Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The range of the function, $$\displaystyle f\left ( x \right ) = \left ( 1 + \sec^{-1} x \right ) \left ( 1 + \cos^{-1} x \right )$$ is

A
$$ \displaystyle \left ( -\infty ,\: \infty \right )$$

B
$$\displaystyle \left (-\infty, \: 0 \right ] \cup \left [4, \: \infty \right )$$

C
$$\displaystyle \left \{ 0, \: \left ( 1 + \pi \right )^{2} \right \}$$

D
$$\displaystyle \left [ 1, \: \left ( 1 + \pi \right )^{2} \right ]$$

Solution

$$f(x)=(1+sec^{-1}(x))(1+cos^{-1}(x))$$
Here the limiting component is $$cos^{-1}(x)$$, since the domain of $$cos^{-1}(x)$$ is $$[-1,1]$$.
Therefore,
$$f(1)=(1+0)(1+0)$$
$$=1$$
$$f(-1)=(1+\pi)(1+\pi)$$
$$=(1+\pi)^{2}$$
Hence range of $$f(x)=[1,(1+\pi)^{2}]$$
Question 2
1 / -0

Number of real value of $$x$$ satisfying the equation, $$\displaystyle \arctan \sqrt{x \left ( x+1 \right )} + \arcsin \sqrt{x \left ( x+1 \right ) + 1} = \frac{\pi}{2}$$ is

A
0

B
1

C
2

D
more than 2

Solution

Let
$$f(x)=tan^{-1}(\sqrt{x(x+1)})+sin^{-1}(\sqrt{x(x+1)+1})-\dfrac{\pi}{2}$$
Hence $$f(x)=0$$ for x={-1,0}.
Where {} denotes singleton set.
Hence we have two solutions for the above given equation.
Question 3
1 / -0

The value of $$x$$ for which $$\sin { \left( \cot ^{ -1 }{ \left( 1+x \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$ is:

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

B
$$1$$

C
$$0$$

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

Solution

We have, $$\sin { \left( \cot ^{ -1 }{ \left( 1+x \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$
$$\displaystyle \Rightarrow \cos { \left( \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$
$$\displaystyle \Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =2n\pi \pm \tan ^{ -1 }{ x } $$
Substitute $$\displaystyle n=0\Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =\pm \tan ^{ -1 }{ x } =\tan ^{ -1 }{ \left( \pm x \right) } $$
$$\displaystyle \Rightarrow \frac { \pi }{ 2 } =\tan ^{ -1 }{ \left( \pm x \right) } +\cot ^{ -1 }{ \left( x+1 \right) } $$
$$\displaystyle \Rightarrow x+1=\pm x\Rightarrow 2x+1=0\Rightarrow x=-\frac { 1 }{ 2 } $$
Question 4
1 / -0

Range of $$\displaystyle f\left( x \right)=\sin ^{ -1 }{ x } +\tan ^{ -1 }{ x } +\cos ^{ -1 }{ x } $$ is

A
$$\displaystyle \left[ 0,\pi \right] $$

B
$$\displaystyle \left[ \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right] $$

C
$$\displaystyle \left[ -\pi ,2\pi \right] $$

D
None of these

Solution

Given $$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$$

First we will calculate domain of function $$f(x)$$
function $$\sin^{−1}(x)$$ is defined in $$x∈[−1,1]$$

Similarly function $$\cos^{−1}(x)$$ is defined in $$x∈[−1,1]$$
and function $$\tan^{−1}(x)$$ is defined in $$x∈(−∞,+∞)$$

So $$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$$ is defined in $$x∈[−1,1]$$

So $$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$$ is defined in
$$x∈[−1,1]$$

So $$f(x)=\dfrac π2+\tan^{−1}(x)$$

Now $$f′(x)=\dfrac {1}{1+x^2}>0 \, ∀x∈[−1,1]$$

So $$f(x)$$ is Strictly Increasing function.

So $$f(−1)=\dfrac π2+\tan^{−1}(−1)=\dfrac π2−\dfrac π4=\dfrac π4$$

and $$f(+1)=\dfrac π2+\tan^{−1}(1)=\dfrac π2+\dfrac π4=\dfrac {3π}4$$
So $$f(x)∈[\dfrac π4,\dfrac {3π}4]$$
Question 5
1 / -0

$$\displaystyle \alpha = \sin^{-1} \left ( \cos \left ( \sin^{-1} x \right ) \right )$$ and $$\displaystyle \beta = \cos^{-1} \left ( \sin \left ( \cos^{-1} x \right ) \right )$$ then:

A
$$\displaystyle \tan \alpha = \cot \beta$$

B
$$\displaystyle \tan \alpha = - \cot \beta$$

C
$$\displaystyle \tan \alpha = \tan \beta$$

D
$$\displaystyle \tan \alpha = - \tan \beta$$

Solution

$$\alpha=sin^{-1}(cos(sin^{-1}(x))))$$
$$=\dfrac{\pi}{2}-cos^{-1}(cos(\dfrac{\pi}{2}-cos^{-1}(x)))$$
$$=\dfrac{\pi}{2}-cos^{-1}(sin(cos^{-1}(x)))$$
$$=\dfrac{\pi}{2}-\beta$$
Hence $$\alpha=\dfrac{\pi}{2}-\beta$$
Hence $$tan(\alpha)=cot(\beta)$$
Question 6
1 / -0

Number of value $$x$$ satisfying the equation $$\displaystyle \sin^{-1} \left ( \frac{5}{x} \right ) + \sin^{-1} \left ( \frac{12}{x} \right ) = \frac{\pi}{2}$$ is

A
0

B
1

C
2

D
more than 2

Solution

$$\sin^{-1}(\frac{5}{x})+\sin^{-1}(\frac{12}{x})=\frac{\pi}{2}$$
$$\sin^{-1}(\frac{5}{x})=\frac{\pi}{2}-\sin^{-1}(\frac{12}{x})$$
$$\sin^{-1}(\frac{5}{x})=\cos^{-1}(\frac{12}{x})$$
$$\cos^{-1}(\frac{\sqrt{x^2-25}}{x})=\cos^{-1}(\frac{12}{x})$$
$$x^2-25=144$$
$$x^2=169$$
$$x=\pm13$$
However for -13, we wouldn't get $$\dfrac{-\pi}{2}$$ on the RHS.
Hence there is only one solution, $$x=13$$
Question 7
1 / -0

There exists a positive real number $$x$$ satisfying $$\displaystyle \cos \left ( \tan^{-1} x \right ) = x$$. 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

$$\cos(\tan^{-1}(x))=x$$
$$\cos(\cos^{-1}(\frac{1}{\sqrt{1+x^{2}}}))=x$$
Therefore
$$\frac{1}{\sqrt{1+x^{2}}}=x$$
Squaring on both the sides we get
$$x^2(x^2+1)=1$$
$$x^4+x^2-1=0$$
Hence $$x^2=\frac{-1+\sqrt{5}}{2}$$
$$\frac{x^2}{2}=\frac{\sqrt{5}-1}{4}$$
$$\cos^{-1}(\frac{x^2}{2})$$ $$=\cos^{-1}(\frac{\sqrt{5}-1}{4})$$
$$=72^{0}$$
$$=\frac{2\pi}{5}$$
Question 8
1 / -0

The greatest and least values of $${ \left( \sin ^{ -1 }{ x } \right) }^{ 3 }+{ \left( \cos ^{ -1 }{ x } \right) }^{ 3 }$$ are

A
$$-\cfrac { \pi }{ 2 } ,\cfrac { \pi }{ 2 } $$

B
$$-\cfrac { { \pi }^{ 3 } }{ 8 } ,\cfrac { { \pi }^{ 3 } }{ 8 } $$

C
$$\cfrac { { \pi }^{ 3 } }{ 32 } ,\cfrac { {7 \pi }^{ 3 } }{ 8 } $$

D
none of these

Solution
Question 9
1 / -0

The value of $$\sin ^{ -1 }{ \left( \sin { 10 } \right) } $$ is

A
$$10$$

B
$$10-3\pi $$

C
$$3\pi -10$$

D
none of these

Solution

Here $$\theta =10 $$ rad doesnt lie between $$ -\dfrac \pi 2 $$ and$$ \dfrac \pi 2 $$

But , $$ 3\pi -\theta $$ lies between $$ -\dfrac \pi 2 $$ and$$ \dfrac \pi 2 $$

Also , $$ \sin (3\pi -10 )=\sin 10 $$

$$\implies \sin ^ {-1} (\sin 10) =\sin ^ {-1} (\sin (3\pi -10))=3\pi -10 $$
Question 10
1 / -0

Which one of the following statement is meaningless?

A
$$\displaystyle \cos^{-1} \left ( ln \left ( \frac{2e + 4}{3} \right ) \right )$$

B
$$\displaystyle cosec^{-1} \left ( \frac{\pi}{3} \right )$$

C
$$ \displaystyle \cot^{-1} \left ( \frac{\pi}{2} \right )$$

D
$$\displaystyle \sec^{-1} \left ( \pi \right )$$

Solution

The domain for $$f(x)=cos^{-1}(x)$$ is $$[-1,1]$$

However, clearly $$ln \left(\dfrac{2e+4}{3}\right)>1$$

Since $$f(x)$$ for $$|x|>1$$ is not possible.

Option A is meaningless.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 26

Result

Inverse Trigonometric Functions Test - 26

Score

-

Rank

-

SHARING IS CARING

Question 1

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

$$\displaystyle \left (-\infty, \: 0 \right ] \cup \left [4, \: \infty \right )$$

$$\displaystyle \left \{ 0, \: \left ( 1 + \pi \right )^{2} \right \}$$

$$\displaystyle \left [ 1, \: \left ( 1 + \pi \right )^{2} \right ]$$

Solution

Question 2

0

1

2

more than 2

Solution

Question 3

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

$$1$$

$$0$$

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

Solution

Question 4

$$\displaystyle \left[ 0,\pi \right] $$

$$\displaystyle \left[ \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right] $$

$$\displaystyle \left[ -\pi ,2\pi \right] $$

None of these

Solution

Question 5

$$\displaystyle \tan \alpha = \cot \beta$$

$$\displaystyle \tan \alpha = - \cot \beta$$

$$\displaystyle \tan \alpha = \tan \beta$$

$$\displaystyle \tan \alpha = - \tan \beta$$

Solution

Question 6

0

1

2

more than 2

Solution

Question 7

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

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

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

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

Solution

Question 8

$$-\cfrac { \pi }{ 2 } ,\cfrac { \pi }{ 2 } $$

$$-\cfrac { { \pi }^{ 3 } }{ 8 } ,\cfrac { { \pi }^{ 3 } }{ 8 } $$

$$\cfrac { { \pi }^{ 3 } }{ 32 } ,\cfrac { {7 \pi }^{ 3 } }{ 8 } $$

none of these

Solution

Question 9

$$10$$

$$10-3\pi $$

$$3\pi -10$$

none of these

Solution

Question 10

$$\displaystyle \cos^{-1} \left ( ln \left ( \frac{2e + 4}{3} \right ) \right )$$

$$\displaystyle cosec^{-1} \left ( \frac{\pi}{3} \right )$$

$$ \displaystyle \cot^{-1} \left ( \frac{\pi}{2} \right )$$

$$\displaystyle \sec^{-1} \left ( \pi \right )$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.