Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The domain of the function $$y=\sin^{-1}(-x^2)$$ is

A
$$[0, 1]$$

B
$$(0, 1)$$

C
$$[-1, 1]$$

D
$$\phi$$

Solution

(C) is the correct answer,

Given $$y=\sin^{-1}(-x^2) \Rightarrow \sin y =-x^2$$
$$\therefore\ -1 \le -x^2 \le 1$$ ( since $$-1 \le \sin y \le 1)$$
$$\Rightarrow 1 \ge x^2 \ge -1$$
$$\Rightarrow 0 \le x^2 \le 1$$
$$\Rightarrow |x| \le 1$$
$$\Rightarrow-1 \le x \le 1$$
Hence the domain is $$[-1, 1]$$
Question 2
1 / -0

The value of $$\sin (2\sin^{-1}(0.6))$$ is

A
$$0.48$$

B
$$0.96$$

C
$$1.2$$

D
$$\sin 1.2$$

Solution

(B) is the correct answer.

Let $$\sin^{-1}(0.6) =\theta \Rightarrow\sin \theta =0.6$$.
$$\therefore\ \sin (2\sin^{-1}(0.6))\\=\sin ( 2\theta )\\=2\sin \theta \cos \theta \\=2 \sin\theta\sqrt{1-\sin^2\theta}\\=2(0.6)(\sqrt{1-{0.6}^2})\\=2(0.6)(0.8)=0.96$$
Question 3
1 / -0

The value of $$\tan^2 ( \sec^{-1}2)+\cot^2 (\csc^{-1}3)$$ is

A
$$5$$

B
$$11$$

C
$$13$$

D
$$15$$

Solution

(B) is the correct answer.

We have
$$\tan^2 ( \sec^{-1}2) \cot^2 ( \csc^{-1}3) $$
$$=\sec^2 (\sec^{-1}2)-1+\csc^2 (\csc^{-1}3)-1\qquad[\because \tan^2\theta=\sec^2\theta-1,\ \ \ \cot^2\theta=\csc^2\theta-1]$$
$$=(2)^2-1+(3)^2-1=11$$
Question 4
1 / -0

$$\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2} , $$ then x

A
$$0 , \dfrac{1 }{2}$$

B
$$ 1, \dfrac{1}{2}$$

C
$$ 0 $$

D
$$ \dfrac{1 }{2}$$

Solution

$$\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2} $$
$$ -2\sin ^{-1} x = \dfrac{\pi }{2} - \sin^{-1} (1 -x)$$
$$-2 \sin ^{-1} x = \cos^{-1} (1 -x)$$
$$\cos ( -2 \sin^{-1} x) = 1 -x$$
$$\cos ( 2 \sin^{-1} x) = 1 -x$$
$$\cos (2 sin ^{-1} x) = 1 -x $$
$$ 1 - 2 \sin^{2} (sin^{-1} x) = 1 - x $$
$$1 - 2x^{2} = 1- x\Rightarrow $$
$$2x^{2} -x = 0 \\x(2x -1) = 0\\ \Rightarrow x = 0 , x = \dfrac{1}{2}$$
$$\dfrac{1}{2}$$ does not satisfy the equation $$\therefore x = 0$$ is the only solution
Question 5
1 / -0

Choose the correct answer :
$$ \sin \left ( \dfrac{\pi }{3} - \sin^{-1}\left ( -\dfrac{1}{2} \right ) \right ) $$ is equal to

A
$$ \pi $$

B
$$ - \dfrac{\pi }{2} $$

C
$$ 1 $$

D
$$ 2 \sqrt{3}$$

Solution

$$ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] $$
$$ \left [ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] \right ] $$
$$ \sin \left ( \dfrac{\pi }{3} + \sin^{-1}\left ( \sin \dfrac{\pi }{6} \right ) \right ) $$
$$\left ( as \sin^{-1} \left ( \dfrac{-1}{2} \right ) = \sin^{-1}\dfrac{1}{2} and \sin \dfrac{\pi }{6} = \dfrac{1}{2} \right ) $$

$$ = \sin \left ( \dfrac{\pi }{3} + \dfrac{\pi }{6} \right ) $$

$$ = \sin\dfrac{\pi }{2} = 1 $$
Question 6
1 / -0

Choose the correct answer
$$ \cos ^{-1} ( \cos \dfrac{7\pi }{6}) $$ is equal to

A
$$ \dfrac{7\pi }{6} $$

B
$$ \dfrac{5\pi }{6} $$

C
$$ \dfrac{\pi }{3} $$

D
$$ \dfrac{\pi }{6} $$

Solution

$$ \cos^{-1} \cos\left ( \dfrac{7\pi }{6} \right )$$
$$ \cos^{-1} \left ( \cos\left ( \pi + \dfrac{\pi }{6} \right ) \right ) = \cos^{-1} \left ( - \cos\dfrac{\pi}{6} \right ) $$
$$ =\pi- \cos^{-1} \left ( \cos\dfrac{\pi }{6} \right ) $$
$$=\pi-\dfrac{\pi}{6}$$
$$ = \dfrac{5\pi }{6} $$
Question 7
1 / -0

If $$ \sin ^{-1} x = y $$ , then

A
$$ 0 \leq y \leq \pi $$

B
$$ - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2} $$

C
$$ 0 < y < \pi $$

D
$$ - \dfrac{\pi}{2} < y < \dfrac{\pi }{2} $$

Solution

pv of $$ \sin^{-1} x$$ is $$\left [ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right ] $$
$$\therefore (B) - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2} $$ is correct
Question 8
1 / -0

$$ \tan^{-1} \sqrt{3} - \sec^{-1} (-2) $$ is equal to

A
$$ \pi $$

B
$$ - \dfrac{\pi }{3} $$

C
$$ \dfrac{\pi }{3} $$

D
$$ - \dfrac{2\pi }{3} $$

Solution

$$ \tan^{-1} \sqrt{3} = \dfrac{\pi }{3} \epsilon \left [ \dfrac{-\pi }{2} ,\dfrac{\pi }{2} \right ]$$
$$ \sec^{-1} (-2) = \pi - \dfrac{\pi }{3} = \dfrac{2 \pi}{3} \epsilon [0, \pi] $$
$$\therefore \tan^{-1} \sqrt{3} - sec^{-1} (-2) = \dfrac{\pi}{3} - \dfrac{2 \pi}{3} $$
$$ = -\dfrac{\pi}{3} $$
$$ \therefore $$ Option (B) is correct .
Question 9
1 / -0

Multiple choice Questions :
$$ 2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$$

A
$$ \dfrac{24}{25}$$

B
$$ - \dfrac{24}{25}$$

C
$$ - \dfrac{6}{25}$$

D
$$ \dfrac{- 6}{25}$$

Solution

$$ 2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$$
$$ = 2 \sqrt{1-\left ( \dfrac{-4}{5} \right )^{2}} \times\left ( \dfrac{-4}{5} \right )$$
$$= -\dfrac{8}{5} \sqrt{\dfrac{9}{25}} = \dfrac{-8}{5} \times\dfrac{3}{5} = \dfrac{-24}{25}$$
Question 10
1 / -0

Multiple choice Questions :
The value of $$ \tan^{-1} (\tan 5)$$

A
$$ 2 \pi $$

B
$$ 5 - 2 \pi $$

C
$$ 5 $$

D
$$ \dfrac{2 \pi}{3} $$

Solution

Let $$ 2\pi - 5 = \theta $$
$$ \Rightarrow 5 = 2 \pi - \theta $$
$$ \tan 5 = \tan ( 2 \pi - \theta )$$
$$ \tan^{-1} (\tan 5) = \tan^{-1} \tan ( 2 \pi - \theta)$$
$$ = \tan^{-1}[-\tan \theta]=-\theta$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 49

Result

Inverse Trigonometric Functions Test - 49

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SHARING IS CARING

Question 1

$$[0, 1]$$

$$(0, 1)$$

$$[-1, 1]$$

$$\phi$$

Solution

Question 2

$$0.48$$

$$0.96$$

$$1.2$$

$$\sin 1.2$$

Solution

Question 3

$$5$$

$$11$$

$$13$$

$$15$$

Solution

Question 4

$$0 , \dfrac{1 }{2}$$

$$ 1, \dfrac{1}{2}$$

$$ 0 $$

$$ \dfrac{1 }{2}$$

Solution

Question 5

$$ \pi $$

$$ - \dfrac{\pi }{2} $$

$$ 1 $$

$$ 2 \sqrt{3}$$

Solution

Question 6

$$ \dfrac{7\pi }{6} $$

$$ \dfrac{5\pi }{6} $$

$$ \dfrac{\pi }{3} $$

$$ \dfrac{\pi }{6} $$

Solution

Question 7

$$ 0 \leq y \leq \pi $$

$$ - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2} $$

$$ 0 < y < \pi $$

$$ - \dfrac{\pi}{2} < y < \dfrac{\pi }{2} $$

Solution

Question 8

$$ \pi $$

$$ - \dfrac{\pi }{3} $$

$$ \dfrac{\pi }{3} $$

$$ - \dfrac{2\pi }{3} $$

Solution

Question 9

$$ \dfrac{24}{25}$$

$$ - \dfrac{24}{25}$$

$$ - \dfrac{6}{25}$$

$$ \dfrac{- 6}{25}$$

Solution

Question 10

$$ 2 \pi $$

$$ 5 - 2 \pi $$

$$ 5 $$

$$ \dfrac{2 \pi}{3} $$

Solution

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