Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

$$\sec^{-1}x+\tan^{-1}x$$ is real if $$\displaystyle x\epsilon \left ( -\infty ,-a \right ]\cup \left [ b,\infty \right )$$.
Find the value of $$a+b$$.

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

For $$\sec^{-1}x+\tan^{-1}x$$ to be real.

$$|x|\geq1 \quad \quad \dots (sec^{-1}x=tan^{-1}\sqrt{x^2-1} \Rightarrow x^2-1\ge 0 \Rightarrow |x|\ge 1)$$

$$\displaystyle \Rightarrow x\in \left ( -\infty ,-1 \right ]\cup \left [ 1,\infty \right )$$

Comparing with $$\displaystyle x\epsilon \left ( -\infty ,-a \right ]\cup \left [ b,\infty \right )$$

We get to know that, $$a=1$$ and $$b=1$$

$$\therefore a+b=1+1=2$$
Question 2
1 / -0

If $$\cos^{-1}\lambda +\cos^{-1}\mu +\cos^{-1}\gamma =3\pi $$, then find the value of $$\lambda \mu +\mu \gamma +\gamma \lambda $$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

We know that $$0\leq \cos^{-1}x\leq \pi $$
Hence, from the question
$$\cos^{-1}\lambda =\pi $$, $$\cos^{-1}\mu =\pi $$, $$\cos^{-1}\gamma =\pi $$
[$$\because \cos^{-1}\lambda +\cos^{-1}\mu +\cos^{-1}\gamma =3\pi $$ is possible only when each term attains its maximum]
$$\Rightarrow $$ $$\lambda =\mu =\gamma =-1$$ $$\Rightarrow $$ $$\lambda \mu +\mu \gamma +\gamma \lambda =3$$
Question 3
1 / -0

Find the value of $$x$$ for which; $${cosec}^{-1} \left( \cos x \right)$$ is real

A
$$x =-\pi$$

B
$$x =\pi$$

C
$$x =2\pi$$

D
All of the above

Solution

Given

Range of $$ \cos x $$ is $$[-1,1]$$

Domain of $$ \csc ^ {-1} x $$ is $$R-(-1,1)$$

Their intersection is $$ \{-1,1\}$$

$$ \cos x $$ can be $$ \pm 1 $$ only when $$ x=n \pi $$
Question 4
1 / -0

Find the number of solutions of the equation $$\cos \left ( \cos^{-1}x \right )=cosec\:\left ( cosec^{-1}\:x \right )$$

A
0

B
1

C
2

D
3

Solution

$$\cos \left ( \cos^{-1}x \right )=x$$ for $$x\epsilon \left [ -1, 1 \right ]$$
$$cosec\:\left ( cosec^{-1}\:x \right )=x$$ for $$c\epsilon \left (
-\infty , -1 \right ]\cup \left [ 1, \infty \right )$$
$$\Rightarrow $$ $$\cos \left ( \cos^{-1}x \right )=cosec\:\left ( cosec^{-1}\:x \right )$$ for $$x=\pm 1$$ only.
Hence, there are two roots only.
Question 5
1 / -0

Value of $$x$$ for which $$\displaystyle \cos^{-1}\left ( \frac{1-x^2}{1+x^2} \right )= 2\tan^{-1}x$$ satisfied is $$x\epsilon \left [ a,\infty \right )$$.
Find the value of $$a.$$

A
$$a=-\infty $$

B
$$a=-1$$

C
$$a=0$$

D
$$a=1$$

Solution

Given

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

As $$ \cos ^{-1} \left( \dfrac{1-x^2}{1+x^2}\right) \in [0,\pi]$$

then $$ 2\tan ^{-1} x \in [0,\pi]$$

$$ \tan ^{-1} x \in \left[0,\dfrac \pi 2\right]$$

$$ \implies x \in \left[0,\infty\right)$$

So $$a=0$$
Question 6
1 / -0

If the range for $$y=\left ( \cot^{-1}x \right )\left ( \cot^{-1}\left ( -x \right ) \right )$$ is
$$0< y\leq \cfrac{\pi ^{a}}{b}$$.
Find the value of $$a+b$$

A
$$2$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Given, $$y=\left ( \cot^{-1}x \right )\left ( \cot^{-1}\left ( -x \right ) \right )$$
$$=\cot^{-1}\left ( x \right )\left ( \pi -\cot^{-1}\left ( x \right ) \right )$$
Now $$\cot^{-1}\left ( x \right )$$ and $$\left ( \pi -\cot^{-1}\left ( x \right ) \right )> 0$$
Using A.M.$$\geq $$G.M., we get
$$\displaystyle \frac{\cot^{-1}x+\left ( \pi
-\cot^{-1}\left ( x \right ) \right )}{2}\geq \sqrt{\left ( \cot^{-1}x

\right )\left ( \pi -\cot^{-1}\left ( x \right ) \right )}$$
$$\Rightarrow

$$ $$\displaystyle 0< \sqrt{\cot^{-1}\left ( x \right )\left ( \pi

-\cot^{-1}\left ( x \right ) \right )}\leq \frac{\cot^{-1}x+\left ( \pi

-\cot^{-1}\left ( x \right ) \right )}{2}=\frac{\pi }{2}$$
$$\Rightarrow $$ $$0< y\leq \cfrac{\pi ^{2}}{4}$$
Question 7
1 / -0

Evaluate the following:
i. $$\displaystyle \sin \left ( \cos^{-1}\frac{3}{5} \right )$$
ii. $$\displaystyle \cos \left ( \tan^{-1}\frac{3}{4} \right )$$
iii. $$\displaystyle \sin \left ( \frac{\pi }{2}-\sin^{-1}\left ( \frac{1}{2} \right ) \right )$$

A
i. $$3/5$$ ii. $$3/5$$ iii. $$\frac{\sqrt{3}}{2}$$

B
i. $$3/5$$ ii. $$4/5$$ iii. $$-\frac{\sqrt{3}}{2}$$

C
i. $$-4/5$$ ii. $$-4/5$$ iii. $$\frac{\sqrt{3}}{2}$$

D
i. $$4/5$$ ii. $$4/5$$ iii. $$\frac{\sqrt{3}}{2}$$

Solution

i. Let $$\cos^{-1}3/5=\theta $$
   Then, $$\cos^{-1}3/5$$   $$\Rightarrow $$   $$\sin \theta =4/5$$
   $$\therefore $$   $$\sin \left ( \cos^{-1}3/5 \right )=\sin \theta =4/5$$

ii. Let $$\tan^{-1}3/4=\theta $$
    Then, $$\tan \theta =3/5$$   $$\Rightarrow $$   $$\cos \theta =4/5$$          ($$\because $$ $$\displaystyle \cos ^{2}\theta =\frac{1}{1+\tan ^{2}\theta }$$)
    $$\therefore $$   $$\cos \left ( \tan^{-1}\left ( 3/4 \right ) \right )=\cos \theta =4/5$$

iii.
$$\displaystyle \sin \left ( \frac{\pi }{2}-\sin^{-1}\left (

\frac{1}{2} \right ) \right )=\sin \left ( \frac{\pi }{2}-\left (

-\frac{\pi }{6} \right ) \right )=\sin \frac{2\pi

}{3}=\frac{\sqrt{3}}{2}$$
Question 8
1 / -0

Find the value of $$\displaystyle \sin^{-1}x+\sin^{-1}\frac{1}{x}+\cos^{-1}x+\cos^{-1}\frac{1}{x}$$

A
$$-\pi $$

B
$$+\pi $$

C
$$-2\pi $$

D
$$+2\pi $$

Solution

The value of $$\displaystyle

\sin^{-1}x+\sin^{-1}\frac{1}{x}+\cos^{-1}x+\cos^{-1}\frac{1}{x}$$ is defined only when $$\displaystyle x, \frac{1}{x}\epsilon \left [ -1, 1

\right ]$$

which is possible only when $$x=\pm 1$$

For which $$\displaystyle \sin^{-1}x+\sin^{-1}\frac{1}{x}+\cos^{-1}x+\cos^{-1}\frac{1}{x}=\pi $$
Question 9
1 / -0

If the range of $$f\left ( x \right )=\sin^{-1}x+\tan^{-1}x+\cos^{-1}x$$ is
$$\displaystyle =\left [ \frac{\pi }{a}, \frac{b\pi }{c} \right ]$$,
Find the value of $$a+b+c$$

A
$$7$$

B
$$8$$

C
$$10$$

D
$$11$$

Solution

Clearly, the domain of the function is $$[-1, 1]$$.
Also, $$\displaystyle

\tan^{-1}x\in \left [ -\frac{\pi }{4}, \frac{\pi }{4} \right ]$$ For $$x\in \left [ -1, 1 \right ]$$
Now, $$\displaystyle \sin^{-1}x+\cos^{-1}x=\frac{\pi }{2}$$for $$x\in \left [ -1, 1 \right ]$$
Thus, $$f\left ( x \right )=\tan^{-1}x+\dfrac{\pi }{2}$$, where $$x\in \left [ -1, 1 \right ]$$
Hence, the range is $$\displaystyle \left [ -\frac{\pi }{4}+\frac{\pi }{2}, \frac{\pi }{4}+\frac{\pi }{2} \right ]=\left [ \frac{\pi }{4}, \frac{3\pi }{4} \right ]$$
Question 10
1 / -0

If $$\cos^{-1}x=\left\{\begin{matrix}
a\pi -b\cos^{-1}\left ( 2x^{2}-1 \right ), if\:-1\leq x< 0\\
c\cos^{-1}\left ( 2x^{2}-1 \right ), if\:0\leq x\leq 1
\end{matrix}\right.$$.
Find the value of $$a+b+c$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let $$\cos^{-1}x=\theta $$, where $$\theta \epsilon \left [ 0, \pi \right ]$$   or   $$\cos \theta =x$$

Now,

$$\cos^{-1}\left ( 2x^{2}-1 \right )=\cos^{-1}\left ( 2\cos ^{2}\theta

-1 \right )=\cos^{-1}\left ( \cos 2\theta \right )=\cos^{-1}\left (

\cos \theta \right )$$, where $$\alpha \epsilon \left [ 0, 2\pi \right

]$$

Refer to the graph of $$y=\cos^{-1}\left ( \cos \alpha \right )$$, $$\alpha \epsilon \left [ 0, 2\pi \right ]$$

From the graph, we have
$$\cos^{-1}\left ( 2x^{2}-1 \right )=\left\{\begin{matrix}
\alpha , if\:0\leq \alpha < \pi \\
2\pi -\alpha , if\:\pi \leq \alpha \leq 2\pi
\end{matrix}\right.$$

     $$=\left\{\begin{matrix}
2\cos^{-1}x , if\:0\leq 2\cos^{-1}x\leq \pi \\
2\pi -\cos^{-1}x, if\:\pi < 2\cos^{-1}x\leq 2\pi
\end{matrix}\right.$$

     $$=\left\{\begin{matrix}
2\cos^{-1}x , if\:0\leq \cos^{-1}x\leq \left ( \pi /2 \right )\\
2\pi -\cos^{-1}x, if\:\left ( \pi /2 \right )< \cos^{-1}x\leq \pi
\end{matrix}\right.$$

     $$=\left\{\begin{matrix}
2\cos^{-1}x , if\:0\leq x\leq 1\\
2\pi -\cos^{-1}x, if\:-1\leq x< 0
\end{matrix}\right.$$

$$\Rightarrow $$   $$2\cos^{-1}x=\left\{\begin{matrix}
2\pi -\cos^{-1}\left ( 2x^{2}-1 \right ), if\:-1\leq x< 0\\
\cos^{-1}\left ( 2x^{2}-1 \right ), if\:0\leq x\leq 1
\end{matrix}\right.$$

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

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 31

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Inverse Trigonometric Functions Test - 31

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SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

$$x =-\pi$$

$$x =\pi$$

$$x =2\pi$$

All of the above

Solution

Question 4

0

1

2

3

Solution

Question 5

$$a=-\infty $$

$$a=-1$$

$$a=0$$

$$a=1$$

Solution

Question 6

$$2$$

$$4$$

$$5$$

$$6$$

Solution

Question 7

i. $$3/5$$ ii. $$3/5$$ iii. $$\frac{\sqrt{3}}{2}$$

i. $$3/5$$ ii. $$4/5$$ iii. $$-\frac{\sqrt{3}}{2}$$

i. $$-4/5$$ ii. $$-4/5$$ iii. $$\frac{\sqrt{3}}{2}$$

i. $$4/5$$ ii. $$4/5$$ iii. $$\frac{\sqrt{3}}{2}$$

Solution

Question 8

$$-\pi $$

$$+\pi $$

$$-2\pi $$

$$+2\pi $$

Solution

Question 9

$$7$$

$$8$$

$$10$$

$$11$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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