Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \:\sum_{i= 1}^{10}\cos^{-1}x_{i}= 0$$, then $$\displaystyle \:\sum_{i= 1}^{10}x_{i}$$ is

A
$$0$$

B
$$10$$

C
$$5$$

D
none of these

Solution

The above expression can be written as
$$cos^{-1}(x_{1})+cos^{-1}(x_{2})+cos^{-1}(x_{3})+...cos^{-1}(x_{10})=0$$
We know that $$cos^{-1}(1)=0$$
Hence
$$x_{1}=x_{2}=...=x_{10}=1$$
Therefore
$$x_{1}+x_{2}+...+x_{10}$$
$$=1+1+1+1...+1$$
$$=10(1)$$
$$=10$$
Question 2
1 / -0

If $$\displaystyle \:\sum_{i= 1}^{2n}\sin^{-1}x_{i}= n\pi$$, then $$\displaystyle \:\sum_{i= 1}^{2n}x_{i}$$ is equal to

A
$$n/2 $$

B
$$2n $$

C
$$\displaystyle \frac{n\left ( n+1 \right )}{2}$$

D
none of these

Solution

As $$\displaystyle -\frac { \pi }{ 2 } \le \sin ^{ -1 }{ x } \le \frac { \pi }{ 2 } $$

Hence from question
$$\displaystyle \sin ^{ -1 }{ { x }_{ i } } =\frac { \pi }{ 2 } $$ for all $$i$$

$$\therefore { x }_{ i }=1$$ for all $$i$$

$$\displaystyle \therefore \sum _{ i=1 }^{ 2n }{ { x }_{ i } } =2n$$
Question 3
1 / -0

If both $$\displaystyle \:\alpha, \beta $$ and $$\displaystyle \:6x^{2}+11x+3= 0$$ are real roots of the equation then

A
both $$\displaystyle \:\cos ^{-1}\alpha $$ and $$\displaystyle \:\cos ^{-1}\beta $$ are real

B
both $$\displaystyle cosec ^{-1}\alpha $$ and $$\displaystyle \cos ^{-1}\beta $$ are real

C
both $$\displaystyle \:\cot ^{-1}\alpha $$ and $$\displaystyle \:\cot ^{-1}\beta $$ are real

D
none of these

Solution

given, $$6x^2+11x+3=0$$

$$6x^2+9x+2x+3=0$$

$$3x(2x+3)+1(2x+3)=0$$

$$(3x+1)(2x+3)=0$$

Hence $$\alpha=\dfrac{-1}{3}$$ and $$\beta=\dfrac{-3}{2}$$

Hence $$cot^{-1}(\alpha)$$ and $$cot^{-1}(\alpha)$$ are real, since $$cos^{-1}(\alpha)\in[-1,1]$$ and $$cosec^{-1}(\alpha)\notin[-1,1]$$.
Question 4
1 / -0

The range of the the function, $$f(x)=(\cot ^{-1}x+\sec ^{-1}x+cosec ^{-1}x)$$ is

A
$$\left ( \frac{\pi }{2}, \frac{3\pi }{2} \right )$$

B
$$\left ( \frac{\pi }{2}, \frac{3\pi }{4} \right ]\cup \left [ \frac{5\pi }{4}, \frac{3\pi }{2} \right )$$

C
$$\left [ \frac{\pi }{2}, \pi \right )\cup \left ( \pi ,\frac{3\pi }{2} \right ]$$

D
$$\left ( \frac{\pi }{2}, \pi \right )\cup \left ( \pi ,\frac{3\pi }{2} \right )$$

Solution

Given,
$$f(x)=(\cot ^{-1}x+\sec ^{-1}x+cosec ^{-1}x)$$

Now,
$$0<cot^{ -1 }x<\pi $$ ..(1)
$$0\le { sec }^{ -1 }x<\dfrac { \pi }{ 2 } $$
and
$$\dfrac { \pi }{ 2 } <{ sec }^{ -1 }x\le \pi $$ ...(2)

$$-\dfrac { \pi }{ 2 } \le { cosec }^{ -1 }x<0$$ and $$0<{ cosec }^{ -1 }\le \dfrac { \pi }{ 2 } $$ ..(3)

Adding (1) , (2) and (3), we get
$$\dfrac { \pi }{ 2 } <cot^{ -1 }x+{ sec }^{ -1 }x+{ cosec }^{ -1 }x\le \dfrac { 3\pi }{ 4 } $$ and $$\dfrac { 5\pi }{ 4 } \le cot^{ -1 }x+{ sec }^{ -1 }x+{ cosec }^{ -1 }x<\dfrac { 3\pi }{ 2 } $$
Question 5
1 / -0

If $$\displaystyle \:\cos ^{-1}x-\sin ^{-1}x= 0$$ , then $$x$$ is equal to

A
$$\displaystyle \:\pm \frac{1}{\sqrt{2}}$$

B
$$1$$

C
$$\displaystyle \: \sqrt{2}$$

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

Solution

Given, $$\sin^{-1}(x)-\cos^{-1}(x)=0$$
$$\sin^{-1}x+ \cos^{-1}x= \pi/2$$
$$\dfrac{\pi}{2}=2\cos^{-1}(x)$$
$$\cos^{-1}(x)=\frac{\pi}{4}$$
$$x=\cos(\dfrac{\pi}{4})$$
$$x=\dfrac{1}{\sqrt{2}}$$
Question 6
1 / -0

If $$\displaystyle \:\cos ^{-1}\lambda +\cos^{-1}\mu +\cos ^{-1}v= 3\pi $$ then $$\lambda \mu +\mu v+v\lambda $$ is equal to

A
$$-3$$

B
$$0$$

C
$$3$$

D
$$-1$$

Solution

Given, $$\cos^{-1}(\lambda)+\cos^{-1}(\mu)+\cos^{-1}(v)=3\pi$$
Now we know that
$$\cos^{-1}(x)\epsilon[0,\pi]$$
Hence the above equation is only possible if
$$\cos^{-1}(\lambda)=\cos^{-1}(\mu)=\cos^{-1}(v)=\pi$$
$$\lambda=\mu=v=\cos(\pi)$$
$$=-1$$
Hence, $$\lambda\mu+\mu v+\lambda v$$
$$=3\lambda^{2}$$
$$=3\mu^{2}$$
$$=3v^{2}$$
$$=3(-1)^{2}$$
$$=3$$.
Question 7
1 / -0

If $$A=\cot ^{ -1 }{ \sqrt { \tan { \theta } } } -\tan ^{ -1 }{ \sqrt { \tan { \theta } } }$$, then $$\displaystyle \tan { \left( \frac { \pi }{ 4 } -\frac { A }{ 2 } \right) } $$ is equal to

A
$$\sqrt { \cot { \theta } } $$

B
$$\tan { \theta } $$

C
$$\sqrt { \tan { \theta } } $$

D
None of these

Solution

Let $$\sqrt { \tan { \theta } } =\tan { \alpha } $$
$$\displaystyle \therefore A=\cot ^{ -1 }{ \left( \tan { \alpha } \right) } -\tan ^{ -1 }{ \left( \tan { \alpha } \right) } $$
$$\displaystyle =\cot ^{ -1 }{ \left( \cot { \left( \frac { \pi }{ 2 } -\alpha \right) } \right) } -\tan ^{ -1 }{ \left( \tan { \alpha } \right) } $$
$$\displaystyle =\frac { \pi }{ 2 } -\alpha -\alpha =\frac { \pi }{ 2 } -2\alpha $$
$$\displaystyle \Rightarrow 2\alpha =\frac { \pi }{ 2 } -A\Rightarrow \alpha =\frac { \pi }{ 4 } -\frac { A }{ 2 } $$
$$\displaystyle \therefore \sqrt { \tan { \theta } } =\tan { \alpha } =\tan { \left( \frac { \pi }{ 4 } -\frac { A }{ 2 } \right) } $$
Question 8
1 / -0

The principal value of $$\displaystyle \:\sin ^{-1}\left \{ \cos \left ( \sin ^{-1}\dfrac{\sqrt{3}}{2} \right ) \right \}$$ is

A
$$\displaystyle \: \frac{\pi }{6}$$

B
$$\displaystyle \: \frac{\pi }{3}$$

C
$$\displaystyle \: -\frac{\pi }{3}$$

D
none of these

Solution

The principle value of $$\sin^{-1}(\cos(\sin^{-1}(\dfrac{\sqrt{3}}{2})))$$

$$=\dfrac{\pi}{2}-\cos^{-1}(\cos(\sin^{-1}(\dfrac{\sqrt{3}}{2})))$$

$$=\dfrac{\pi}{2}-\sin^{-1}(\dfrac{\sqrt{3}}{2})$$

$$=\cos^{-1}(\dfrac{\sqrt{3}}{2})$$

$$=\dfrac{\pi}{6}$$
Question 9
1 / -0

A function $$f(x)=\sqrt{1-2x}+x$$ is defined from $$D_{1}\rightarrow D_{2}$$ and is onto. If the set $$D_{1}$$ is its complete domain then the set $$D_{2}$$ is

A
$$\left ( -\infty ,\frac{1}{2} \right ]$$

B
$$\left ( -\infty ,2 \right )$$

C
$$\left ( -\infty ,1 \right )$$

D
$$\left ( -\infty ,1 \right ]$$

Solution

We know, $$1-2x\geq0$$
$$1\geq 2x$$
$$x\leq \frac{1}{2}$$
Hence domain is $$(-\infty,\frac{1}{2}]$$
Now
$$\frac{df(x)}{dx}$$
$$=\frac{-1}{\sqrt{1-2x}}+1$$
$$=0$$ for maxima minima.
$$\sqrt{1-2x}=1$$
$$1-2x=1$$
$$\therefore x=0$$
Hence $$f(0)=1$$ which is the maxima.
Thus range is $$(-\infty,1]$$
Question 10
1 / -0

The set of values of $$x$$ for which $$\displaystyle \:\tan ^{-1}\frac{x}{\sqrt{1-x^{2}}}= \sin ^{-1}x$$ holds is

A
$$R$$

B
$$\displaystyle \:\left ( -1, 1 \right )$$

C
$$\displaystyle \:\left [ 0, 1 \right ]$$

D
$$\displaystyle \:\left [-1, 0 \right ]$$

Solution

Since, x is the argument of inverse function of sin ,

$$x \in (-1,1)$$ $$[ x \neq \ ^{+}_{-}1$$, can be oberved from tan inverse function]

Let $$\displaystyle x = sin(t)$$ for $$t \in (\dfrac{-\pi}{2}, \dfrac{\pi}{2})$$

Then, we can rewrite the given equation as,

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

$$\rightarrow tan^{-1} tant = sin^{-1}sint$$

Since $$t \in (\frac{-\pi}{2},\frac{\pi}{2})$$, Therefore,

$$tan^{-1}tant = t$$ and $$sin^{-1}sint = t$$

$$\rightarrow t = t$$ which is always true.

So, the given equation is true for every allowed value of x i.e, $$(-1,1).$$

Thus, B is the correct answer.

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 27

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Inverse Trigonometric Functions Test - 27

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SHARING IS CARING

Question 1

$$0$$

$$10$$

$$5$$

none of these

Solution

Question 2

$$n/2 $$

$$2n $$

$$\displaystyle \frac{n\left ( n+1 \right )}{2}$$

none of these

Solution

Question 3

both $$\displaystyle \:\cos ^{-1}\alpha $$ and $$\displaystyle \:\cos ^{-1}\beta $$ are real

both $$\displaystyle cosec ^{-1}\alpha $$ and $$\displaystyle \cos ^{-1}\beta $$ are real

both $$\displaystyle \:\cot ^{-1}\alpha $$ and $$\displaystyle \:\cot ^{-1}\beta $$ are real

none of these

Solution

Question 4

$$\left ( \frac{\pi }{2}, \frac{3\pi }{2} \right )$$

$$\left ( \frac{\pi }{2}, \frac{3\pi }{4} \right ]\cup \left [ \frac{5\pi }{4}, \frac{3\pi }{2} \right )$$

$$\left [ \frac{\pi }{2}, \pi \right )\cup \left ( \pi ,\frac{3\pi }{2} \right ]$$

$$\left ( \frac{\pi }{2}, \pi \right )\cup \left ( \pi ,\frac{3\pi }{2} \right )$$

Solution

Question 5

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

$$1$$

$$\displaystyle \: \sqrt{2}$$

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

Solution

Question 6

$$-3$$

$$0$$

$$3$$

$$-1$$

Solution

Question 7

$$\sqrt { \cot { \theta } } $$

$$\tan { \theta } $$

$$\sqrt { \tan { \theta } } $$

None of these

Solution

Question 8

$$\displaystyle \: \frac{\pi }{6}$$

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

$$\displaystyle \: -\frac{\pi }{3}$$

none of these

Solution

Question 9

$$\left ( -\infty ,\frac{1}{2} \right ]$$

$$\left ( -\infty ,2 \right )$$

$$\left ( -\infty ,1 \right )$$

$$\left ( -\infty ,1 \right ]$$

Solution

Question 10

$$R$$

$$\displaystyle \:\left ( -1, 1 \right )$$

$$\displaystyle \:\left [ 0, 1 \right ]$$

$$\displaystyle \:\left [-1, 0 \right ]$$

Solution

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