Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 23

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Question 1
1 / -0

$$\sin^{-1}{0}$$ is equal to:

A
$$0$$

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

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

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

Solution

As we know that,
$$\sin{0} = 0$$
$$\Rightarrow 0 = \sin^{-1}{\left( 0 \right)}$$
Hence the value of $$\sin^{-1}{\left( 0 \right)}$$ is $$0$$.
Question 2
1 / -0

The value of $$x$$ for which $$\sin { \left( \cot ^{ -1 }{ \left( 1+x \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$ is

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

B
$$1$$

C
$$0$$

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

Solution

We have, $$\displaystyle \sin { \left( \cot ^{ -1 }{ \left( x+1 \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$
$$\displaystyle \Rightarrow \cos { \left( \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$
$$\displaystyle \Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =2n\pi \pm \tan ^{ -1 }{ x } $$
Put $$\displaystyle n=0\Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =\pm \tan ^{ -1 }{ x } =\tan ^{ -1 }{ \left( \pm x \right) } $$
$$\displaystyle \Rightarrow \frac { \pi }{ 2 } =\tan ^{ -1 }{ \left( \pm x \right) } +\cot ^{ -1 }{ \left( x+1 \right) } $$
$$\displaystyle \Rightarrow x+1=\pm x\Rightarrow 2x+1=0$$
$$\displaystyle \therefore x=-\frac { 1 }{ 2 } $$
Question 3
1 / -0

Assertion (A) : The maximum value of $$f(x)=\sin^{-1}x+\cos^{-1}x+\tan^{-1}x$$ is $$\displaystyle \frac{3\pi}{4}$$
Reason (R) : $$\sin ^{-1} x>\cos^{-1}x$$ for all $$x$$ in $$R$$

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true and R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$f(x)=sin^{-1}(x)+cos^{-1}(x)+tan^{-1}(x)$$
Now
$$sin^{-1}(x)+cos^{-1}(x)=\dfrac{\pi}{2}$$
Hence
$$f(x)=tan^{-1}(x)+\dfrac{\pi}{2}$$
Substituting x=1, we get the maximum as
$$\dfrac{\pi}{4}+\dfrac{\pi}{2}$$

$$=\dfrac{3\pi}{4}$$
Reason.
The intersection point of sin(x) and cos(x) is $$x=\dfrac{\pi}{4}$$ between $$[0,\dfrac{\pi}{2}]$$.
Now cos(x) is a decreasing function in the above interval while sin(x) is an increasing function in the above interval.
Also.
$$cos(x)>sin(x)$$ between $$[0,\dfrac{\pi}{4}]$$ while

$$sin(x)>cos(x)$$ between $$[\dfrac{\pi}{4},\dfrac{\pi}{2}]$$
Hence
$$sin^{-1}(x)>cos^{-1}(x)$$ between $$[\dfrac{1}{\sqrt{2}},1]$$.
Hence reason is false.
Question 4
1 / -0

The number of solutions of the equation

$$2(Sin^{-1}x)^{2}-5Sin^{-1}x+2=0$$ is

A
0

B
1

C
2

D
3

Solution

Let $$sin ^{-1}x=t , t\epsilon [-\frac{\pi }{2} ,\frac{\pi }{2}]$$
$$\therefore 2(sin ^{-1}x)^{2}-5 sin ^{2}x+2=0$$
$$2t^{2}-5t+2=0$$
$$\displaystyle t=\frac{5\pm \sqrt{25-16}}{4}$$
$$\displaystyle =\frac{5\pm 3}{4}$$
$$t=2, \frac{1}{2}$$
$$sin ^{-1}x=2, \frac{1}{2}$$
$$sin ^{-1}x=2 or sin ^{-1}x=\frac{1}{2}$$
$$x=sin (\frac{1}{2}) and x=sin2$$ satisfy the equations.
Question 5
1 / -0

The number of triplets $$(x,y,z)$$ satisfying $$\sin^{-1}x+\sin^{-1}y+\cos^{-1}z=2\pi$$ is

A
$$1$$

B
$$0$$

C
$$2$$

D
$$\infty$$

Solution

$$\sin ^{-1}x+\sin ^{-1}y+\cos ^{-1}z=2\pi $$

We have,

$$\dfrac{-\pi}{2}\leq \sin^{-1} x \leq \dfrac {\pi}{2} $$

$$\dfrac{-\pi}{2}\leq \sin^{-1} y \leq \dfrac {\pi}{2} $$

and $$ 0 \leq\cos^{-1} z \leq \pi $$

thus for $$RHS$$ to become $$2\pi$$ we need to take the maximum values only where,
$$\sin^{-1}x=\sin^{-1}y=\dfrac{\pi}{2}$$
$$\cos^{-1}z=\pi$$

Hence,
$$x = y = 1 $$
$$z = -1 $$

$$\therefore (x,y,z)=(1,1,-1)$$

There is only one triplet satisfying the given equation.
Hence, option A is correct.
Question 6
1 / -0

The value of $$x$$ where $$x>0$$ $$\displaystyle \tan(\sec^{-1}\frac{1}{x})=\sin(\tan^{-1}2)$$ is

A
$$\sqrt{5}$$

B
$$\displaystyle \frac{\sqrt{5}}{3}$$

C
$$1$$

D
$$\displaystyle \frac{2}{3}$$

Solution

$$tan(sec^{-1}{\dfrac{1}{x}})=sin(tan^{-1}(2))$$
$$tan(tan^{-1}(\dfrac{\sqrt{1-x^2}}{x}))=sin(sin^{-1}(\dfrac{2}{\sqrt{5}}))$$
$$\dfrac{\sqrt{1-x^2}}{x}=\dfrac{2}{\sqrt{5}}$$
$$\dfrac{1-x^2}{x^2}=\dfrac{4}{5}$$
$$\dfrac{1}{x^2}-1=\dfrac{4}{5}$$
$$\dfrac{1}{x^2}=\dfrac{9}{5}$$
$$x^2=\dfrac{5}{9}$$
$$x=\pm\dfrac{\sqrt{5}}{3}$$
Since $$x>0$$
$$x=\dfrac{\sqrt{5}}{3}$$
Question 7
1 / -0

The ascending order of $$A=\sin^{-1}(\log_{3}{2})$$ , $$B=\displaystyle \cos^{-1}\left(\log_{3}\left(\frac{1}{2}\right)\right)$$ , and $$C=\tan^{-1}\left(\log_{1/3} 2 \right)$$ is

A
$$\text{C, B, A}$$

B
$$\text{B, A, C}$$

C
$$\text{C, A, B}$$

D
$$ \text {B, C, A}$$

Solution

We have, $$A = \displaystyle \frac{1}{2} < \ log_{3}{2} < 1, \quad \because \sqrt 3 < 2 <3$$

$$\Rightarrow \displaystyle \frac{\pi}{6} < \sin^{-1} \left( \log_3 2\right) < \frac{\pi}{2}$$

$$\displaystyle \because \sin^{-1}{\frac{1}{2}} = \frac{\pi}{6}$$, $$\displaystyle \sin^{-1} {1} = \frac{\pi}{2}$$

$$\displaystyle B=\cos^{-1}(\log_{3}{1/2})=\cos^{-1}\left(-\log_{3}{2}\right)$$

$$=\pi -cos ^{-1}(\log _{3}{2}) \quad [\cos^{-1}1/2 =60^{o} , \cos^{-1}(1)=0^{\circ}]$$

$$C=\tan^{-1}(\log _{1/3}{2})=\tan^{-1}(-\log_{3}{2})=- \tan^{-1}(\log_{3}{2}) \quad [\tan^{-1}(1/2) > 0^{\circ} , \tan^{-1}(1)=45^{\circ}]$$

$$C \text{ is } negative , 30^{o} < A < 90^{0} , \dfrac{2\pi }{3} < B < \pi $$
Arrange By asscending order

$$\Rightarrow \text{C, A, B}$$
Question 8
1 / -0

Assertion ($$A$$) lf $$0<\displaystyle x<\frac{\pi}{2}$$ then $$\sin^{-1}(cosx)+\cos^{-1}(sinx)=\pi-2x$$
Reason (R) $$\displaystyle \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\forall x\in[0,1]$$

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but Ris false

D
A is false but R is true

Solution

Given

$$ \sin ^ {-1} (\cos x) +\cos ^{-1} (\sin x)$$

$$ \sin ^ {-1} \left(\sin \left(\dfrac\pi 2- x\right)\right) +\cos ^{-1} \left(\cos \left(\dfrac\pi 2- x\right)\right)$$

$$\bigg( \sin \theta =\cos \left(\dfrac\pi 2- \theta \right) \bigg)$$

$$ \dfrac \pi 2-x +\dfrac \pi 2-x $$

$$ \pi -2x $$

We know

$$ \sin ^ {-1} x+\cos ^ {-1} x=\dfrac \pi 2 $$

$$ \cos ^ {-1} x=\dfrac \pi 2 -\sin ^ {-1} x$$
Question 9
1 / -0

The smallest and the largest values of
$$\displaystyle \tan^{-1}\left (\dfrac{1-x}{1+x}\right)$$ , $$0\leq x\leq 1$$ are.

A
$$ 0,\pi$$

B
$$0,\displaystyle \dfrac{\pi}{4}$$

C
$${-\dfrac{\pi}{4},\dfrac{\pi}{4}}$$

D
$$\displaystyle \dfrac{\pi}{4},\dfrac{\pi}{2}$$

Solution

$$\displaystyle tan ^{-1}(\frac{1-x}{1+x}) $$

$$0 \leq x\leq 1$$

$$1\leq x+1 \leq 2$$

$$1\geq \frac{1}{1+x} \geq \frac{1}{2}$$

$$\displaystyle 0 \leq \frac{1-x}{1+x}\leq 1$$

$$\therefore x\epsilon [-1,1] \space by\space x\epsilon [0,1]$$

So, $$x\epsilon [0,1]$$

$$\displaystyle tan ^{-1}(\frac{1-x}{1+x}) \space \epsilon [0,\frac{\pi }{4}]$$
Question 10
1 / -0

The equation 2$$\displaystyle \cos^{-1}x+\sin^{-1}x=\frac{11\pi}{6}$$ has

A
No solution

B
One solution

C
Two solutions

D
Three solutions

Solution

$$\displaystyle 2 cos^{-1}x+sin ^{-1}x=\frac{11\pi }{6}$$
$$\displaystyle \therefore cos ^{-1}x=\frac{11\pi }{6}-\frac{\pi }{2}$$ Since $$[sin ^{-1}x+cos ^{-1}x=\frac{\pi }{2}]$$
$$\displaystyle cos ^{-1}x=\frac{8\pi }{6}=\frac{4\pi }{3}$$
But range of $$cos^{-1}x\epsilon [0,\pi ]$$
So, this equation has no solution.

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Inverse Trigonometric Functions Test - 23

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Inverse Trigonometric Functions Test - 23

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Question 1

$$0$$

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

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

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

Solution

Question 2

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

$$1$$

$$0$$

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

Solution

Question 3

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 4

0

1

2

3

Solution

Question 5

$$1$$

$$0$$

$$2$$

$$\infty$$

Solution

Question 6

$$\sqrt{5}$$

$$\displaystyle \frac{\sqrt{5}}{3}$$

$$1$$

$$\displaystyle \frac{2}{3}$$

Solution

Question 7

$$\text{C, B, A}$$

$$\text{B, A, C}$$

$$\text{C, A, B}$$

$$ \text {B, C, A}$$

Solution

Question 8

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but Ris false

A is false but R is true

Solution

Question 9

$$ 0,\pi$$

$$0,\displaystyle \dfrac{\pi}{4}$$

$${-\dfrac{\pi}{4},\dfrac{\pi}{4}}$$

$$\displaystyle \dfrac{\pi}{4},\dfrac{\pi}{2}$$

Solution

Question 10

No solution

One solution

Two solutions

Three solutions

Solution

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