MCQ 511 Mark
Solve for $x :\{\text{x}\cos(\cot^{-1}\text{x})+\sin(\cot^{-1}\text{x})\}^2=\frac{51}{50}$
- A
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{5\sqrt{2}}$
- C
$2\sqrt{2}$
- D
$5\sqrt{2}$
AnswerCorrect option: B. $\frac{1}{5\sqrt{2}}$
View full question & answer→MCQ 521 Mark
If $\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta,$ then $\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2$ is equal to:
- A
$36$
- B
$-36\sin^2\theta$
- ✓
$36\sin^2\theta$
- D
$-36\cos^2\theta$
AnswerCorrect option: C. $36\sin^2\theta$
We know
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big[\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big]$
Now,
$\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta$
$\Rightarrow\cos^{-1}\Big[\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}\Big]=\theta$
$\Rightarrow\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}=\cos\theta$
$\Rightarrow\text{xy}-\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=6\cos\theta$
$\Rightarrow\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=\text{xy}-6\cos\theta$
$\Rightarrow(4-\text{x}^2)(9-\text{y}^2)=\text{x}^2\text{y}^2+36\cos^2\theta-12\text{xy}\cos\theta$ (Squaring both the sides)
$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^{2}\theta-12\text{xy}\cos\theta$
$\Rightarrow36-4\text{y}^2-9\text{x}^2=36\cos^2\theta-12\text{xy}\cos\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36-36\cos^2\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36\sin^2\theta$
View full question & answer→MCQ 531 Mark
$\Bigg(\cot\Bigg(\sin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}+\cos^{-1}\frac{\sqrt{-12}}{4}+\sec^{\sqrt{2}}\Bigg)\Bigg)$ is:
AnswerThe above expression can be rewritten as
$\sin^{-1}(\cot(15^{0}+30^{0}+45^{0}))$
$ =\sin^{-1}(\cot(90^{0}))$
$ =\sin^{-1}(0)$
$ = 0$
View full question & answer→MCQ 541 Mark
Choose the correct answer from the given four options. If $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi,$ then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ equals:
AnswerThe domain of $\cos^{-1} x$ is $[0,\pi]$
We are given that, $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi$
Which is possible only when $\alpha=\beta=\gamma=\cos\pi\ $ or $-1$
Now, $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$
$=-1(-1-1)-1(-1-1)-1(-1-1)$
$=2+2+2$
$=6$
View full question & answer→MCQ 551 Mark
$\cos^{-1}[\cos(2\cot^{-1}(\sqrt2-1))]= ..............$
- A
$\sqrt2-1$
- B
$1+\sqrt2$
- C
$\frac{\pi}{4}$
- ✓
$\frac{3\pi}{4}$
AnswerCorrect option: D. $\frac{3\pi}{4}$
View full question & answer→MCQ 561 Mark
The equation $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$ has:
- ✓
- B
- C
Infinitely many solution.
- D
View full question & answer→MCQ 571 Mark
What is the value of $ \sin-1(\sin 6)?$
AnswerWe know that $\sin(\text{x}) = \sin(2\text{A} * π + \text{x})$ where A can be positive or negative integer.
If A is -1, then $ \sin(6) = \sin(-2π + 6);$
If A is 1, then $ \sin(6) = \sin(2π + 6).$
View full question & answer→MCQ 581 Mark
The value of $\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)=$
- ✓
$\frac{19}{8}$
- B
$\frac{8}{19}$
- C
$\frac{19}{2}$
- D
$\frac{3}{4}$
AnswerCorrect option: A. $\frac{19}{8}$
$\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\Bigg(\tan^{-1}\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Bigg(\tan^{-1}\frac{\frac{4}{5}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Big(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\bigg(\tan^{-1}\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\bigg)$
$=\frac{\frac{16+3}{12}}{\frac{2}{3}}$
$=\frac{19}{8}$
Hence, the correct answer is option (a).
View full question & answer→MCQ 591 Mark
$ \tan^{−1}\sqrt{3}+\sec−12–\cos−^{1}1$ is equal to ________.
- A
- ✓
$ \frac{2}{π3}$
- C
$ \frac{\pi}{3}$
- D
$ \frac{\pi}{4}$
AnswerCorrect option: B. $ \frac{2}{π3}$
$\tan^{-1}\sqrt{3}=\frac{\pi}{3},\sec^{-1}2,\cos^{-1}1=0$
$ ∴\tan^{−1}\sqrt{13}+\sec^{−1}2−\cos^{−1}1=\frac{π}{3}+\frac{π}{3}−0$
$ =\frac{2π}{3}$
View full question & answer→MCQ 601 Mark
Consider $ \text{x} = 4\tan^{-1}\left (\frac {1}{5}\right ), \text{y} = \tan^{-1} \left (\frac {1}{70}\right )\text{and } \text{z} = \tan^{-1}\bigg (\frac {1}{99}\bigg)$ .What is xx equal to?
- A
$ \tan^{-1}\left (\frac {60}{119}\right )$
- ✓
$ \tan^{-1}\left (\frac {120}{119}\right )$
- C
$ \tan^{-1}\left (\frac {90}{119}\right )$
- D
$ \tan^{-1}\left (\frac {170}{119}\right )$
AnswerCorrect option: B. $ \tan^{-1}\left (\frac {120}{119}\right )$
$\text{x}=4{ \tan }^{ -1 }\left(\dfrac { 1 }{ 5 } \right)\\$
$ =2{ \tan }^{ -1 }\left(\frac { 1 }{ 5 } \right)+2{ \tan }^{ -1 }\left(\frac { 1 }{ 5 } \right)\\$
$ =2{ \tan }^{ -1 }\left(\frac { \frac { 1 }{ 5 } +\frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 5 } \times \frac { 1 }{ 5 } } \right)\\$
$ =2{ \tan }^{ -1 }\left(\frac { 5 }{ 12 } \right)\\$
$ =2{ \tan }^{ -1 }\left(\frac {120 }{ 119 } \right)\\$
View full question & answer→MCQ 611 Mark
$3\tan^{-1} a$ is equal to:
- A
$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1+3\text{a}^2}\Big)$
- B
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1+3\text{a}^2}\Big)$
- C
$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1-3\text{a}^2}\Big)$
- ✓
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$
AnswerCorrect option: D. $\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$
View full question & answer→MCQ 621 Mark
Consider the following statements:
- $\tan^{-1} 1+ \tan^{-1} (0.5) = \dfrac {\pi}2$
- $\sin^{-1}{\cfrac{1}{3} }+ \cos^{-1}{\cfrac{1}{3}} =\cfrac{\pi}{2}$
Which of the above statements is/are correct ?
- A
$1$ only
- ✓
$2$ only
- C
Both $1$ and $2$
- D
Neither $1$ nor $2$
AnswerCorrect option: B. $2$ only
We know that $\tan^{-1} { \text{x} } + \cot^{-1} { \text{x} } =\frac { \pi }{ 2 } \text{x} \in \text{R and}\sin^{-1}{\text{x}} + \cos^{-1}{\text{x}} =\frac{\pi}{2}$,
and $ \sin-1\frac{1}{3}+\cos-1{\frac{1}{3}} =\cfrac{\pi}{2}$
Hence, only second statement is correct.
View full question & answer→MCQ 631 Mark
The range of $\sin^{-1}\text{x}+\cos^{-1}\text{x}+\tan^{-1}\text{x}$ is:
AnswerCorrect option: B. $[\frac{\pi}{4},\frac{3\pi}{4}]$
View full question & answer→MCQ 641 Mark
Choose the correct answer from the given four options.Which of the following is the principal value branch of $\cos^{-1}x?$
- A
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
- B
$(0,\pi)$
- ✓
$[0,\pi]$
- D
$(0,\pi)-\Big\{\frac{\pi}{2}\Big\}$
AnswerCorrect option: C. $[0,\pi]$
The principal value branch of $\cos^{-1}x$ is $[0,\pi].$
View full question & answer→MCQ 651 Mark
$\cos^{-1}\frac{1}{2}+2\sin^{-1}\frac{1}{2}$ is equal to:
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{6}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{2\pi}{3}$
AnswerCorrect option: D. $\frac{2\pi}{3}$
View full question & answer→MCQ 661 Mark
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$ is
AnswerThe value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$
$ =\cos^{−1}(\cos(4π−12))−\sin^{−1}(−\sin(4π−14))$
$ =4π − 12 − 14 + 4π = 8π − 26$
View full question & answer→MCQ 671 Mark
The value of $ \cos \left( \sin^{-1} \left( \frac {2}{3} \right) \right)$ is equal to :
- A
$ \frac {\sqrt4}{8}$
- B
$ \frac {\sqrt4}{3}$
- C
$ \frac {\sqrt5}{4}$
- ✓
$ \frac {\sqrt5}{3}$
AnswerCorrect option: D. $ \frac {\sqrt5}{3}$
The value of $ \cos \left( \sin^{-1} \left( \frac {2}{3} \right) \right)$ is
$=\cos\Bigg(\cos^{-1}\sqrt{1-\bigg(1-\frac{2}{3}\bigg)^2}\Bigg)$
$ = \cos \left( \cos^{-1}\left( \sqrt{1 - \frac{4}{9} } \right) \right)$
$ = \cos \left( \cos^{-1} \left( \frac {\sqrt5}{3} \right) \right)$
$ = \frac {\sqrt5}{3}$
View full question & answer→MCQ 681 Mark
$ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$ is equal to
- ✓
$ \frac{\pi}{2}$
- B
$ \frac{\pi}{3}$
- C
$ \frac{\pi}{4}$
- D
$ \frac{\pi}{6}$
AnswerCorrect option: A. $ \frac{\pi}{2}$
Given, $ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$
$⇒\sin^{−1}\text{x}+\sin^{−1}\text{y}=\sin ^{ -1 } \Big(\text{x}\sqrt { 1-{ \text{y} }^{ 2 } } +\text{y}\sqrt { 1-\text{x}^{ 2 } }\Big)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { 1-\left(\frac { 4 }{ 5 } \right) } +\frac { 4 }{ 5 } \sqrt { 1-\left(\frac { 3 }{ 5 } \right)^{ 2 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { \frac { 25-16 }{ 25 } ) } +\frac { 4 }{ 5 } \sqrt { \frac { 25-9 }{ 25 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \times \frac { 3 }{ 5 } +\frac { 4 }{ 5 } \times \frac { 4 }{ 5 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 16 }{ 25 } +\frac { 9 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 25 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } (1)$
$ \Rightarrow \cfrac { \pi }{ 2 }$
View full question & answer→MCQ 691 Mark
In a $\triangle\text{ABC},$ if C is a right angle, then $\tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{b}}\Big)=$
- A
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
- C
$\frac{5\pi}{2}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: B. $\frac{\pi}{4}$
We know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
View full question & answer→MCQ 701 Mark
The number of real solution of the equation
$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$ is:
Answer$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$
$\Rightarrow\sqrt{2\cos^2\text{x}}=\sqrt2(-\pi-\text{x})$
$\Rightarrow|\cos\text{x}|=\text{x}$
If $\cos\text{x}$ is positive then $\cos\text{x}=-\pi-\text{x}$
It does not satisfy any value in the interval $\Big(-\pi,-\frac{\pi}{2}\Big)$
For the interval $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
$\cos\text{x}=\text{x}$
It gives the value of x in the $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
For the interval $\Big[-\frac{\pi}{2},\pi\Big]$
$-\cos\text{x}=\pi-\text{x}$
$\cos\text{x}=\text{x}-\pi$
It gives one value of x in the interval $\Big[\frac{\pi}{2},\pi\Big].$
Two real solution in the interval $[-\pi,\pi]$
View full question & answer→MCQ 711 Mark
The value of expression $2\sec^{-1}0+\sin^{-1}\Big(\frac{1}{2}\Big)$
- A
$\frac{\pi}{6}$
- ✓
$\frac{5\pi}{6}$
- C
$\frac{7\pi}{6}$
- D
$1$
AnswerCorrect option: B. $\frac{5\pi}{6}$
View full question & answer→MCQ 721 Mark
Solve for $x :\sin^{-1}2\text{x}+\sin^{-1}3\text{x}=\frac{\pi}{3}$
- A
$\sqrt{\frac{76}{3}}$
- ✓
$\sqrt{\frac{3}{76}}$
- C
$\frac{3}{\sqrt{76}}$
- D
$\frac{\sqrt{3}}{76}$
AnswerCorrect option: B. $\sqrt{\frac{3}{76}}$
View full question & answer→MCQ 731 Mark
$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$ is equal to:
Answer$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{2}{11}\times\frac{1}{11}}\Bigg)$
$=\tan^{-1}\Bigg(\frac{\frac{3}{11}}{1-\frac{2}{121}}\Bigg)$
$=\tan^{-1}\Big(\frac{33}{119}\Big)$
View full question & answer→MCQ 741 Mark
$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$ is equal to:
- ✓
$\text{x}$
- B
$\sqrt{1-\text{x}^2}$
- C
$\frac{1}{\text{x}}$
- D
$\text{none of these}$
AnswerCorrect option: A. $\text{x}$
Put $\cos^{-1}\text{x}=\text{u}$
$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$
$=\sin\big[\cot^{-1}\{\tan(\text{u})\}\big]$
$=\sin\Big[\cot^{-1}\Big\{\cot\Big(\frac{\pi}{2}-\text{u}\Big )\Big\}\Big]$
$=\sin\Big[\frac{\pi}{2}-\text{u}\Big]$
$=\cos\text{u}$
$=\text{x}$ $\big(\therefore\ \cos^{-1}\text{x}=\text{u}\Rightarrow\text{x}=\cos\text{u}\big)$
View full question & answer→MCQ 751 Mark
If $ \text{x} \in \left ( \frac{3\pi}{2}, 2\pi \right )$ then the value of the expression $ \sin^{-1}[\cos({\cos^{-1}(\cos \, \text{x})}+\sin^{-1}(\sin \, \text{x}))]$ is:
AnswerCorrect option: B. $ \frac{\pi}{2}$
$\text{x}\in \left ( \frac{3\pi}{2}, 2\pi \right )$
Now, $ \cos^{-1}(\cos \,\text{ x})=2π−\text{x}$
and $ \sin^{-1}(\sin \, \text{x})=\text{x}-2\pi$
$ ∴\cos^{−1}(\cos\text{x})+\sin−1(\sin\text{x})=0$
$\sin−1[\cos{\cos−1(\cos\text{x})+\sin−1(\sin\text{x})}]$
$ =\sin^{−1}{\cos(0)}=\sin^{−1}(1)=\frac{π}{2}$
View full question & answer→MCQ 761 Mark
Number of triplets (x, y, z) satisfying $ \sin ^{-1}\text{x}+\sin ^{-1}\text{y}+\cos ^{-1}\text{z}=2\pi$ is:
AnswerLet f(x, y, z) $ =\sin ^{-1}\text{x}+\sin ^{-1}\text{y}+\cos^{-1}\text{z}$
It will attain the value 2π only if $ \sin ^{-1}\text{x}=\sin ^{-1}\text{y}=\frac{\pi }{2} \text{and } \cos^{-1}\text{z}=\pi$
This ispossible only if x = y = 1and z = -1
Hence there is only one solutionf $ (1, 1, -1)= 2π.$
View full question & answer→MCQ 771 Mark
If $\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of $x$ is:
- A
$\frac{3}{2}$
- B
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{\sqrt{3}}{2}$
- D
$\frac{2}{\sqrt{3}}$
AnswerCorrect option: C. $\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 781 Mark
The value of $\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)$ is:
- A
$\frac{5}{17}$
- ✓
$\frac{6}{17}$
- C
$\frac{3}{17}$
- D
$\frac{4}{17}$
AnswerCorrect option: B. $\frac{6}{17}$
View full question & answer→MCQ 791 Mark
$2\cos^{-1}\text{x}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$ is true for:
AnswerCorrect option: D. $\frac{1}{\sqrt{2}}\leq\text{x}\leq1$
View full question & answer→MCQ 801 Mark
$2\tan^{-1}(\cos\text{x})=\tan^{-1}(2\ \text{cosec x})$
- A
$0$
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: C. $\frac{\pi}{4}$
View full question & answer→MCQ 811 Mark
If $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8,$ then x =
- A
$5$
- ✓
$\frac{1}{5}$
- C
$\frac{5}{14}$
- D
$\frac{14}{5}$
AnswerCorrect option: B. $\frac{1}{5}$
We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}$
Now,
$\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8$
$\Rightarrow\tan^{-1}\Big(\frac{3+\text{x}}{1-3\text{x}}\Big)=\tan^{-1}8$
$\Rightarrow\frac{3+\text{x}}{1-3\text{x}}=8$
$\Rightarrow3+\text{x}=8-24\text{x}$
$\Rightarrow3-8=-24\text{x}-\text{x}$
$\Rightarrow-5=-25\text{x}$
$\Rightarrow\text{x}=\frac{5}{25}=\frac{1}{5}$
View full question & answer→MCQ 821 Mark
The value of the $\tan\Big(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\Big)=$
- A
$\frac{6}{17}$
- B
$\frac{7}{16}$
- C
$\frac{16}{7}$
- ✓
View full question & answer→MCQ 831 Mark
The value of $\tan^{-1}\Big(\frac{1}{2}\Big)+\tan^{-1}\Big(\frac{1}{3}\Big)+\tan^{-1}\Big(\frac{7}{8}\Big)$ is:
AnswerCorrect option: C. $\tan^{-1}(15)$
View full question & answer→MCQ 841 Mark
If $6\sin^{-1}(\text{x}^2-6\text{x}+8.5)=\pi,$ then $x$ is equal to:
View full question & answer→MCQ 851 Mark
If $3\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)-4\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+2\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{\pi}{3}$ is equal to:
- ✓
$\frac{1}{\sqrt3}$
- B
$-\frac{1}{\sqrt3}$
- C
$\sqrt3$
- D
$-\frac{\sqrt3}{4}$
AnswerCorrect option: A. $\frac{1}{\sqrt3}$
Let $\text{x}=\tan\text{y}$
Then,
$3\sin^{-1}\Big(\frac{\tan2\text{y}}{1+\tan^2\text{y}}\Big)-4\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)+2\tan^{-1}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)=\frac{\pi}{3}$
$\Rightarrow3\sin^{-1}(\sin2\text{y})-4\cos^{-1}(\cos2\text{y})+2\tan^{-1}(\tan2\text{y})=\frac{\pi}{3}$
$\Big[\because\ \sin2\text{y}=\Big(\frac{2\tan\text{y}}{1+\tan^2\text{y}}\Big),\cos2\text{y}=\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)\text{ and }\tan2\text{y}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)\Big]$
$\Rightarrow3\times2\text{y}-4\times2\text{y}+2\times2\text{y}=\frac{\pi}{3}$
$\Rightarrow6\text{y}-8\text{y}+4\text{y}=\frac{\pi}{3}$
$\Rightarrow2\text{y}=\frac{\pi}{3}$
$\Rightarrow\text{y}=\frac{\pi}{6}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$ $\big[\because\ \tan^{-1}\text{x}=\text{y}\big]$
$\Rightarrow\text{x}=\tan\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$
View full question & answer→MCQ 861 Mark
The domain of $\cos^{-1}\big(\text{x}^2-4\big)$ is:
- A
$[3,5]$
- B
$[-1,1]$
- ✓
$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
- D
$\Big[-\sqrt5,-\sqrt3\Big]\cap\Big[\sqrt3,\sqrt5\Big]$
AnswerCorrect option: C. $\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
Let, $\cos^{-1}\big(\text{x}^2-4\big)=\text{y}$
$\Rightarrow\cos\text{y}=\text{x}^2-4$
$\Rightarrow-1\leq\text{x}^2-4\leq1$
$\Rightarrow3\leq\text{x}^2\leq5$
$\Rightarrow\pm\sqrt3\leq\text{x}\pm\sqrt5$
$\text{x}\in\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
View full question & answer→MCQ 871 Mark
If $\text{x }\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),$ then the value of $\tan^{-1}\Big(\frac{\tan\text{x}}{4}\Big)+\tan^{-1}\Big(\frac{3\sin2\text{x}}{5+3\cos2\text{x}}\Big)$ is:
- A
$\frac{\text{x}}{2}$
- B
$2x$
- C
$3x$
- ✓
$x$
View full question & answer→MCQ 881 Mark
$ \sin^{-1}\text{x}+\cos^{1}\text{x}= $
AnswerCorrect option: A. $ \frac{π}{2}$
$ \sin-1\text{x}+\cos-1\text{x}=π2; \text{x} ∈ [-1,1] $
View full question & answer→MCQ 891 Mark
Solve:$\sin { \left( { \tan }^{ -1 }\text{x} \right) } ,\left| \text{x} \right| <1$ is equal to:
- A
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
- B
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
- C
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
- ✓
$\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$
AnswerCorrect option: D. $\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$
We need to find value of $ \sin (\tan^{-1}\text{x})\text{Put } \text{y}=\tan^{-1}\text{x}$
$ \Rightarrow \displaystyle \tan {\text{ y} }$
$ \therefore \tan \text{y}=\frac {\sin \text{y}}{\cos \text{y}}$
$\Rightarrow \sin \text{y}=\frac{\tan \text{y}}{\sec \text{y}}$
$ \Rightarrow \sin { \text{y} } =\frac { \text{x} }{ \sqrt { 1+{\text{ x }}^{ 2 } } }$
View full question & answer→MCQ 901 Mark
The domain of the function defind by $\text{f(x)}=\sin^{-1}\sqrt{\text{x}-1}$ is:
- ✓
$[1, 2]$
- B
$[-1, 1]$
- C
$[0, 1]$
- D
AnswerCorrect option: A. $[1, 2]$
View full question & answer→MCQ 911 Mark
If $ \text{x}=\cos^{-1}(\cos 4): \text{y}=\sin^{-1}(\sin 3)$ then which of the following holds?
AnswerGiven, $\text{x}=\cos^{-1}(\cos 4)$
$ = 2π - 4$
and $\text{y}=\sin^{-1}(\sin 3)\text{y}$
$ π - 3$
$ \tan(\text{x}+\text{y})$
$ =\tan(3\pi-4-3)$
$ =\tan(3\pi-7)$
$ =-\tan(7)$
View full question & answer→MCQ 921 Mark
If $ 3\cos ^{ -1 }{ \text{x} } +\sin ^{ -1 }{\text{ x} } =π$ then x:
- A
$\frac { 4 }{ \sqrt { 2 } }$
- B
$ -\frac { 1 }{ \sqrt { 2 } }$
- ✓
$\frac { 1 }{ \sqrt { 2 } }$
- D
$\frac { 1 }{ \sqrt { 4 } }$
AnswerCorrect option: C. $\frac { 1 }{ \sqrt { 2 } }$
$ \sin^{-1}\text{x} +\cos^{-1}\text{x}=\frac{\pi}{2}$
$ =3\cos^{-1}\text{x}+\sin^{-1}\text{x}$
$ =2\cos^{1}\text{x}+\cos^{-1}\text{x}+\sin^{-1}\text{x}$
$=\sin^{−1}\text{x}=π$
$ = 2\cos^{-1}\text{x}+\frac{\pi}{2}$
$ =\pi=2\cos^{−1}\text{x}=\frac{\pi}{2}$
$= \cos^{-1}\text{x}=\frac{\pi}{4}\text{x}$
$=\cos (\frac{\pi}{4})=\frac{1}{\sqrt 2}$
View full question & answer→MCQ 931 Mark
If $\alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$ and $\beta=\tan^{-1}\Big(-\tan\frac{2\pi}{3}\Big),$ then:
AnswerCorrect option: A. $4\alpha=3\beta$
We know that $\tan^{-1}(\tan\text{x})=\text{x}$
$\therefore\ \alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$
$=\tan^{-1}\Big\{\tan\Big(\pi+\frac{\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
and
$\beta=\tan^{-1}\Big\{-\tan\Big(\frac{2\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\pi-\frac{\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}$
$\therefore\ 4\alpha=\pi$
$3\beta=\pi$
$\therefore\ 4\alpha=3\beta$
View full question & answer→MCQ 941 Mark
Choose the correct answer from the given four options.
If $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$ then $\cot^{-1}\text{x}+\cot^{-1}\text{y}$ equals to:
- ✓
$\frac{\pi}{5}$
- B
$\frac{2\pi}{5}$
- C
$\frac{3\pi}{5}$
- D
$\pi$
AnswerCorrect option: A. $\frac{\pi}{5}$
We have, $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$
$\Rightarrow\ \frac{\pi}{2}-\cot^{-1}\text{x}+\frac{\pi}{2}-\cot^{-1}\text{y}=\frac{4\pi}{5}$
$\Rightarrow\ -(\cot^{-1}\text{x}+\cot^{-1}\text{y})=\frac{4\pi}{5}-\pi$
$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=-\Big(-\frac{\pi}{5}\Big)$
$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=\frac{\pi}{5}$
View full question & answer→MCQ 951 Mark
The given graph is for which equation?
- A
$\text{y}= \sin\text{x}$
- ✓
$\text{y} = \sin-1\text{x}$
- C
$\text{y} = \text{cosec }\text{x}$
- D
$\text{y} = \sec\text{x}$
AnswerCorrect option: B. $\text{y} = \sin-1\text{x}$
The following graph represents 2 equations.
The pink curve is the graph of $\text{y} = \sin\text{x}$
The blue curve is the graph for $\text{y} = \sin^{-1}{\text{x}}$
This curve passes through the origin and approaches to infinity in both positive and negative axes. View full question & answer→MCQ 961 Mark
What is the value of $ \cos (2 \cos^{-1} 0.8)\cos(2\cos−10.8)?$
- A
$0.81$
- ✓
$0.56$
- C
$0.48$
- D
$0.28$
AnswerCorrect option: B. $0.56$
View full question & answer→MCQ 971 Mark
$\tan^{-1}1+\cos^{-1}\Big(\frac{-1}{2}\Big)+\sin^{-1}\Big(\frac{-1}{2}\Big)$
- A
$\frac{2\pi}{3}$
- ✓
$\frac{3\pi}{4}$
- C
$\frac{\pi}{2}$
- D
$6\pi$
AnswerCorrect option: B. $\frac{3\pi}{4}$
View full question & answer→MCQ 981 Mark
The value of $\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$ is:
- A
$\frac{1}{\sqrt2}$
- B
$\frac{1}{\sqrt3}$
- ✓
$\frac{1}{2\sqrt2}$
- D
$\frac{1}{3\sqrt3}$
AnswerCorrect option: C. $\frac{1}{2\sqrt2}$
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
Let, $\sin^{-1}\frac{\sqrt{63}}{8}=\text{x}$
$\sin\text{x}=\frac{\sqrt{63}}{8}$
$\cos\text{x}\sqrt{1-\sin^2\text{x}}$
$\cos\text{x}=\sqrt{1-\frac{63}{64}}$
$\cos\text{x}=\frac{1}{8}$
Consider,
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
$=\sin\Big(\frac{1}{4}\text{x}\Big)$
$=\sqrt{\frac{1-\cos\frac{\text{x}}{2}}{2}}$ $\Big(\because\ \sin\text{x}=\frac{1-\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{\frac{1+\cos\text{x}}{2}}}{2}}$ $\Big(\because\ \cos\text{x}=\frac{1+\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{1-\frac{1}{8}}}{2}}$
$=\sqrt{\frac{1-\frac{3}{4}}{2}}$
$=\sqrt{\frac{1}{8}}$
$=\frac{1}{2\sqrt2}$
View full question & answer→MCQ 991 Mark
If the polynomial equation $\text{a}_0\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$ n positive integer,has two different real roots $\alpha$ and $\beta,$ then between $\alpha$ and $\beta,$ the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ has:
AnswerWe observe that, $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ is the derivative of the polynomial $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$Polynomial function is continuous everywhere in R and concequently derivative in R.
Therefore, $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0$ is continuous on $\alpha,\beta$ and derivative on $\alpha,\beta.$
Hence, it is satisfies the both the conditions of Rolle's theorem.
By algebric interpretation of Roll's theorem, we know that between any two roots of a function f(x), there exists atleast one root of its derivative.
Hence, the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ will have atleast one root between $\alpha$ and $\beta.$
View full question & answer→MCQ 1001 Mark
If x takes negative permissible value, then $\sin−1\text{x} $ is equal to:
- A
$\cos^{-1}\sqrt{1-\text{x}^2}$
- ✓
$-\cos^{-1}\sqrt{1-\text{x}^2}$
- C
$\cos^{-1}\sqrt{\text{x}^2-1}$
- D
$\pi-\cos^{-1}\sqrt{1-\text{x}^2}$
AnswerCorrect option: B. $-\cos^{-1}\sqrt{1-\text{x}^2}$
$\sin^{-1}(\text{x})$
$-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big),$ for x > 0
Since x takes negative permissible value.
$\sin^{-1}\text{x}=-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
View full question & answer→