Multiple Choice Questions

Choose and write the correct option in the following questions.

1. The value of sin- 1 ( cos l~7r) is (CBSE 2021-22 (Term-1)]

(b) - ~
3 1[
(a) - 11: (c) 3rr (d) 10
5 10 5
1
2. If sin- x > cos-1 x, then x should lie in the interval [CBSE 2021-22 (Term-1))

(a) (- 1, - }i) (b) ( 0, - }i) (c) ( 1, 1) (d) (}i,o)

3. sin[~ -sin-
1
(-½)] is equal to [CBSE Sample Paper 2021-22 (Tenn-1)]

(a) ; (b) ; (c) - 1 (d) 1

4. The value of cot (sin-1 x) is [NCERT Exemplar)

(a) ./1+?- X 1
(c) - (d)
/1-x2
x (b) /1 + x2 X X

5. The value of sin-1 ( cos ; ) is [NCERT Exemplar)

rr Srr 771:
(a) 9 (b) 9 (c) -~TI
(d) 18
6. Let 8 = sin-1 (sin (-600°)), then value of 8 is
rr 1[ -2TI
(a) - (b) - (c) 211: (d) 3
3 2 3

7. The value of the expression 2 sec - J 2 + sin - l ( ½) is [NCERT Exemplar)

n: Srr 7TI
(a) - (b) - (c) - (d) 1
6 6 6

8. The value of cot[ cos-1( ; ) ] is [NCERT Exemplar)

5
25 25 24 7
(a) 24 (b) 7 (c) 25 (d) 24

. 1 [ cos ( -337!:
9. The value o f sm- - )] 1s
.
5
1[
(a) 3n: (b) -7rr (c) 10 (d) - TI
5 5 10

10. The value of cot[½ sin-1 ~ ] is

1
(a) 1 (b) - (c) fj (d) O
fj
11. sin (cot-1 x) is eq ual to
(a) /I + x 2 (b) x (c) (1 + :?r 1/ 2

12. If tan-1 x = y, then [CBSE Sample Paper 2021 -22 (Ter111-1)]

(a) -1 < y < l (c) -271: < y < i (d) y E { -t, ; }

lnv(•r<t• Tr i(_Jonor1 1dric rumt1L111• 49
 13. If y = coC1 x, x < 0, lhen [CBSE 2021-22 (Te
7t 7t 7t rr. l'n1-1i1
(a) 2 < Y ~ 7t (/J) 2 < !/ < rr. (c) - - < y < 0
2
(d) -2 ~ y < 0
14. sin (tan- 1 x), lxl < 1 is equal to
X 1 1 X
(a)
✓i - x2 /1 + x1-
r:-'j
vl - r (/J) (c) (d) Ji +x2
1
15. If cos- a+ cos- 1 S + cos- 1 -y = 31t, then a (S + -y) + S (r +a)+ 'Y (a + S) equals
[NCERT Exetn I
(a) 0 (b) 1 (c) 6 (d) 12 p llr]

·
16. Simples t form of tan-1 (Ji+
J /+
cos x h- cos x )
, 1t < x < T37t 1s.
1 + cos x- 1- cosx
[CBSE Sample Paper 2021-22 (Te
Tnz-1)]
(a) E.. - .:!.
4 2 (b) 3; -½. (c) -2
X
(d) n-½
l7. The number of real solutions of the equation h + cos 2.x = fi. cos-I (cos x) in [ ~ , 1t] is
(a) 0 (b) 1 (c) 2 (d) oo
18. The value of cos-1 (2x2-1), Q ~ x ~ 1 is equal to
(a) 2 cos-1 x (b) 2 sin-1 x (c) 1t -2 cos-
1
x (d) 1t + 2 cos- 1 x
19. If cos ( sin-1 ; + cos-1 x) = O, then x is equal to
(a) .!_
5
(b) ~ (c) 0 (d) 1
5
20. Which of the followin g is the principal value branch of cosec- x? 1

(a) (-;,;) (b) (0, 1t) -g} (c) {- ; ,; } (d) [ 2-7t '27t l
-{O}
21. The domain of the function cos-1 (2.x-1) is
(a) [O, 1] (b) [- 1, 1] (c) (- 1, 1) (d) [O, 1t]
22. The domain of the function defined by f (x) = sin - l Jx -1 is
[NCERT Exemplar)
(a) [l, 2] (b) [- 1, 1] (c) [O, 1] (d) none of these
23. sin (tan-1 x), where Ix I < 1, is equal to [CBSE Sample Paper 2021-22 (Tenn-1))
X l l x
(a) Ji - x1- (b) /1 - xi (c) Ji + xi (d) [l+;i

24. sin[; +sin- 1 (½)] is equal to [CBSE 2023 (65/511))

(a) 1 1 1
(c) - (d) 4
3
Answers
1. (b) 2. (c) 3. (d) 4. (d) 5. (d) 6. (a) 7. (b)
8. (d) 9. (d) 10. (c) 11. (d) 12. (c) 13. (b) 14. (d)
15. (c) 16. (a) 17. (a) 18. (a) 19. (b) 20. (d) 21. (a)
22. (a) 23. (d) 24. (a)

50 Xam idea Mathematics-XII

solutions of Selected Multiple Choice Questions
1. We have,
1 1
sin- ( cos l~rr) = sin- [cos( 3rr - ~It )I
⇒
. -, [- cos 2rr ] =sin
sin
5
. ( rr
. - t [ -sm - 2rr )]
2 5
⇒ SITI , l()
, - I [ -Sm l , l·(10
IT =sin- 1 Sm - lt )] - lt
=1()

:. Option (b) is correct.

2. We have,
sin_, x > cos-1x
1[ . -I lt . -1 lt
sin_, x >- -sin-1 x ⇒ 2 sin x > ⇒ sin x >
⇒
2 2 4
⇒

1
As we know that sin-1x and cos- x exist for xe[-1, l].

xE(1,l)
⇒ ( 1, 1) C( ½,1)
:. Option (c) is correct.

3. We have, sm[;-sm-
1
(-½)]
• [ 1[ - ( -
3 - sin
. [
= sin IT . ( - 7[ ))]
. - I ( sm = sm 7t )]
6 3 6
. 7[ = 1
. [7[ + 7[1 =sm
=sin
3 6 2
: . Option (d) is correct.
4. Let sin- 1 x = 0, then sin 0 = x.
1 1
⇒ cosec0 = - ⇒ cosec 2 0 = -
X xi

✓1- .r2 I - .r~

⇒ 1 + cot2 0 = _!__ cot 0 = - - - ⇒ cot(sin- 1.r) = ' - - -
x x
x2
:. Option (d) is correct'.

5. sin - J ( cos ; ) = sin - J ( sin ( I -f))

• _ 1 ( • 7rr ) 7rr
=sm sm
18 =18
:. Option (d) is correct.
7. We have,

2 sec- 1 (2) + sin - 1 ( ½)

rr re 2n n 5n
::.- 2 X I- = + --
] 6 1 6 6

:. Option (b) is correct.

11 ·r 'Trit 0'10 t k r Jr'-.tr l'J 51

 :. Option (d) is correct.
1
10.
1
cot[½ sin- (~)] =cot[½x;] = cot- (;) = /3
:. Optio n (c) is correct.
b
11. L e t cot - I x = a ==> cot a = x = p

. p l
⇒ sma= - = - --
h + x2/1
2
⇒ Sin (cot-IX) = sin(a) = l = (x 2 + 1r1/
/1 +xi

: . Option (d) is correct.

12. ·: tan- 1 x = y (Given)

:. Option (c) is correct.

1
13. We h ave, y = cot- x, x < 0
⇒ x = coty < 0
th quadr ant, i.e. cot y lie in Ilnd o
As we know that cotangent (i.e. cot y) is negative in Ilnd and IV
IVth quadra nt.
rr 3rr
⇒ < y < 7t or < y < 2rr
2 2
:. Option (b) is correct.
1
15. We have, cos- a + cos- p + cos- y = 3n
1 1

We know that, 0 ~ cos- x ~ n

1

1 1
cos-1 a+ cos- p + cos- y = 3n
1 1
If and only if, cos- 1 a = cos- p = cos- y = n
⇒ cos n = a = p = y ⇒ - 1 = a = p= y
⇒ a = P=y= -1
a (P + y) + P(Y+a)+ y(a + P) = - 1(-1 - 1) -1(-1 - 1) -1(- 1 -1)
=2 + 2+2= 6
:. Option (c) is correc t.
. - I ( / 1 + COS= + /1 - COS
X-""""r== ==
X)
16 G1ven, tan ~ == COS X - / 1- COS X
. / 1+
r,;2 r,;2 . xl
x + v L. sm-
-v L. cos-
= tan - 1 2 2 h . 3rr
r,; x r,; . x , w ere rr < ., < 2
I - vL.cos - v 2sm
2 2

52 Xam idea Ma thematics-X II

 X , X X
-sm cos 1 - tan
= tan- 1 - ~2- -=-2 = tan- 1 2
X , X X
cos + sm 1 + tan 2
2 2
1
=tan- [tan( ~ - ~)]= ~ -½
Option (a) is correct.
17. /1 + cos 2.x = .fi. cos -1 (cos x)
y

Y'

⇒ /2cos 2 x=.fi.x,xE[;,rc]c[O,rc]
⇒ .fj_ COS X = .fj_ X
⇒ COS X =X ( · .' X E [ ; , TC])

As in the given figure graph of y = x and y = cos x does not intersect in [; , rr.] so the number of
solutions is zero.
: . Option (a) is correct.
18. cos- 1 (2x 2 -l)
Put x=cosa ⇒ a=cos-1 x
cos-1(2x2-l) = cos-1 (2 cos 2 a- l) ⇒ cos (- f )S cos a S cos 0
1 1
= cos- (cos 2 a)= 2a = 2 cos- x rr
⇒ -
:. Option (a) is correct. 2 S a S0
. -1 3 -] ) ⇒ - rr S 2a S 0
19.
5 +cos X = 0
COS ( Stn

⇒ sin-1 l + cos-1 x = cos- 1 (0) = ~ -1 TT . -I 3 -l 3

5 2 COS X =
2 - Stn S = COS S
⇒ x=cos(cos-
1
(¾))=¾ [·:¾<1]
⇒ x=s3
.-. Option (b) is correct.
20. Principal value branch of cosec- 1 x is [-; , ; ] - (0).
:. Option (d) is correct.
21. Let a = cos -I (2x - 1) ⇒ cos a = 2t - 1
- 1 S cos a S 1 ⇒ - 1 S 2t - 1 5- 1
⇒ 0 S 2x S 2 ⇒ 0 S x S ] ⇒ x E [0, 1]
Hence domain of cos- 1 (2x - 1) is [O, 1].
.-. Option (n) is correct.

l11vc1::.c TligonomctriL Functions 53

 22. /(x) = sin- 1 Jx - 1
⇒ O~x -1 ~1
⇒ l ~x ~ 2
X E [1, 2]
Option (a) is correct.
23. Let tan-1 x =0 => x =tan 0 => cot O= !
X
sin B_ 1 _ 1 _ 1 = X
- COSec O- /1 + cot2 0 - ✓l + 1 /1 + x2
x2
=> sin(tan- 1 x)= x
h+x2
Option (d) is correct.

Assertion-Reason Questions

The following questions consist of two statements-Assertion(A) and Reaso11(R). Answer thes,
questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A
(b) Both A and R are true but R is not tire correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A) : Domain offix)= sin-1 x + cos xis [-1, 1).
Reason (R) : Domain of a function is the set of all possible values for which function will b
defined.
2. Assertion (A) : Function f: R ➔ R given byf(.x) = sin x is not a bijection.
Reason (R) : A function f: A ➔ B is said to be bijection if it is one-one and onto.
3. Assertion (A) : Principal value of tan-1 (-/3) is - ; .

Reason (R) : tan- 1 : R ➔ (-f, ; )so for any x e R, tan- (x) represents an angle in (- ;, ; ) .
1

4. Assertion (A) : sin-I (-x) =-sin-Ix; x e [-1, 1]

Reason (R) : sin-I: [- 1, l] ➔ r ; , ; ] is a bijection map.

5. Assertion (A) : A function f : R ➔ R given by fix) = cos x is a bijection.
Reason (R) : A function g: A ➔ B is a bijection then 3 a function /1 : B ➔ A such that
goh = IO and hog = IA •
6. Assertion (A) : Range of [sin- 1 x + 2 cos-1 x] is [O, rr].

Reason (R) : Principlcvaluebranchofsin- 1 xhasrange [-f, f]. [CBSE 2023 (65/3/2)]

7. Assertion (A) : All trigonometric functions have their inverses over their respective domains.
Reason (R) : The inverse of tan- 1 .r exists for some x E R. [CBSE 2023 (65/411)]

54 y J Iil 1dc.i Muth1:111J t1c<..-XII

 3
8. Assertion (A) : The range of the function J(x) = 2 sin-1 x + ; , where x E [-1, 1 ], is [; , s; ].
Reason (R) : The range of the principal value branch of sin-1 (x) is [O, 1t]. [CBSE 2023 (65/111))
9. Assertion (A) : Maximum value of (cos- 1 x)2 is rr2.
Reason (R) : Range of the principal value branch of cos-1x is [- rr , ; ]. [CBSE 2023 (65/211)]
2
Answers
1. (a) 2. (a) 3. (a) 4. (b) 5. (d) 6. (d) 7. (d)
8. (c) 9. (c)

Solutions of Assertion-Reason Questions

1. The domain of sin- 1 xis [-1, 1] and that of cos xis JR .
1
:. Domain off(x) = sin- x + cos x is [-1, 1] n R = [-1, l].
Here, A and R are true and R gives correct explanation A.
: . Option (a) is correct.
2. For 0, 1t E R such that /(0) = 0 = f(n) . But 0 * n
So, f is not one-one.
R (f(x))= [-1, 1] * JR (co-domain)
⇒ f is not onto.
Therefore f is not a bijection map.
Here A and Rare true and R gives the correct explanation of A.
Option (a) is correct.

3. tan-1 : JR ➔ (-;,;) suchthattan-1 (x)=0,where 8 E ( -; , ;) & XE JR

tan- 1 (-/3)=8 => tan8=- /3=tan(- ; ) => 8= -;

Here, A and R are correct and R gives correct explanation of statement A.

: . Option (a) is correct.
4. Letsin-1 (- x)=y => -x=siny
⇒ x=-siny=sin(-y) ⇒ , -1
sm · -1 ( )
x=-y=-sm -x
⇒ sin-1 (-x) = - sin-1 x
A and R are correct but R does not give correct explanation of A.
:. Option (b) is correct.
re 3rr
5. For - - E JR
2' 2

rr = ~ 2rr) but ; i= 32rr

3 3
~;)=cos; = 0 = cos
2
:. f is not a one-one function. Also f is not onto. So, it is not a bijection.
Here, A is false and R is true.
: . Option (d) is correct.
7[ 7[

2 - cos- x + 2 cos- x
6. Let/(x) = sin- 1 x + 2 cos-1 x = 1 1
⇒ j(x) = + cos- 1 x
2
As O::; cos- 1 x :;; n
re 3rr
⇒ - <f(x) < -
2- - 2

lnve1 se Trigonometric Functi on s 55

 3
Range of/(x) = [; , ; ] .
Here, Assertion (A) is false but Reason (R) is true.
:. Option (d) is correct.
7. A is false but R is true as tan- 1 x exists for xE (-;,I)•
:. Option (d) is correct.
3rr
8. We have, f(x)=2sin- 1x+
2
2· 2rr
As -~<s•n- 1
2 - u , .,,· -<~
2 ⇒ -~<2sin-
2 -
1
x~ -2
2rr 3rr 3rr 2n 3rr
⇒
-2+2~ 2sin-'x+2~2+ 2
rr Srr
⇒ 2~f(x)~2

⇒ Range of [;, s; ].
Here, Assertion (A) is true but Reason (R) is false.
:. Option (c) is correct.
9. Let cos-1 x = 0 ⇒ x = cos 0
⇒ 0 e [0, 1t] ⇒ 0 :,; 0 :,; 1t
⇒ 0:,; 92:,; 7t2

⇒ Maximum value of 02 is 1r.2.

⇒ Maximum.value of (cos -\)2 is rr2•
⇒ Here, Assertion (A) is true but Reason (R) is false.
Option (c) is correct.

Case-based/Data-based Questions

Each of tlie following questions are of 4 marks.

1. Read the following passage and answer the following questions.

Two men on either side of a temple 30 meters high observe its top at the angles of elevation a an
prespectively. (as shown in the figure above). The distance between the two men is 40/3 metn
and the distance between the first person A and the temple is 30/3 metres.
[CBSE Q11estio11 811111

56 Xam idea Mathematics-XII

j
 (i) Find LCAB = a in tem,s of sin-1.
(ii) Find L CAB = a in tenns of cos-1.
(iii) (11) Find L BCJ\ =pin terms of tan-1.
OR
(iii) (I.,) Find the domain and range of cos-1 x.
Sol. We have, B

30 m

A C
30\IJm D 10\IJm

(1) Now in MBD (right angled)

BD 30 I I
tan a = - = - - = - ⇒ tan a= - =tan 30° ⇒ a =30°
AD 30 ./3 .fj ./3
⇒ sin a -- sin 30° = .!.
2

⇒ a=sin-1 ( ; )

./3
(i1) We have from (1) a = 30° ⇒ cos a =cos 30° = -
2
⇒ a= cos-1 ( ~ )
(iii) (11) In right t:.BCD, we have
BO
tanS= DC ⇒
.
tan a=_lQ_ = _l_ = ./3
10./3 ./3
⇒ S= tan-I ( ./3)
OR
1
(iii) (b) Let cos- x = y ⇒ x =cosy
-1 5 cosy 5 l ⇒ -1 5 x 5 1 ⇒ Domain = [-1, 1]
05 y5 1t ⇒ Range= [O, n].

2. Read the following passage and answer the following questions.

In a school project Manish was asked to construct a triangle ABC in which rn•o angles B and C are
given by tan-1 ( ½) and tan- 1
( !)
respectively.

(i) Find the value of sin B.

(ii) Find the value of cos C.
(iii) (a) Find the value of B + C.
OR
(iii) (b) Find the value of cos (B + C)

½) = B ⇒
1
Sol. We have, tan-1 ( - = tan B
2
1
tan-
1
( ½) = C ⇒ - = tan C
3

Im c1 se T1 i9on0rnetr11.. Funct1011c; 57
 (,) 11111 If I
I I J I ~} • l!,
(11) ,' 11•1 l'
.1 .I
I I J I _1 J 1/ l(l

I ,
(iii) (,1 ) ·: l,111 /I
..
') / li) ll (

l,111 /I I 11111 ( '

... l,111 (/1 I C)
I l,lll /11,lll C
I !'i ~
2
I
I 1
'
I
.. r, - I (1 It
(1
!i
, I

2 )( ;l (1 (i
/(
:- II I (' II II l I ( I ) ,,
OR
(iii) (/•) · · l-'rll111 (i) nnd (II) Wl' lrnv1•
I J
:-1111 /I C
l:i ' . l ' llS
/ 10

ms II .- / 1- sln J II O ,/ 1- !.1
tJ 7r.
v :1

sin c- / 1- ('llS~C / 1- ll:) a ✓:o O

ms( /l ,. C) - n1s /J cos C - sin /J aln C

2 3 I I 6 I
C / 5 )( / 10 - / 5 )( / 10 /50- /50 Cl

/ 25 ✓ I
5
= /§() = \ 50 ::, 2 a J2I
CONCEPTUAL QUESTIONS
I. \\' ri l,• lh,· pr indp .1' , .rhrl• of co-. (~ )+2sl1
1 '(~) . 1 ICBSE (F) 201

Sul. Wl• h,lVL', cus 1(~) =cos·' (cos ; )= f {-:;E IO,rrl/

Also, S Ill ·· ·'(··SIil !!)
·'( 2')-- Sll1 6
-~
-
6 /·: ~ El-f,; II
ms
1
( ~ 2si n ~ =; + 2(Ti) =;
) -,
1
( ) ·I· ; = ~rr
INot<·: l'rind p,1I v.1l uc br,1nclll's o( s1r1·
1
., · ond cos· ' .r ore /-;, ; / ond fO, rr] respectively. I

!. · \\' rik lh,· pri rr d p.rl v,iltrl 'o( l.111

1
J I- cos'(-~). ICBS£ Dc/1,i 20l

0....11 1. 1,111
1
1, co, '(-~) =1.111 '(1,111J) +cos· (cos(rr -; )) 1

- 1,1 11 I ( l,lll ~ ) I- COS - I ( COS ~IT)

1t
.,
27f
J I ( 7t 7() 2n I
·: ,7t1 E - 2, f .i nd J E l0, rr]
Jn I Hrr I Irr
12 12

58 111/ r f lllh lll.i f rc II