Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 40

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

$$\sec\ h^{-1}\left(\dfrac{1}{5}\right)$$=

A
$$\log( {\sqrt{24}+5})$$

B
$$\log {5+\sqrt{27}}$$

C
$$\log {26+\sqrt{5}}$$

D
$$\log {27+\sqrt{5}}$$

Solution

We have,
$$\sec h^{-1}\left (\dfrac {1}{5}\right)$$
We know that,
$$\sec h^{-1}x=\log \left (\dfrac {1}{x}+\sqrt {\dfrac {1}{x^{2}}1}\right)$$
Now,
$$\sec ^{ -1 }{ \dfrac { 1 }{ 5 } } =\log { \left( \dfrac { 1 }{ \dfrac { 1 }{ 5 } } +\sqrt { \dfrac { 1 }{ { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 } } -1 } \right) }$$
$$=\log { \left( 5+\sqrt { 25-1 } \right) } $$
$$=\log { \left( 5+\sqrt { 24 } \right) } $$
Hence this is the answer.
Question 2
1 / -0

If $$\cot \dfrac {2x}{3}+\tan \dfrac {x}{3}=\csc \dfrac {x}{3}$$ then value of $$\tan^{-1(\tan k)}$$ equals

A
$$2$$

B
$$2-\pi$$

C
$$\pi-2$$

D
$$2\pi-2$$

Solution
Question 3
1 / -0

If $$f ( x ) = \sqrt { \sec ^ { - 1 } \left( \frac { 2 - | x | } { 4 } \right) } ,$$ then the domain of $$f ( x )$$ is ______________.

A
$$[ - 2,2 ]$$

B
$$[ - 6,6 ]$$

C
$$( - \infty , - 6 ] \cup [ 6 , \infty )$$

D
None Of These
Question 4
1 / -0

The value of $${ sin }^{ -1 }(sin12)+{ cos }^{ -1 }(cos12)$$ is equal to :

A
$$Zero$$

B
$$24-2\pi $$

C
$$4\pi -24$$

D
None of these

Solution

Given,

$$\sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$$

The principle values,

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

$$\cos ^{-1}x\in \left [0,\pi \right ],\forall x\in [-1,1]$$

$$\therefore \sin ^{-1}(\sin 12)\neq 12\notin \left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$$

$$\therefore \cos ^{-1}(\cos 12)\neq 12\notin \left [ 0,\pi \right ]$$

$$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$$

$$=\sin ^{-1}(\sin (12-4\pi ))+\cos ^{-1}(\cos (4\pi -12))$$

$$=(12-4\pi ) + (4\pi -12)$$

$$=0$$

$$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)=0$$
Question 5
1 / -0

$$\displaystyle \sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left(\dfrac { { 2 }^{ r-1 } }{ 1+{ 2 }^{ 2r-1 } } \right)$$ is equal to :

A
$${ tan }^{ -1 }({ 2 }^{ n })$$

B
$${ tan }^{ -1 }({ 2 }^{ n })-\frac { \pi }{ 4 } $$

C
$${ tan }^{ -1 }({ 2 }^{ n+1 })$$

D
$${ tan }^{ -1 }({ 2 }^{ n+1 })-\frac { \pi }{ 4 } $$

Solution

$$\displaystyle\sum ^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}}{1+2^{2 r-1}}\bigg)=\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}(2-1)}{1+2^r.2^{r-1}}\bigg)$$

$$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^r-2^{r-1}}{1+2^r.2^{r-1}}\bigg)$$

$$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}{2^r}-\text{tan}^{-1}(2^{r-1})$$

$$=\text{tan}^{-1} 2-\text{tan}^{-1}1+\text{tan}^{-1}2^2-\text{tan}^{-1}2^1+\cdots+\text{tan}^{-1}{2^n}-\text{tan}^{-1}2^{n-1}$$

$$=\text{tan}^{-1}(2^n)-\dfrac{\pi}{4}$$
Question 6
1 / -0

If $$3{\tan ^{ - 1}}\left( {\frac{1}{{2 + \sqrt 3 }}} \right) - {\tan ^{ - 1}}\frac{1}{x} = {\tan ^{ - 1}}\frac{1}{3}$$ then $$x = $$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\sqrt 3 $$

Solution

Let $$y=\text{tan}^{-1} \bigg(\dfrac{1}{2+\sqrt{3}}\bigg)\implies \tan y=\dfrac{1}{2+\sqrt{3}}$$

$$\tan 3 y=\dfrac{3\tan y-\tan^3 y}{1+3\tan^2 y}=\dfrac{\frac{1}{2+\sqrt{3}}(3-\frac{1}{(2+\sqrt{3})^2})}{1+\frac{3}{(2+\sqrt{3})^2}}=\dfrac{\frac{1}{2+\sqrt{3}}(20+12\sqrt{3})}{4+4\sqrt{3}}=\dfrac{20+12\sqrt{3}}{20+12\sqrt{3}}=1$$

$$\implies 3\text{tan}^{-1}\bigg(\dfrac{1}{2+\sqrt{3}}\bigg)=\text{tan}^{-1}(1)$$

$$\text{tan}^{-1}\bigg(\dfrac{1}{x}\bigg)=\text{tan}^{-1}(1)-\text{tan}^{-1}\dfrac{1}{3}=\text{tan}^{-1}\dfrac{1-\dfrac{1}{3}}{1+\dfrac{1}{3}}=\dfrac{1}{2}$$

$$\implies x=2$$
Question 7
1 / -0

For $$0<x<1$$ the value of $$\cos^{-1}x+\cos^{-1}(-x)=?$$

A
$$0$$

B
$$\pi$$

C
$$-\pi$$

D
none of these

Solution

We've for $$|x|\le 1$$, $$\cos^{-1} (-x)=\pi -\cos^{-1}x$$.

It also holds in this case.

So for $$0<x<1$$ we've,

$$\cos^{-1}x+\cos^{-1} (-x)$$

$$=\cos^{-1} x+\pi-\cos^{-1}x$$

$$=\pi$$.
Question 8
1 / -0

The value of $$\sin\ h(\cos\ h^{-1}x)$$ is

A
$$\sqrt{x^{2}+1}$$

B
$$1/\sqrt{x^{2}+1}$$

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

D
$$none of these$$
Question 9
1 / -0

The value of $$\displaystyle \sum_{r=0}^{\infty}{ tan}^{1}\left(\dfrac{1}{1+r+{r}^{2}}\right)$$ is equal to

A
$$\dfrac { \pi } { 4 }$$

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

C
$$\pi$$

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

Solution

given, $$\displaystyle\sum^{\infty}_{r=0}\tan^{-1}\left(\dfrac{1}{1+r+r^2}\right)$$

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

$$=\tan^{-1}\left(\dfrac{(r+1)-r}{1+(r+1)r}\right)=\tan^{-1}(r+1)-\tan^{-1}r$$

G.E$$=\displaystyle\sum^{\infty}_{r=0}\tan^{-1}(r+1)-\tan^{-1}(r)$$

$$=\tan^{-1}1-\tan^{-1}0+\tan^{-1}2-\tan^{-1}1+...…+\tan^{-1}\infty$$

$$=\tan^{-1}\infty -\tan^{-1}0$$

$$=\dfrac{\pi}{2}-0$$

$$=\dfrac{\pi}{2}$$.
Question 10
1 / -0

The simplified form of $${ cos }^{ -1 }\left( \frac { 3 }{ 5 } { cos }x+\frac { 4 }{ 5 } { sin }x \right) $$ is :

A
$${ tan }^{ -1 }\frac { 4 }{ 3 } -x$$

B
$${ tan }^{ -1 }\frac { 1 }{ 3 } -x$$

C
$${ tan }^{ -1 }\frac { 4 }{ 3 } +x$$

D
None of these

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 40

Result

Inverse Trigonometric Functions Test - 40

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\log( {\sqrt{24}+5})$$

$$\log {5+\sqrt{27}}$$

$$\log {26+\sqrt{5}}$$

$$\log {27+\sqrt{5}}$$

Solution

Question 2

$$2$$

$$2-\pi$$

$$\pi-2$$

$$2\pi-2$$

Solution

Question 3

$$[ - 2,2 ]$$

$$[ - 6,6 ]$$

$$( - \infty , - 6 ] \cup [ 6 , \infty )$$

None Of These

Question 4

$$Zero$$

$$24-2\pi $$

$$4\pi -24$$

None of these

Solution

Question 5

$${ tan }^{ -1 }({ 2 }^{ n })$$

$${ tan }^{ -1 }({ 2 }^{ n })-\frac { \pi }{ 4 } $$

$${ tan }^{ -1 }({ 2 }^{ n+1 })$$

$${ tan }^{ -1 }({ 2 }^{ n+1 })-\frac { \pi }{ 4 } $$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$\sqrt 3 $$

Solution

Question 7

$$0$$

$$\pi$$

$$-\pi$$

none of these

Solution

Question 8

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

$$1/\sqrt{x^{2}+1}$$

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

$$none of these$$

Question 9

$$\dfrac { \pi } { 4 }$$

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

$$\pi$$

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

Solution

Question 10

$${ tan }^{ -1 }\frac { 4 }{ 3 } -x$$

$${ tan }^{ -1 }\frac { 1 }{ 3 } -x$$

$${ tan }^{ -1 }\frac { 4 }{ 3 } +x$$

None of these

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.