Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Number of solutions of the equation $$ \sin \left ( \displaystyle \frac{1}{3}\cos^{-1}x \right )=1 $$ are

A
only one

B
no solution

C
only three

D
at least two

Solution

$$\displaystyle \sin \left ( \frac{1}{3}\cos ^{-1}x \right )=1$$
$$\displaystyle \Rightarrow \frac { 1 }{ 3 } \cos ^{ -1 } x=\frac { \pi }{ 2 } $$
$$\displaystyle \Rightarrow \cos ^{-1}x=\frac{3\pi }{2}$$
$$\displaystyle \Rightarrow \exists $$ no solution, as $$\displaystyle x \epsilon \left [ 0,\pi \right ]$$
Question 2
1 / -0

If value of $$x$$ which satisfy equation $$\displaystyle \sin^{-1}\frac{2x}{1+x^{2}}=\tan^{-1}\frac{2x}{1-x^{2}}$$, is $$-a< x< b$$.
Find the value of $$a+b$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

We know, $$-1\leq \dfrac{2x}{1+x^{2}}\leq 1$$

Now, $$-(1+x^{2})\leq 2x$$

$$1+x^{2}+2x\geq 0$$

$$(1+x)^{2}\geq 0$$

$$1+x\geq 0$$

$$x\geq-1$$ ...(i)

And $$\dfrac{2x}{1+x^{2}}\leq 1$$

$$0\leq 1+x^{2}-2x$$

$$0\leq (1-x)^{2}$$

$$0\leq 1-x$$

$$x\leq 1$$

Hence, $$-1\leq x\leq 1$$.

Hence, $$a=-1$$ and $$b=1$$

$$\therefore a+b=0$$
Question 3
1 / -0

If $$\displaystyle \sin ^{-1}x-\sin ^{-1}y= \frac{\pi }{2}$$ ,then

A
$$x^{2}+y^{2}= 1$$

B
$$y= -\sqrt{1-x^{2}}, \:0\leq x\leq 1,-1\leq y\leq 0$$

C
$$y= \sqrt{1-x^{2}}, \: \left | x \right |< 1$$

D
None of these

Solution

$$\sin ^{ -1 }{ x } -\sin ^{ -1 }{ y } =\frac { \pi }{ 2 } \\ \Rightarrow \sin ^{ -1 }{ y } =-\left( \frac { \pi }{ 2 } -\sin ^{ -1 }{ x } \right) \\ \Rightarrow \sin { \sin ^{ -1 }{ y } } =-\sin { \left( \frac { \pi }{ 2 } -\sin ^{ -1 }{ x } \right) } \\ \Rightarrow y=-\cos { \sin ^{ -1 }{ x } } =-\cos { \cos ^{ -1 }{ \sqrt { 1-{ x }^{ 2 } } } } $$

$$for\quad 0\le x\le 1$$
$$y=-\sqrt { 1-{ x }^{ 2 } } $$
$$\Rightarrow -1\le y\le 0$$

Ans: B
Question 4
1 / -0

$$\displaystyle \sin ^{-1}\left ( a-\frac{a^{2}}{3}+\frac{a^{3}}{9}\cdots \infty \right )+\cos ^{-1}\left ( 1+b+b^{2}+b^{3}+\cdots \infty \right )= \frac{\pi }{2}$$ , when

A
$$\displaystyle a= 1, b= -\frac{1}{3}$$

B
$$\displaystyle a= -\frac{1}{6}, b= \frac{1}{2}$$

C
$$\displaystyle a= \frac{1}{6}, b= \frac{1}{2}$$

D
None of these

Solution

$$a-\dfrac{a^{2}}{3}+\dfrac{a^{3}}{9}...\infty $$ is an infinite G.P with a common ration of $$\dfrac{-a}{3}$$.

Hence
$$S_{a}=\dfrac{a}{1+\dfrac{a}{3}}$$ $$=\dfrac{3a}{a+3}$$ ...(i)

Similarly $$S_{b}=1+b+b^{2}...\infty$$.

$$S_{b}=\dfrac{1}{1-b}$$ ...(ii)

Now $$sin^{-1}(A)+cos^{-1}(B)=\dfrac{\pi}{2}$$ implies $$A=B$$, then

$$i=ii$$

$$\dfrac{3a}{a+3}=\dfrac{1}{1-b}$$

$$3a-3ab=a+3$$

$$2a-3ab=3$$

$$2a-3=3ab$$

$$b=\dfrac{2a-3}{3a}$$

Substituting $$b=\dfrac{1}{2}$$, we get

$$\dfrac{3a}{2}-2a=-3$$

$$\dfrac{a}{2}=3$$

$$a=6$$.

However $$|a|<1$$ and $$|b|<1$$ for the above infinite G.P.

Hence answer is none of the above.
Question 5
1 / -0

The number of solutions of the equation $$ 2\sin^{-1}\left ( \sqrt{x^{2}-x+1} \right )+\cos^{-1}\left ( \sqrt{x^{2}-x} \right )= \displaystyle \frac{3\pi }{2} $$ is

A
$$0$$

B
Infinite

C
$$2$$

D
$$4$$

Solution

The maximum value of the function is obtained when the value of $$\sqrt {x^2-x+1}$$ is equal to $$1$$
and value of $$\sqrt {x^2-x}$$ is equal to $$0$$

$$\displaystyle \sin ^{-1}\sqrt{x}, \cos ^{-1}\sqrt{x}$$ are defined for $$\displaystyle x\leq 1$$ and $$\displaystyle x\geq 0$$
$$\displaystyle \therefore \sqrt{x^{2}-x+1}\leq 1$$ and $$\displaystyle \sqrt{x^{2}-x}\geq 0$$
$$\displaystyle \Rightarrow x^{2}-x\leq 0$$ and $$\displaystyle x^{2}-x\geq 0$$

Thus, reducing the equation to

$$\displaystyle \Rightarrow x^{2}-x=0$$
$$\displaystyle \Rightarrow x=1, 0$$
$$\displaystyle \therefore $$ there are two solutions, both satisfies the equation.
Question 6
1 / -0

The equation $$\sin ^{-1}x=2\sin ^{-1}a$$ , has a solution for

A
$$\forall \:R$$

B
$$\displaystyle \left | a \right |< \frac{1}{2}$$

C
$$\displaystyle \left | a \right |\geq \frac{-1}{2}$$

D
$$\displaystyle \frac{-1}{\sqrt{2}}\leq a\leq \frac{1}{\sqrt{2}}$$

Solution

Given equation $$\sin ^{-1}x= 2\sin ^{-1}a$$

$$\displaystyle \Rightarrow -\frac{\pi}{2}\leq 2\sin ^{-1}a\leq \frac{\pi}{2}\left \{ \because \sin ^{-1}x= 2\sin ^{-1}a \:and -\frac{\pi}{2}\leq 2\sin ^{-1}x\leq \frac{\pi}{2} \right\}$$

$$\displaystyle \Rightarrow -\frac{\pi}{4}\leq \sin ^{-1}a\leq \frac{\pi}{4}$$

$$\displaystyle \Rightarrow \sin \left ( -\frac{\pi}{4}\right )\leq a\leq \sin \left ( \frac{\pi}{4} \right )$$

$$\displaystyle \Rightarrow -\frac{1}{\sqrt{2}}\leq a\leq \frac{1}{\sqrt{2}}$$
Question 7
1 / -0

Let $$f(x) =e^{\displaystyle\cos^{-1}\sin \left ( x+\displaystyle\frac{\pi }{3} \right )}$$, then $$f\left (\displaystyle \frac{8\pi }{9} \right )$$ equals

A
$$e^{\displaystyle\frac{7\pi }{12}}$$

B
$$e^{\displaystyle\frac{13\pi }{18}}$$

C
$$e^{\displaystyle\frac{5\pi }{18}}$$

D
$$e^{\displaystyle\frac{\pi }{12}}$$

Solution

$$f(x)=e^{\cos ^{-1}\sin\left ( x+\displaystyle \frac{\pi }{3} \right )}$$
$$\therefore f\left (\displaystyle \frac{8\pi }{9} \right )$$
$$f(x)=e^{\cos ^{-1}\sin\left ( \dfrac{\pi}{3}+\displaystyle \frac{8\pi }{9} \right )}$$
$$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{11\pi }{9} \right )}$$
$$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{22\pi }{18} \right )}$$
$$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{9\pi }{18}+\displaystyle \frac{13\pi }{18} \right )}$$
$$=e^{\cos ^{-1}\sin\left (\displaystyle \frac{\pi }{2}+\displaystyle \frac{13\pi }{18} \right )}$$
$$=e^{\cos ^{-1}\cos\left (\displaystyle \frac{13\pi }{18} \right )}$$
$$=e^{\displaystyle \frac{13\pi }{18}}$$
Question 8
1 / -0

If $$\displaystyle \cot ^{-1}\left [ \left ( \cos \alpha \right )^{1/2} \right ]+\left [ \tan ^{-1}\left ( \cos \alpha \right )^{1/2} \right ]=x$$ , then $$\sin x$$ equals

A
$$1$$

B
$$\displaystyle \cot ^{2}\left ( \frac{\alpha }{2} \right )$$

C
$$\tan \alpha $$

D
$$\displaystyle \cot\left ( \frac{\alpha }{2} \right )$$

Solution

As $${ \tan }^{ -1 }\theta +{ \cot }^{ -1 }\theta =\cfrac { \pi }{ 2 } $$
$$\therefore x=\cfrac { \pi }{ 2 } $$
Hence,
$$\sin x=\sin\cfrac { \pi }{ 2 } =1$$
Question 9
1 / -0

The domain of $$\sin ^{-1}[x]$$, where $$[x]$$ is greatest integer function, given by

A
$$\left [ -1,1 \right ]$$

B
$$\left [ -1,2 \right )$$

C
$$\left \{ -1,0,1 \right \}$$

D
None of these

Solution

Domain of $$\sin^{-1}(x)$$ is $$[-1,1]$$

Now $$[x]$$ is greatest integer function, lesser than or equal to $$x.$$

Hence
$$-1\leq [x]\leq 1$$

Hence
$$x\in[-1,2)$$
Question 10
1 / -0

If $$\cos^{-1}x+\cos^{-1}y+\cos^{-1}z= 3\pi $$, then value of $$\displaystyle \sum xy$$ equals

A
-3

B
0

C
3

D
-1

Solution

As $$0\le cos^{ -1 }\theta \le \pi $$
For $$cos^{ -1 }x+cos^{ -1 }y+cos^{ -1 }z=3\pi $$ this to true
$$\Rightarrow cos^{ -1 }x=cos^{ -1 }y=cos^{ -1 }z=\pi \\ \Rightarrow x=y=z=-1\\ \therefore \sum { xy } =xy+yz+zx=(-1)^2+(-1)^2+(-1)^2=3$$

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Inverse Trigonometric Functions Test - 32

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Inverse Trigonometric Functions Test - 32

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SHARING IS CARING

Question 1

only one

no solution

only three

at least two

Solution

Question 2

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 3

$$x^{2}+y^{2}= 1$$

$$y= -\sqrt{1-x^{2}}, \:0\leq x\leq 1,-1\leq y\leq 0$$

$$y= \sqrt{1-x^{2}}, \: \left | x \right |< 1$$

None of these

Solution

Question 4

$$\displaystyle a= 1, b= -\frac{1}{3}$$

$$\displaystyle a= -\frac{1}{6}, b= \frac{1}{2}$$

$$\displaystyle a= \frac{1}{6}, b= \frac{1}{2}$$

None of these

Solution

Question 5

$$0$$

Infinite

$$2$$

$$4$$

Solution

Question 6

$$\forall \:R$$

$$\displaystyle \left | a \right |< \frac{1}{2}$$

$$\displaystyle \left | a \right |\geq \frac{-1}{2}$$

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

Solution

Question 7

$$e^{\displaystyle\frac{7\pi }{12}}$$

$$e^{\displaystyle\frac{13\pi }{18}}$$

$$e^{\displaystyle\frac{5\pi }{18}}$$

$$e^{\displaystyle\frac{\pi }{12}}$$

Solution

Question 8

$$1$$

$$\displaystyle \cot ^{2}\left ( \frac{\alpha }{2} \right )$$

$$\tan \alpha $$

$$\displaystyle \cot\left ( \frac{\alpha }{2} \right )$$

Solution

Question 9

$$\left [ -1,1 \right ]$$

$$\left [ -1,2 \right )$$

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

None of these

Solution

Question 10

-3

0

3

-1

Solution

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