Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 34

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Question 1
1 / -0

If $$2\sinh ^{ -1 }{ \left( \dfrac { a }{ \sqrt { 1-{ a }^{ 2 } } } \right) } =\log { \left( \dfrac { 1+x }{ 1-x } \right) }$$, then $$x=$$

A
$$a$$

B
$$\dfrac { 1 }{ a } $$

C
$$\sqrt { 1-{ a }^{ 2 } } $$

D
$$\dfrac { 1 }{ \sqrt { 1-{ a }^{ 2 } } } $$

Solution

As we know that
$$\sinh^{-1} x=\log(x+\sqrt{x^2+1})$$
$$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=2\log\bigg(\dfrac{a}{\sqrt{1-a^2}}+\sqrt{\dfrac{a^2}{1-a^2}+1}\bigg)$$
$$=2\log \bigg(\dfrac{a+\sqrt{a^2+1-a^2}}{\sqrt{1-a^2}}\bigg)$$
$$=2\log \bigg(\dfrac{a+1}{\sqrt{1-a^2}}\bigg)$$
$$=\log\bigg(\dfrac{(1+a)^2}{(1-a)(1+a)}\bigg)$$
$$=\log \bigg(\dfrac{1+a}{1-a}\bigg)$$
But given $$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=\log \bigg(\dfrac{1+x}{1-x}\bigg)$$
So $$x=a$$
Question 2
1 / -0

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

A
$$\pi$$

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

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

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

Solution

$$E = \pi - cos^{-1} (1) - sin^{-1} (1) = \pi - \dfrac{\pi}{2} = \dfrac{\pi}{2}$$

$$\because cos^{-1} (-x) = \pi - cos^{-1} x$$

and $$cos^{-1} x + sin^{-1} x = \dfrac{\pi}{2}$$
Question 3
1 / -0

Range of the function $$f(x)=4{ tan }^{ -1 }x+3{ sin }^{ -1 }x+{ sec }^{ -1 }x$$ is

A
$$\left\{ \frac { -3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\} $$

B
$$\left\{ \frac { 3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\} $$

C
$$\left\{ \frac { 3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\} $$

D
$$\left\{ \frac { -3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\} $$

Solution
Question 4
1 / -0

If $$\sin \left( \sin^{-1} \dfrac{1}{5} + \cos^{-1} x \right) = 1$$, then x is equal to

A
$$1$$

B
$$0$$

C
$$\dfrac{4}{5}$$

D
$$\dfrac{1}{5}$$

Solution

$$\sin\left(\sin^{-1}\dfrac{1}{5}+\cos^{-1}x\right)=1...........(1)$$
$$\because \sin\dfrac{\pi }{2}=1$$
$$\therefore \sin^{-1}\dfrac{1}{5}+\cos^{-1}x=\dfrac{\pi }{2}........$$ from $$(1)$$
$$\therefore \cos^{-1}x =\dfrac{\pi }{2}- \sin^{-1}\dfrac{1}{5}$$
$$\therefore \cos^{-1}x= \cos^{-1}\dfrac{1}{5}......\left(\dfrac{\pi }{2}- \sin^{-1}x= \cos^{-1}x\right)$$
$$\therefore x=\dfrac{1}{5}$$
Question 5
1 / -0

Number of solution of the equation $$\cos^{-1}(1-x)-2\cos^{-1}x=\dfrac{\pi}{2}$$ is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution
Question 6
1 / -0

If $$\sin ^{ -1 }{ \left( x-\displaystyle\frac { { x }^{ 2 } }{ 2 } +\displaystyle\frac { { x }^{ 3 } }{ 4 } -....\infty \right) } +\cos ^{ -1 }{ \left( { x }^{ 2 }-\displaystyle\frac { { x }^{ 4 } }{ 2 } +\displaystyle\frac { { x }^{ 6 } }{ 4 } -....\infty \right) } =\displaystyle\frac { \pi }{ 2 } $$ and $$0<x<\sqrt { 2 } $$ then $$x$$ =

A
$$\displaystyle\frac { 1 }{ 2 } $$

B
$$1$$

C
$$\displaystyle\frac{- 1 }{ 2 } $$

D
$$-1$$

Solution

$$x-\displaystyle\frac { { x }^{ 2 } }{ 2 } +\displaystyle\frac { { x }^{ 3 } }{ 4 } -....\infty $$ = $${ x }^{ 2 }-\displaystyle\frac { { x }^{ 4 } }{ 2 } +\displaystyle\frac { { x }^{ 6 } }{ 4 } -....\infty $$
$$\quad x={ x }^{ 2 }\Rightarrow x=1$$
Question 7
1 / -0

If $$ a sin^{-1} x -b cos^{-1}x =c$$, then the value of $$ a sin ^{-1} x + b cos^{-1} x$$ (whenever exists) is equal to

A
$$0$$

B
$$\frac{\pi ab + c(b-a)}{a+b}$$

C
$$\frac{\pi}{2}$$

D
$$\frac{\pi ab + c(a-b)}{a+b}$$

Solution

As we know that
$$\sin^{-1} x+\cos^{-1} x=\dfrac{\pi}{2}$$

Given $$a\sin^{-1} x-b\cos^{-1} x=c$$

$$\implies a\sin^{-1} x-b(\frac{\pi}{2}-\sin^{-1} x)=c$$

$$\implies (a+b)\sin^{-1} x=c+\dfrac{b\pi}{2}$$

$$\implies \sin^{-1} x=\dfrac{2 c+b\pi}{2(a+b)}$$

$$a\sin^{-1} x+b\cos^{-1} x=a\sin^{-1} x+b(\frac{\pi}{2}-\sin^{-1} x)$$

$$=(a-b)\sin^{-1} x+b\dfrac{\pi}{2}$$

$$=\dfrac{(a-b)(2 c+b\pi)}{2(a+b)}+\dfrac{b\pi}{2}$$

$$=\dfrac{2 c(a-b)+b\pi(a-b+a+b)}{2(a+b)}$$

$$=\dfrac{c(a-b)+a b \pi}{(a+b)}$$
Question 8
1 / -0

The set of values of $$'x'$$ for which the formula $$2 \sin^{-1}x=\sin^{-1} (2x\sqrt{1-x^{2}})$$ is true, is

A
$$(-1,0)$$

B
$$[0,1]$$

C
$$\left[-\dfrac{\sqrt{3}}{2},\dfrac{\sqrt{3}}{2}\right]$$

D
$$\left[-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right]$$

Solution

Consider the given equation,
$$2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$$

Put, $$x=\sin \theta \,\,\Rightarrow \theta ={{\sin }^{-1}}x$$
$$ 2{{\sin }^{-1}}\sin \theta ={{\sin }^{-1}}\left( 2\sin \theta \sqrt{1-{{\sin }^{2}}\theta } \right) $$
$$ 2\theta ={{\sin }^{-1}}\left( 2\sin \theta \sqrt{{{\cos }^{2}}\theta } \right)\,\,\,\,\,\,\,\,\left( \because \,{{\sin }^{2}}A+{{\cos }^{2}} \right) $$
$$ 2\theta ={{\sin }^{-1}}\left( 2\sin \theta \cos \theta \right)\,\, $$
$$ 2\theta ={{\sin }^{-1}}\sin 2\theta $$
$$ 2\theta =2\theta \,\,\,\,\,\, $$
$$\sin x=\sin x$$

Now,
$$ \because \,-1\le\sin \ x\ge1 $$
$$ \therefore \,\,-\dfrac{\pi }{2}\le\ x \ge\dfrac{\pi }{2} $$
$$ x \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] $$

Hence, this is the answer.
Question 9
1 / -0

If $$x=\cos^{ - 1}\left( \dfrac{2}{3} \right) + \tan^{ - 1}\left( \dfrac{1}{7} \right)$$ then $$x$$=

A
$$ {{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\} $$

B
$$ {{\cos }^{-1}}\left\{ \dfrac{10-\sqrt{5}}{3\sqrt{50}} \right\} $$

C
$$ {{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{15}}{3\sqrt{50}} \right\} $$

D
None of these

Solution

We have,

$${{\cos }^{-1}}\dfrac{2}{3}+{{\tan }^{-1}}\dfrac{1}{7}$$

We know that,

$${{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$$

Therefore,

$$ ={{\cos }^{-1}}\dfrac{2}{3}+{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{1+{{\left( \dfrac{1}{7} \right)}^{2}}}} \right) $$

$$ ={{\cos }^{-1}}\dfrac{2}{3}+{{\cos }^{-1}}\left( \dfrac{7}{\sqrt{50}} \right) $$

We know that,

$${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left\{ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right\}$$

Therefore,

$$ ={{\cos }^{-1}}\left\{ \dfrac{2}{3}\times \dfrac{7}{\sqrt{50}}-\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{7}{\sqrt{50}} \right)}^{2}}} \right\} $$

$$ ={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\sqrt{\dfrac{5}{9}}\sqrt{\dfrac{1}{50}} \right\} $$

$$ ={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\dfrac{\sqrt{5}}{3\sqrt{50}} \right\} $$

$$ ={{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\} $$

Hence, this is the answer.
Question 10
1 / -0

The set for which $$2{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)$$ is valid is

A
$$x \in \left[ {0,\,1} \right]$$

B
$$x \in \left( {0,\,1} \right)$$

C
$$x \in \left[ {0,\,1} \right)$$

D
$$x \in \left( {0,\,1} \right]$$

Solution

$$2\cos ^{-1}x =\cos^{-1} (2x^2 -1)$$
Range of $$\cos^{-1}f(x)$$ is $$[0, \pi]$$

$$0\le\cos^{-1}f(x)\le \pi$$
$$0\le\cos^{-1}(2x^2-1)\le\pi$$
$$0\le 2\cos^{-1}{x}\le\pi$$
then $$0 \le \cos^{-1}x < \dfrac {\pi}{2}$$
$$x \in [0, 1]$$
$$D$$ is correct

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Inverse Trigonometric Functions Test - 34

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Inverse Trigonometric Functions Test - 34

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Question 1

$$a$$

$$\dfrac { 1 }{ a } $$

$$\sqrt { 1-{ a }^{ 2 } } $$

$$\dfrac { 1 }{ \sqrt { 1-{ a }^{ 2 } } } $$

Solution

Question 2

$$\pi$$

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

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

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

Solution

Question 3

$$\left\{ \frac { -3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\} $$

$$\left\{ \frac { 3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\} $$

$$\left\{ \frac { 3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\} $$

$$\left\{ \frac { -3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\} $$

Solution

Question 4

$$1$$

$$0$$

$$\dfrac{4}{5}$$

$$\dfrac{1}{5}$$

Solution

Question 5

$$3$$

$$2$$

$$1$$

$$0$$

Solution

Question 6

$$\displaystyle\frac { 1 }{ 2 } $$

$$1$$

$$\displaystyle\frac{- 1 }{ 2 } $$

$$-1$$

Solution

Question 7

$$0$$

$$\frac{\pi ab + c(b-a)}{a+b}$$

$$\frac{\pi}{2}$$

$$\frac{\pi ab + c(a-b)}{a+b}$$

Solution

Question 8

$$(-1,0)$$

$$[0,1]$$

$$\left[-\dfrac{\sqrt{3}}{2},\dfrac{\sqrt{3}}{2}\right]$$

$$\left[-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right]$$

Solution

Question 9

$$ {{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\} $$

$$ {{\cos }^{-1}}\left\{ \dfrac{10-\sqrt{5}}{3\sqrt{50}} \right\} $$

$$ {{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{15}}{3\sqrt{50}} \right\} $$

None of these

Solution

Question 10

$$x \in \left[ {0,\,1} \right]$$

$$x \in \left( {0,\,1} \right)$$

$$x \in \left[ {0,\,1} \right)$$

$$x \in \left( {0,\,1} \right]$$

Solution

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