) + f (1)
 18 RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS

20
79. The value of tan–1 (1) + tan–1 (0) + tan–1 (2) + tan–1 (3) is
-1
Sol. å sin x i = 10p, equal to kp. Then the value of k is
i =1
Ans. (1)

p
£ sin -1 x £
p Sol. tan -1 (1) + tan -1 (0) + tan -1 (2) + tan -1 3
We know that -
2 2
since tan -1 (0) = 0
p
So only possibility if sin -1 x i = "i Î [1, 20]
2 æ 2+3ö
tan -1 2 + tan -1 3 = tan -1 ç ÷
So x i = 1. è 1- 6 ø

20 æ 5 ö
= tan -1 ç ÷ + p
å x i = 20 è -5 ø
i =1

3p
77.
3 4
If cos -1 - sin -1 = cos -1 x, then x is equal to = p + tan -1 (-1) =
5 5 4

Ans. (1) p 3p
tan -1 1 + tan -1 2 + tan -1 3 = + =p
4 4
3 4
Sol. cos -1 - sin -1 = cos -1 x
5 5 æ p pö
80. If x Îç- , ÷ , then the value of
è 2 2ø
æ4ö æ4ö
Þ sin -1 ç ÷ - sin -1 ç ÷ = cos-1 x
5
è ø è5ø æ tan x ö -1 æ 3sin 2x ö
tan -1 ç ÷ + tan ç ÷ is kx. Then the value
è 4 ø è 5 + 3cos 2x ø
0 = cos -1 x
of k is
x = cos 0 = 1 Ans. (1)
78. If k £ sin–1 x + cos–1 x + tan–1 x £ K and k + K = mp. Then the æ tan x ö æ 3sin 2x ö
value of m is. Sol. tan -1 ç ÷ + tan -1 ç ÷
è 4 ø è 5 + 3cos 2x ø
Ans. (1)

Sol. k £ sin -1 x + cos -1 x + tan -1 x £ K é tan x 3sin 2x ù

ê + ú
= tan -1 ê 4 5 + 3cos 2x ú
-1 -1 p ê æ tan x ö æ 3sin 2x ö ú
Since sin x + cos x = " x Î [-1, 1] ê1 - ç ÷ç ÷ú
2
ëê è 4 ø è 5 + 3cos 2x ø ûú
Now sin -1 & cos -1 x are defined only when
é -1 -1 -1 æ a + b ö ù
-1 £ x £ 1 êQ tan (a) + tan (b) = tan ç 1 - ab ÷ ú
ë è øû
p p
Þ- £ tan -1 x £
4 4

p p 3p
Þ £ + tan -1 x £
4 2 4

Þ k =p /4

3p
&K =
4
RELATION , FUNCTION & INVERSE TRIGONOMETRIC FUNCTIONS 19

é æ 2 tan x ö ù
ê tan x 3ç ÷ ú
ê + è 1 + tan 2 x ø ú
ê 4 æ 1 - tan 2 x ö ú
ê 5 + 3 çç 2 ÷ ÷ú
= tan -1 ê è 1 + tan x ø ú
ê æ tan x ö é 6 tan x ù ú
ê1 - ç ÷ê 2 úú
ê è 4 ø ë 8 + 2 tan x û ú
ê ú
ê ú
ëê ûú

é tan x 6 tan x ù
ê + 2 ú
4 8 + 2 tan x ú
=ê
ê1 - æ tan x ö é 6 tan x ù ú
ê çè 4 ÷ø ëê 8 + 2 tan 2 x ûú ú
ë û

é 8 tan x + 2 tan 3 x + 24 tan x ù

= tan -1 ê 2 2 ú
ë 32 + 8 tan x - 6 tan x û

é 2 tan x [tan 2 x + 16] ù

= tan -1 ê 2 ú
ë 2 [tan x + 16] û

= tan -1 (tan x)
=x