Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

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

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

B
$$\displaystyle 1$$

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

D
$$\displaystyle -1$$

Solution

$$\sin^{-1}(x-\frac{x}{2}+\frac{x^{2}}{4}-..\infty)+cos^{-1}(x^{2}-\frac{x^{4}}{2}+\frac{x^{3}}{6}+...\infty)$$

$$=\sin^{-1}(x-[\dfrac{\dfrac{x}{2}}{1+\dfrac{x^{2}}{2}}])+cos^{-1}(\dfrac{x^{2}}{1+\dfrac{x^{2}}{2}})$$
$$=\sin^{-1}(x-\dfrac{x}{x^{2}+2})+cos^{-1}(\dfrac{2x^{2}}{x^{2}+2})$$
$$=\dfrac{\pi}{2}$$
Therefore
$$x-\dfrac{x}{x^{2}+2}=\dfrac{2x^{2}}{x^{2}+2}$$
$$x^{3}+2x-x=2x^{2}$$
$$x^{3}-2x^{2}+x=0$$
$$x(x^{2}-2x+1)=0$$
$$x((x-1)^{2})=0$$
$$x=0$$ and $$x=1$$
But $$|x|>0$$
Hence
$$x=1$$
Question 2
1 / -0

The value of $$\displaystyle sin^{-1}(cos(cos^{-1}(cos\:x)+sin^{-1}(sin\:x)))$$, where $$\displaystyle x\:\epsilon\:\left ( \frac{\pi}{2},\pi \right )$$, is equal to

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

B
$$-\pi$$

C
$$\pi$$

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

Solution

$$sin^{-1}[coscos^{-1}(cos(x)+sin^{-1}(sin(x))]$$
$$=sin^{-1}[cos(x+\pi-x)]$$
$$=sin^{-1}[cos(\pi)]]$$
$$=sin^{-1}(-1)$$
$$=\dfrac{-\pi}{2}$$
Question 3
1 / -0

$$\displaystyle \:\sin \cot ^{-1}\tan \cos ^{-1}x$$ is equal to

A
$$x$$

B
$$\sqrt{1-x^{2}}$$

C
$$\displaystyle \:\frac{1}{x}$$

D
none of these

Solution

Given, $$\sin \cot^{-1}\tan(\cos^{-1}(x))$$
$$=\sin \cot^{-1}\tan(\tan^{-1}(\frac{\sqrt{1-x^2}}{x}))$$
$$=\sin \cot^{-1}(\frac{\sqrt{1-x^2}}{x})$$
$$=\sin(\sin^{-1}(x))$$
$$=x$$
Question 4
1 / -0

If $$\displaystyle \frac{1}{\sqrt{2}}< x< 1$$ then $$\displaystyle\cos ^{-1}x+\cos ^{-1}\left ( \frac{x+\sqrt{1-x^{2}}}{\sqrt{2}} \right )$$ is equal to

A
$$\displaystyle 2\cos ^{-1}x-\frac{\pi }{4}$$

B
$$\displaystyle 2\cos ^{-1}x$$

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

D
0

Solution

$$cos^{-1}(x)+cos^{-1}(\dfrac{1}{\sqrt{2}}x+\dfrac{1}{\sqrt{2}}\sqrt{1-x^{2}})$$
$$=cos^{-1}(x)+cos^{-1}(\dfrac{1}{\sqrt{2}})-cos^{-1}(x)$$
$$=cos^{-1}(\dfrac{1}{\sqrt{2}})$$
$$=\dfrac{\pi}{4}$$
Question 5
1 / -0

If $$\displaystyle \cos \:^{-1}x+\cos \:^{-1}y+\cos ^{-1}z=3\pi$$ then the value of $$xy+yz+zx$$ is

A
$$-3$$

B
$$1$$

C
$$3$$

D
none of these

Solution

If $$y=cos^{-1}(x)$$ $$ R\epsilon[\pi,0]$$ where$$ R$$ is the range of $$ y$$.

Therefore in the above case, which is only possible if
$$cos^{-1}(x)=cos^{-1}(y)=cos^{-1}(z)=\pi$$

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

Therefore $$xy+yz+zx$$
$$=3 (-1)^2=3$$
Question 6
1 / -0

$$\displaystyle sec^{2}(tan^{-1}2)+cosec^{2}(cot^{-1}3)$$ is equal to

A
$$5$$

B
$$13$$

C
$$15$$

D
$$6$$

Solution

Given, $$\sec^{2}(\tan^{-1}(2))+cosec^{2}(\cot^{-1}(3))$$
$$=2+\tan^{2}(\tan^{-1}(2))+\cot^{2}(\cot^{-1}(3))$$
$$=2+2^{2}+3^{2}$$
$$=2+4+9$$
$$=15$$
Question 7
1 / -0

If $$\displaystyle f\left ( x \right ) = \sin^{-1}x + \sec^{-1} x$$ is defined, then which of the following value/values is/are in its range?

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

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

C
$$\displaystyle \pi$$

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

Solution

The domain for $$\sec^{-1}(x)$$ is $$(-\infty,-1]\cup[1,\infty)$$

The domain for $$\sin^{-1}(x)$$ is $$[-1,1]$$

Hence the domain for

$$f(x)=\sin^{-1}(x)+\sec^{-1}(x)$$ will be

$$-1$$ and $$1$$.

Therefore the range of values of $$f(x)$$ will be

$$f(-1)=\dfrac{-\pi}{2}+\pi$$

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

$$f(1)=\dfrac{\pi}{2}+0$$ $$=\dfrac{\pi}{2}$$

Hence the range of $$f(x)$$ includes a single element that is $$\dfrac{\pi}{2}$$
Question 8
1 / -0

The value of $$\displaystyle sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}}\right]$$ is equal to

A
$$sin^{-1}x+sin^{-1}\sqrt{x}$$

B
$$sin^{-1}x-sin^{-1}\sqrt{x}$$

C
$$sin^{-1}\sqrt {x}-sin^{-1}x$$

D
$$0$$

Solution

$$sin^{-1}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}})$$
$$=sin^{-1}(x)-sin^{-1}(\sqrt{x})$$ ...( by applying the formula of $$sin^{-1}(x)-sin^{-1}(y)$$)
Hence answer is Option B
Question 9
1 / -0

The number of integer $$x$$ satisfying $$\displaystyle sin^{-1}|x-2|+cos^{-1}(1-|3-x|)=\frac{\pi}{2}$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

If $$x<2$$
$$sin^{-1}(2-x)+cos^{-1}(1-(3-x))$$
$$=sin^{-1}(2-x)+cos^{-1}(-2+x)$$
$$=sin^{-1}(2-x)+\pi-cos^{-1}(2-x)$$
$$=\frac{\pi}{2}$$
$$-\frac{pi}{2}+sin^{-1}(2-x)+\pi+sin^{-1}(2-x)=\frac{\pi}{2}$$
$$2sin^{-1}(2-x)=0$$
$$x=2$$ ...(i)
For $$2<x<3$$
$$sin^{-1}(x-2)+cos^{-1}(-2+x)$$
$$=sin^{-1}(x-2)+cos^{-1}(x-2)$$
$$=\frac{\pi}{2}$$
For $$x>3$$
$$sin^{-1}(x-2)+cos^{-1}(4-x)=\frac{\pi}{2}$$
$$x-2=4-x$$
$$2x=6$$
$$x=3$$
Hence there are in total 2, solutions.
Question 10
1 / -0

If $$x_{ 1 }=2\: tan^{ -1 }\left( \frac { 1+x }{ 1-x } \right) ,x_{ 2 }=sin^{ -1 }\left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) $$, where $$x_1,x_2\:\epsilon\:(0,1)$$, then $$2\left( x_{ 1 }+x_{ 2 } \right) $$ is equal to

A
$$0$$

B
$$2\pi$$

C
$$\pi$$

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

Solution

$${ x }_{ 1 }=2\tan^{ -1 }\left( \cfrac { 1+x }{ 1-x } \right) =\cos^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) \\ { x }_{ 2 }=\sin^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) $$
Therefore,
$${ x }_{ 1 }+{ x }_{ 2 }=\sin^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) +\cos^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) =\cfrac { \pi }{ 2 } \\ \Rightarrow 2\left( { x }_{ 1 }+{ x }_{ 2 } \right) =\pi $$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 28

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Inverse Trigonometric Functions Test - 28

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Question 1

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

$$\displaystyle 1$$

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

$$\displaystyle -1$$

Solution

Question 2

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

$$-\pi$$

$$\pi$$

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

Solution

Question 3

$$x$$

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

$$\displaystyle \:\frac{1}{x}$$

none of these

Solution

Question 4

$$\displaystyle 2\cos ^{-1}x-\frac{\pi }{4}$$

$$\displaystyle 2\cos ^{-1}x$$

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

0

Solution

Question 5

$$-3$$

$$1$$

$$3$$

none of these

Solution

Question 6

$$5$$

$$13$$

$$15$$

$$6$$

Solution

Question 7

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

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

$$\displaystyle \pi$$

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

Solution

Question 8

$$sin^{-1}x+sin^{-1}\sqrt{x}$$

$$sin^{-1}x-sin^{-1}\sqrt{x}$$

$$sin^{-1}\sqrt {x}-sin^{-1}x$$

$$0$$

Solution

Question 9

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 10

$$0$$

$$2\pi$$

$$\pi$$

$$none\:of\:these$$

Solution

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