Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \tan^{-1}\left [ \cfrac{\cos\:x}{1+\sin\:x} \right ]$$ is equal to

A
$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$$

B
$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$

C
$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)$$

D
$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{3\pi}{2},-\frac{3\pi}{2}\right)$$

Solution

$$ \displaystyle \tan^{-1}\left[\frac{\sin(\frac{\pi}{2}-x)}{1+\cos(\frac{\pi}{2}-x)}\right]$$

$$ \displaystyle = \tan^{-1}\left[\cfrac{2\sin(\frac{\pi}{4}-\frac{x}{2}).\cos(\frac{\pi}{4}-\frac{x}{2})}{2\cos^{2}(\frac{\pi}{4}-\frac{x}{2})}\right]$$

$$ \displaystyle =\tan^{-1}\left[\tan(\frac{\pi}{4}-\frac{x}{2})\right]$$
$$ \displaystyle =\cfrac{\pi}{4}-\frac{x}{2}$$
where,
$$ \displaystyle -\frac { \pi }{ 2 } < \frac { \pi }{ 4 } -\frac { x }{ 2 } < \frac { \pi }{ 2 } \\ \displaystyle \Rightarrow -\frac { \pi }{ 2 } < x < \frac { 3\pi }{ 2 } $$
Question 2
1 / -0

The value of $$\displaystyle tan(sin^{-1}(cos(sin^{-1}x)))tan(cos^{-1}(sin(cos^{-1}x)))$$, where $$x\:\epsilon\:(0,1)$$, is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$none\:of\:these$$

Solution

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

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

Hence, option B is correct.
Question 3
1 / -0

The value of $$\displaystyle sin^{-1}(x^{2}-4x+6)+cos^{-1}(x^{2}-4x+6)$$ for all $$x\:\epsilon\:R$$ is

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

B
$$\pi$$

C
$$0$$

D
none of these

Solution

$$\sin^{-1}(x)\epsilon[-1,1]$$
Therefore
$$-1\leq x^{2}-4x+6\leq 1$$
$$-1\leq (x-2)^{2}-4+6\leq 1$$
$$-1\leq (x-2)^{2}+2\leq 1$$
$$-3\leq (x-2)^{2}\leq -1$$
Now, $$(x-2)^{2}>0$$ for all $$x\epsilon R$$.
Hence the inequality
$$-3\leq (x-2)^{2}\leq -1$$ yields no real real values of x.
Therefore,
$$|x^{2}-4x+6|$$ cannot be less than or equal to 1.
Hence $$\sin^{-1}(x^{2}-4x+6)$$ is no possible.
The, same goes for $$\cos x$$.
Hence the answer is none of these.
Question 4
1 / -0

Which of the following is the solution set of the equation $$\displaystyle 2cos^{-1}x=cot^{-1}\left(\frac{2x^{2}-1}{2x\sqrt{1-x^{2}}}\right)$$

A
$$(0,1)$$

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

C
$$(-1,0)$$

D
$$[-1,1]$$

Solution

$$2{ cos }^{ -1 }x={ cos }^{ -1 }\left( { 2 }x^{ 2 }-1 \right) $$ for $$x\epsilon \left[ 0,1 \right] $$
And $${ cos }^{ -1 }\left( { 2 }x^{ 2 }-1 \right) ={ cot }^{ -1 }\left( \cfrac { { 2 }x^{ 2 }-1 }{ 2x\sqrt { 1-{ x }^{ 2 } } } \right) $$ for $$x\epsilon \left( 0,1 \right) $$
Question 5
1 / -0

If $$\displaystyle \cos^{-1} x = a + \tan^{-1} \frac{\sqrt{1 - x^{b}}}{cx}$$ for $$\displaystyle x < 0$$.
Find the value of a ,b and c.

A
$$a=\pi,b=1,c=1$$

B
$$a=-\pi,b=2,c=2$$

C
$$a=-\pi,b=1,c=2$$

D
$$a=\pi,b=2,c=1$$

Solution

Let $$x = \cos \theta $$

$$\Rightarrow \tan \theta = \cfrac{\sin\theta}{\cos\theta}=\cfrac{\sqrt{1-x^2}}{x}$$ (i)

$$\therefore \cos^{-1}x = \pi+\theta $$ since $$x< 0$$

$$= \pi+\tan^{-1}\cfrac{\sqrt{1-x^2}}{x}$$ using (i)
Comparing with
$$\displaystyle \cos^{-1} x = a + \tan^{-1} \frac{\sqrt{1 - x^{b}}}{cx}$$ for $$x<0$$

$$a=\pi, b=2 ,c=1$$
Question 6
1 / -0

The equation $$\displaystyle 3cos^{-1}x-\pi\:x-\frac{\pi}{2}=0$$ has

A
one negative solution

B
one positive solution

C
no solution

D
more than one solution

Solution

Substituting $$x=\frac{1}{2}$$ , we get
$$3cos^{-1}(\frac{1}{2})-\frac{\pi}{2}-\frac{\pi}{2}$$
$$=3(\frac{\pi}{3})-\pi$$
$$=0$$
Hence there is one positive real solution to the above equation since range of $$cos^{-1}(x)\epsilon[\pi,0]$$.
Question 7
1 / -0

$$\sin ^{ -1 }{ \left( \cos { \left( \sin ^{ -1 }{ x } \right) } \right) } +\cos ^{ -1 }{ \left( \sin { \left( \cos ^{ -1 }{ x } \right) } \right) } $$ is equal to

A
$$\displaystyle \frac { \pi }{ 4 } $$

B
$$\displaystyle \frac { \pi }{ 2 } $$

C
$$\displaystyle \frac { 3\pi }{ 4 } $$

D
$$0$$

Solution

We have $$\sin ^{ -1 }{ \left( \cos { \left( \sin ^{ -1 }{ x } \right) } \right) } +\cos ^{ -1 }{ \left( \sin { \left( \cos ^{ -1 }{ x } \right) } \right) } $$
$$\displaystyle =\sin ^{ -1 }{ \left[ \cos { \left( \frac { \pi }{ 2 } -\cos ^{ -1 }{ x } \right) } \right] } +\cos ^{ -1 }{ \left[ \sin { \left( \frac { \pi }{ 2 } -\sin ^{ -1 }{ x } \right) } \right] } $$
$$\displaystyle =\sin ^{ -1 }{ \left[ \sin { \left( \cos ^{ -1 }{ x } \right) } \right] } +\cos ^{ -1 }{ \left[ \cos { \left( \sin ^{ -1 }{ x } \right) } \right] } $$
$$\displaystyle =\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } =\frac { \pi }{ 2 } $$
Question 8
1 / -0

The number of real solution of the equation $$\displaystyle \sqrt{1+cos2x}=\sqrt{2}sin^{-1}(sin\:x),-\pi< x\leq \pi$$, is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$infinite$$

Solution

Given equation is,
$$\\ \sqrt { 1+\cos 2x } =\sqrt { 2 } sin^{ -1 }\left( \sin\: x \right) \\ \Rightarrow \sqrt { 1+{ \cos }^{ 2 }x-{\sin }^{ 2 }x } =\sqrt { 2 } x\\ \Rightarrow \sqrt { 2{ \cos }^{ 2 }x } =\sqrt { 2 } x\\ \Rightarrow \pm \sqrt { 2 } \cos x=\sqrt { 2 } x\\ \Rightarrow \cos x=\pm 1\\ x=0,\pi $$
Question 9
1 / -0

There exists a positive real number $$x$$ satisfying $$\displaystyle \cos(\tan^{-1}x)=x$$. Then the value of $$\displaystyle \cos^{-1}\left(\frac{x^{2}}{2}\right)$$ is

A
$$\displaystyle \frac{\pi}{10}$$

B
$$\displaystyle \frac{\pi}{5}$$

C
$$\displaystyle \frac{2\pi}{5}$$

D
$$\displaystyle \frac{4\pi}{5}$$

Solution

$$\displaystyle \cos(\tan^{-1}x)=x\Rightarrow \tan^{-1}x = \cos^{-1}x$$
$$\displaystyle \Rightarrow \cos^{-1}\frac{1}{\sqrt{1+x^2}} = \cos^{-1}x\Rightarrow x^2(1+x^2) = 1$$
$$\Rightarrow x^2 = \cfrac{-1+\sqrt{5}}{2}\Rightarrow \cfrac{x^2}{2} = \cfrac{\sqrt{5}-1}{4} = \cos\cfrac{2\pi}{5}$$
$$\therefore \cos\left(\cfrac{x^2}{2}\right) = \cfrac{2\pi}{5}$$
Question 10
1 / -0

The value of $$\displaystyle \lim_{|x|\to\infty}cos(tan^{-1}(sin(tan^{-1}x)))$$ is equal to

A
$$-1$$

B
$$\sqrt{2}$$

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

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

Solution

$$lim_{|x|\rightarrow \infty}cos(tan^{-1}(sin(tan^{-1}(x))))$$
$$=cos(tan^{-1}(sin(\dfrac{\pi}{2})))$$
$$=cos(tan^{-1}(1))$$
$$=cos(\dfrac{\pi}{4})$$
$$=\dfrac{1}{\sqrt{2}}$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 29

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Inverse Trigonometric Functions Test - 29

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SHARING IS CARING

Question 1

$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$$

$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$

$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)$$

$$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{3\pi}{2},-\frac{3\pi}{2}\right)$$

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$none\:of\:these$$

Solution

Question 3

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

$$\pi$$

$$0$$

none of these

Solution

Question 4

$$(0,1)$$

$$(-1,1)-\left\{0\right\}$$

$$(-1,0)$$

$$[-1,1]$$

Solution

Question 5

$$a=\pi,b=1,c=1$$

$$a=-\pi,b=2,c=2$$

$$a=-\pi,b=1,c=2$$

$$a=\pi,b=2,c=1$$

Solution

Question 6

one negative solution

one positive solution

no solution

more than one solution

Solution

Question 7

$$\displaystyle \frac { \pi }{ 4 } $$

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

$$\displaystyle \frac { 3\pi }{ 4 } $$

$$0$$

Solution

Question 8

$$0$$

$$1$$

$$2$$

$$infinite$$

Solution

Question 9

$$\displaystyle \frac{\pi}{10}$$

$$\displaystyle \frac{\pi}{5}$$

$$\displaystyle \frac{2\pi}{5}$$

$$\displaystyle \frac{4\pi}{5}$$

Solution

Question 10

$$-1$$

$$\sqrt{2}$$

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

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

Solution

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