MCQ 1011 Mark
The number of real values of $x$ satisfying the equation $ \tan^{-1}\left(\frac{\text{x}}{1-\text{x}^2}\right)+\tan^{-1}\left(\frac{1}{\text{x}^3}\right)=\frac{3\pi}{4}$, is?
Answer$\tan ^{-1} \frac{\text{x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$
$= ^{-1} \frac{3\pi \text{ x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$
$ =−1\text{x}^4 + 1 -\text{x}^2 + \text{x}^3 - \text{x}^5-\text{x} = 0$
$= 0$ Real roots $= 11$
View full question & answer→MCQ 1021 Mark
If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta=$
- A
$\pm\frac{\pi}{3}$
- B
$\pm\frac{\pi}{4}$
- ✓
$\pm\frac{\pi}{6}$
- D
AnswerCorrect option: C. $\pm\frac{\pi}{6}$
We have, $\tan^{-1}(\cot\theta)=2\theta$
$\Rightarrow\tan2\ \theta=\cot\theta$
$\Rightarrow\frac{2\tan\theta}{1-\tan^2\theta}=\frac{1}{\tan\theta}$
$\Rightarrow2\tan^2\theta=1-\tan^2\theta$
$\Rightarrow3\tan^2\theta=1$
$\Rightarrow\tan^2\theta=\frac{1}{3}$
$\Rightarrow\tan\theta=\pm\frac{1}{\sqrt3}$
View full question & answer→MCQ 1031 Mark
$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)$
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{3}$
- C
$\frac{\pi}{6}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: A. $\frac{\pi}{4}$
View full question & answer→MCQ 1041 Mark
The value of $\cos^{-1}\Big(\cos\Big(\frac{33\pi}{5}\Big)\Big)$ is:
- ✓
$\frac{3\pi}{5}$
- B
$\frac{-3\pi}{5}$
- C
$\frac{\pi}{10}$
- D
$\frac{-\pi}{10}$
AnswerCorrect option: A. $\frac{3\pi}{5}$
View full question & answer→MCQ 1051 Mark
Find the value of $\sec^2(\tan^{-1}2)+\text{cosec}^2(\cot^{-1}3)$
View full question & answer→MCQ 1061 Mark
If two angles of a triangle are $ \tan ^{ -1 }{ (2) }$ and $ \tan ^{ -1 }{ (3) }$, then the third angle is:
- ✓
$ \frac { \pi }{ 4 }$
- B
$ \frac { \pi }{ 5 }$
- C
$ \frac { \pi }{ 6 }$
- D
$ \frac { \pi }{ 8 }$
AnswerCorrect option: A. $ \frac { \pi }{ 4 }$
Given two angles are $ \tan ^{ -1 }{ (2) }$ and $ \tan ^{ -1 }{ (3) }$.
Now$, (2) (3) > 1$
$= \tan ^{ -1 }{ (2) } +\tan ^{ -1 }{ (3) }$
$ =\pi +\tan ^{ -1 }{ \cfrac { 2+3 }{ 1-2\times 3 }}$
$ =\pi +\tan ^{ -1 }{ (-1) } $
$=\pi -\frac { \pi }{ 4 } $
$=\frac { 3\pi }{ 4 }$
Hence the third angle is $ \pi -\frac { 3\pi }{ 4 } =\frac { \pi }{ 4 }\pi $
View full question & answer→MCQ 1071 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. $\frac{1}{5}$
View full question & answer→MCQ 1081 Mark
What is the value of $ \cos^{-1}(-\text{x})$ for all $x$ belongs to $[-1, 1]\ ?$
AnswerCorrect option: B. $\pi- \cos^{-1}(-\text{x})$
Let, $ \theta = \cos-1(\text{-x})$
So, $ 0 \leq \theta \leq \pi $
$ \Rightarrow -\text{x} = \cos\theta $
$ \Rightarrow \text{x} = -\cos\theta $
$ \Rightarrow \text{x} = \cos-\theta $
Also, $ -\pi \leq -\theta \leq 0$
So, $ 0 \leq \pi -\theta \leq \pi $
$ \Rightarrow -\theta = \cos^{-1}(\text{x})$
$ \Rightarrow \theta = \cos^{-1}(\text{x})$
So, $\cos-1(\text{x}) = \pi – \theta $
$ \theta = \pi – \cos-1(\text{x})$
$ \Rightarrow \cos^{-1}(-\text{x}) = \pi – \cos^{-1}(\text{x})$
View full question & answer→MCQ 1091 Mark
Choose the correct answer from the given four options.If $|\text{x}|\leq1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:
- ✓
$4\tan^{-1}\text{x}$
- B
$0$
- C
$\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: A. $4\tan^{-1}\text{x}$
We have, $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Let $\text{x}=\tan\theta$
$\therefore\ 2\tan^{-1}\tan\theta+\sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta}$
$[\because\ \tan^{-1}(\tan\text{x})=\text{x}]$
$=2\theta+\sin^{-1}\sin2\theta$
$\Big[\because\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$=2\theta+2\theta$
$[\because\ \sin^{-1}=(\sin\text{x})=\text{x}]$
$=4\theta\ [\because\ \theta=\tan^{-1}\text{x}]$
$=4\tan^{-1}\text{x}$
View full question & answer→MCQ 1101 Mark
What is the value of $ {\sin}^{-1}(\sin 160^{\circ})?$
- A
$160^{\circ}$
- B
$70^{\circ}$
- C
$-20^{\circ}$
- ✓
$20^{\circ}$
AnswerCorrect option: D. $20^{\circ}$
sinsin of an angle is positive in first and second quadrants.
$\Rightarrow \sin ^{ -1 }{ (\sin { { 160 }^{ \circ } } } )$
$\Rightarrow(\sin ^{ -1 }{ (\sin { { (180-20) }^{ \circ } } })=20^\circ$
View full question & answer→MCQ 1111 Mark
$\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}=$
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- D
$\frac{3\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{4}$
View full question & answer→MCQ 1121 Mark
If $\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$ then $\text{f}\Big(\frac{8\pi}{9}\Big)=$
- A
$\text{e}^{\frac{5\pi}{18}}$
- ✓
$\text{e}^{\frac{13\pi}{18}}$
- C
$\text{e}^{\frac{-2\pi}{18}}$
- D
AnswerCorrect option: B. $\text{e}^{\frac{13\pi}{18}}$
Given
$\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$
Then,
$\text{f}\Big(\frac{8\pi}{9}\Big)=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{8\pi}{9}+\frac{\pi}{3}\big)\big\}}$
$=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{11\pi}{9}\big)\big\}}=\text{e}^{\cos^{-1}\big\{\cos\frac{\pi}{2}+\frac{13\pi}{18}\big\}}$
$\Big[\because\ \cos\Big(\frac{\pi}{2}+\theta\Big)=\sin\theta\Big]$
$=\text{e}^{\cos^{-1}\big\{\cos\big(\frac{13\pi}{18}\big)\big\}}$
$=\text{e}^{\frac{13\pi}{18}}$
View full question & answer→MCQ 1131 Mark
If $\theta=\sin^{-1}\{\sin(-600^\circ)\},$ then one of the possible values of $\theta$ is:
- ✓
$\frac{\pi}{3}$
- B
$\frac{\pi}{2}$
- C
$\frac{2\pi}{3}$
- D
$-\frac{2\pi}{3}$
AnswerCorrect option: A. $\frac{\pi}{3}$
$\theta=\sin^{-1}\{\sin(-600^\circ)\}$
$\theta=\sin^{-1}[\sin(-600^\circ)]$
$\theta=\sin^{-1}[-\sin(180^\circ\times3+60)]$
$\theta=\sin^{-1}[-\{-\sin(60^\circ)\}]$
$\theta=\sin^{-1}(\sin(60^\circ))$
$\theta=\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$
$\theta=\frac{\pi}{3}$
View full question & answer→MCQ 1141 Mark
$\tan\Big(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\Big)=$
- A
$\text{x}$
- B
$\frac{1}{\text{x}}$
- C
$2\text{x}$
- ✓
$\frac{2}{\text{x}}$
AnswerCorrect option: D. $\frac{2}{\text{x}}$
View full question & answer→MCQ 1151 Mark
The value of $\tan^{-1}\Big(\frac{3}{4}\Big)+\tan^{-1}\Big(\frac{1}{7}\Big)$ is:
- A
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{3\pi}{4}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
View full question & answer→MCQ 1161 Mark
The value of $ \left( {{{\tan }^{ - 1}}\pi + {{\tan }^{ - 1}}\left( {\frac{1}{\pi }} \right)} \right) + {\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$ is equal to
- A
$ \frac{{ - \pi }}{3}$
- ✓
$ \frac{{ \pi }}{6}$
- C
$ \frac{{ 2 \pi }}{3}$
- D
$ \pi$
AnswerCorrect option: B. $ \frac{{ \pi }}{6}$
$ \left( {{{\tan }^{ - 1}}\pi + {{\tan }^{ - 1}}\left( {\frac{1}{\pi }} \right)} \right) + {\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$
$=\tan^{-1}\pi+\cot^{-1}\pi+\tan^{-1}\sqrt 3-\sec^{-1}(-2)$
$ [\because \tan^{−1}\frac{1}{\text{y}}=\cot−1\text{y}]$
$ =\cfrac{\pi}{2}+\tan^{-1}\sqrt 3-\sec^{-1}(-2)\Big[\because \tan^{−1}x+\cot^{−1}\text{x}=\frac{\pi }{2}\Big]=\frac{\pi }{2}+\frac{\pi }{3}-\frac{2\pi }{3}$
$ =[\because \tan\frac{\pi }{3}=\sqrt{3};\sec\frac{2\pi }{3}=−2]$
$ =\frac{\pi }{2}−\frac{\pi }{3}$
$ =\frac{\pi}{6}$
None of the given options are correct.
View full question & answer→MCQ 1171 Mark
If $\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$ then, $\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)=$
- ✓
$\sqrt{\tan\theta}$
- B
$\sqrt{\cot\theta}$
- C
$\tan\theta$
- D
$\cot\theta$
AnswerCorrect option: A. $\sqrt{\tan\theta}$
Let $\text{y}=\sqrt{\tan\theta}$
Then,
$\Rightarrow\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$
$\Rightarrow\text{u}=\cot^{-1}\text{y}-\tan^{-1}\text{y}$
$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow2\tan^{-1}\text{y}=\frac{\pi}{2}-\text{u}$
$\Rightarrow\tan^{-1}\text{y}=\frac{\pi}{4}-\frac{\text{u}}{2}$
$\Rightarrow\text{y}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$
$\Rightarrow\sqrt{\tan\theta}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$
$\Big[\because\ \text{y}=\sqrt{\tan\theta}\Big]$
View full question & answer→MCQ 1181 Mark
If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then $x =$
- A
$\frac{1}{2}$
- ✓
$\frac{\sqrt3}{2}$
- C
$-\frac{1}{2}$
- D
AnswerCorrect option: B. $\frac{\sqrt3}{2}$
We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\therefore \sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\frac{\pi}{2}-\cos^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow-2\cos^{-1}\text{x}=\frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow-2\cos^{-1}\text{x}=-\frac{\pi}{3}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{\sqrt3}{2}$
View full question & answer→MCQ 1191 Mark
The value of $\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]:$
- ✓
$ \frac { 1 }{ \sqrt { 5 } }$
- B
$ \frac { 1 }{ \sqrt { 7 } }$
- C
$ \frac { 1 }{ \sqrt { 8 } }$
- D
$ \frac { 1 }{ \sqrt { 9 } }$
AnswerCorrect option: A. $ \frac { 1 }{ \sqrt { 5 } }$
We have, $\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$
Put $ \cos ^{ -1 }{ \frac { 2 }{ 3 } } =θ$
$ \Rightarrow \cos { \theta } =\frac { 2 }{ 3 }$
$\therefore\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$
$ \therefore \ \tan\frac{\theta}{2}=\sqrt {\frac {1-\cos {\theta}}{1+\cos{\theta }}}$
$=\sqrt {\frac {1-\frac{2}{3} }{1+{\frac{ 2}{3}}}}$
$=\sqrt { \frac { \frac{ 1 }{ 3 } }{ \frac{5}{3} } } $
$=\frac { 1 }{ \sqrt { 5 } }$
View full question & answer→MCQ 1201 Mark
Choose the correct answer from the given four options. If $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$ then $x$ is equal to:
- A
$\frac{1}{5}$
- ✓
$\frac{2}{5}$
- C
$0$
- D
$1$
AnswerCorrect option: B. $\frac{2}{5}$
We have, $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$
$\Rightarrow \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\cos^{-1}0$
$\Rightarrow \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\frac{2}{5}$
$\Rightarrow \cos^{-1}\text{x}=\cos^{-1}\frac{2}{5}$
$\Big(\because \cos^{-1}\text{x}+\sin^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\therefore \text{x}=\frac{2}{5}$
View full question & answer→MCQ 1211 Mark
The value of $ \cos { \left( \tan ^{ -1 }{ \tan { 4 } } \right) }$ is$-$
- A
$ \frac { 1 }{ \sqrt { 17 } }$
- B
$ \frac { 1 }{ \sqrt {- 17 } }$
- C
$ \frac { 1 }{ \sqrt {- 14 } }$
- ✓
$ -\cos 4$
AnswerCorrect option: D. $ -\cos 4$
As for $ \displaystyle \tan ^{ -1 }{ \text{x} }; \text{x}\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$
$\cos(\tan^{−1}(\tan4))$
$=\cos(\tan^{−1}(\tan(π−4))$
$ =\cos(π−4)$
$=−\cos4$
View full question & answer→MCQ 1221 Mark
If $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{A},$ then $A$ is equal to:
AnswerCorrect option: C. $\frac{\text{x}-\text{y}}{1+\text{xy}}$
View full question & answer→MCQ 1231 Mark
Find the value of $ {\sin ^{ - 1}}\left( 1 \right):$
- A
$ \dfrac{\pi}{7}$
- B
$ \dfrac{\pi}{6}$
- C
$ \dfrac{\pi}{4}$
- ✓
$ \dfrac{\pi}{2}$
AnswerCorrect option: D. $ \dfrac{\pi}{2}$
Value of $\sin^{-1}(1)\sin\text{x}$ is in vertible form $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ in this range only $\sin\frac{\pi}{2}=1\sin^{-1}(1)\frac{\pi}{2}.$
View full question & answer→MCQ 1241 Mark
Choose the correct answer from the given four options. The value of the expression $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$ is:
- A
$\frac{\pi}{6}$
- ✓
$\frac{5\pi}{6}$
- C
$\frac{7\pi}{6}$
- D
$1$
AnswerCorrect option: B. $\frac{5\pi}{6}$
We have, $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
$=2\sec^{-1}\sec\frac{\pi}{3}+\sin^{-1}\sin\frac{\pi}{6}$
$=2\frac{\pi}{3}+\frac{\pi}{6}$
$[\because \sec^{-1}(\sec\text{x})=\text{x} $ and $ \sin^{-1}(\sin\text{x})=\text{x}]$
$=\frac{4\pi+\pi}{6}$
$=\frac{5\pi}{6}$
View full question & answer→MCQ 1251 Mark
The equation $2\cos^{-1}\text{x}+\sin^{-1}\text{x}=\frac{11\pi}{6}$ has:
View full question & answer→MCQ 1261 Mark
If $ \cos^{-1}\left (\frac {1 - \text{x}^{2}}{1 + \text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right )=\frac{\pi}{2}$, where $xy < 1,$ then:
- A
$x - y - xy = 1$
- B
$x - y + xy = 1$
- C
$x + y - xy = 1$
- ✓
$x + y + xy = 1$
AnswerCorrect option: D. $x + y + xy = 1$
Given, $ \cos^{-1} \left (\frac {1 - \text{x}^{2}}{1 +\text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right ) = \frac {\pi}{2}$
$ \Rightarrow \tan^{-1} \left (\frac {\text{x} + \text{y}}{1 - \text{xy}}\right ) = \frac {\pi}{4}$
$ \Rightarrow \frac {\text{x} + \text{y}}{1 -\text{ xy}} = \tan \frac {\pi}{4}$
$\Rightarrow \text{x} + \text{y} = 1 - \text{xy} = \text{x} + \text{y} + \text{xy}$
View full question & answer→MCQ 1271 Mark
The value of $\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$ is:
- A
$\frac{\pi}{2}$
- B
$\frac{5\pi}{3}$
- C
$\frac{10\pi}{3}$
- ✓
$0$
AnswerWe have
$\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$
$=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{\sin\Big(2\pi-\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{-\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}-\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}-\frac{\pi}{3}$
$=0$
View full question & answer→MCQ 1281 Mark
If $\tan^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{x}}\Big)=\frac{\pi}{2},$ then $x$ is equal to:
- ✓
$\sqrt{\text{ab}}$
- B
$\sqrt{2\ \text{ab}}$
- C
$2\ \text{ab}$
- D
$\text{ab}$
AnswerCorrect option: A. $\sqrt{\text{ab}}$
View full question & answer→MCQ 1291 Mark
The value of $\sin\big(2\big(\tan^{-1}0.75\big)\big)$ is equal to:
- A
$0.75$
- B
$1.5$
- ✓
$0.96$
- D
$\sin ^{-1} 1.5$
AnswerCorrect option: C. $0.96$
$\sin\big(2\big(\tan^{-1}0.75\big)\big)$
$=\sin\big(2\tan^{-1}0.75\big)$
$=\sin\Big(\sin^{-1}\frac{2\times0.75}{1+(0.75)^2}\Big)$
$=\sin\big(\sin^{-1}0.96\big)$
$=0 .96$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 1301 Mark
Choose the correct answer from the given four options. If $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi,$ then $x$ equals:
- A
$0$
- ✓
$1$
- C
$-1$
- D
$\frac{1}{2}$
AnswerWe have, $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+(\tan^{-1}\text{x}+\cot^{-1}\text{x})=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+\frac{\pi}{2}=\pi$
$\Big(\because \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\Rightarrow 2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow \tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow \text{x}=1$
View full question & answer→MCQ 1311 Mark
Choose the correct answer from the given four options.The domain of the function defined by $\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$ is:
- ✓
$[1, 2]$
- B
$[-1, 1]$
- C
$[0, 1]$
- D
AnswerCorrect option: A. $[1, 2]$
$\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$
$\Rightarrow\ 0\leq\text{x}-1\leq1\ [\because\ \sqrt{\text{x}-1}\geq0$ and $-1\leq\sqrt{\text{x}-1}\leq1]$
$\Rightarrow\ 1\leq\text{x}\leq2$
$\therefore\ \text{x}\in[1,2]$
View full question & answer→MCQ 1321 Mark
The number of solution of the equation $ 1+\text{x}^{2}+2\text{x}\:\sin \left ( \cos^{-1}\text{y} \right )= 0$ is:
AnswerGiven, $ 1+\text{x}^2+2\text{x}(\sin(\cos^{-1}(\text{y})))=0$
$ 1+\text{x}^2+2\text{x}(\sin(\sin^{-1}(\sqrt{1-\text{y}^2})))=0$
$ 1+\text{x}^2+2\text{x}(\sqrt{1-\text{y}^2})=0$
$ 2\text{x}(\sqrt{1-\text{y}^2})=-(1+\text{x}^2)$
$ 4\text{x}^2(1-\text{y}^2)=1+\text{x}^4+2\text{x}^2$
$ 4\text{x}^2-4\text{x}^2\text{y}^2=1+\text{x}^4+2\text{x}^2$
$ -4\text{x}^{2}(\text{y}^2)=(1-\text{x}^2)^{2}$
Hence solution will be $x = 1$ and $\text{y}=\frac{1}{2}$
View full question & answer→MCQ 1331 Mark
$\sin^{-1}(1-\text{x})-2\sin^{-1}\text{x}=\frac{\pi}{2}$
- ✓
$0$
- B
$\frac{1}{2}$
- C
$0,\frac{1}{2}$
- D
$-\frac{1}{2}$
View full question & answer→MCQ 1341 Mark
$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=$
AnswerLet $2\cot^{-1}3=\text{y}$
Then, $\cot\frac{\text{y}}{2}=3$
$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)$
$=\cot\Big(\frac{\pi}{4}-\text{y}\Big)$
$=\frac{\cot\frac{\pi}{4}\cot\text{y}+1}{\cot\text{y}-\cot\frac{\pi}{4}}$
$=\frac{\cot\text{y}+1}{\cot\text{y}-1}$
$=\frac{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}+1}{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}-1}$
$=\frac{\cot^2\frac{\text{y}}{2}+2\cot\frac{\text{y}}{2}-1}{\cot^2\frac{\text{y}}{2}-2\cot\frac{\text{y}}{2}-1}$
$=\frac{9+6-1}{9-6-1}$
$=7$
View full question & answer→MCQ 1351 Mark
If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta$ is equal to:
- A
$\frac{\pi}{3 }$
- B
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{6}$
- D
AnswerCorrect option: C. $\frac{\pi}{6}$
View full question & answer→MCQ 1361 Mark
The positive integral solution of the equation $\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$ is:
- ✓
$x = 1, y = 2$
- B
$x = 2, y = 1$
- C
$x = 3, y = 2$
- D
$x = -2, y = -1$
AnswerCorrect option: A. $x = 1, y = 2$
We have,
$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\begin{bmatrix}\frac{\sqrt{1-\bigg(\frac{\text{y}}{\sqrt{1+\text{y}^2}}\bigg)^2}}{\frac{\text{y}}{\sqrt{1+\text{y}^2}}}\end{bmatrix}=\tan^{-1}\begin{bmatrix}\frac{\frac{3}{\sqrt{10}}}{\sqrt{1-\Big(\frac{3}{\sqrt{10}}\Big)^2}}\end{bmatrix}$
View full question & answer→MCQ 1371 Mark
$\sin^{−1}\text{x}+\sin^{−1}\frac{1}{\text{x}}+\cos^{−1}\text{x}+\cos^{−1}\frac{1}{\text{x}}=$
- ✓
$\pi$
- B
$2\pi$
- C
$ \cfrac{3\pi}{2}$
- D
AnswerWe know, $ \displaystyle \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$
$\therefore \sin ^{ -1 }{ \text{x} } +\sin ^{ -1 }{ \frac { 1 }{ \text{x} } } +\cos ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \frac { 1 }{ \text{x} } }$
$ \displaystyle =\frac { \pi }{ 2 } +\frac { \pi }{ 2 } $
$=\pi$
View full question & answer→MCQ 1381 Mark
Simplify $ {\cot ^{ - 1}}\frac{1}{{\sqrt {{\text{x}^2} - 1} }} $ for $\text{ x} < - 1$:
AnswerCorrect option: B. $ \sec ^{-1}\text{x}$
View full question & answer→MCQ 1391 Mark
$\sin[\cot^{-1}\{\cos(\tan^{-1}\text{x})\}]=$
- ✓
$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
- B
$\sqrt{\frac{\text{x}^2-1}{\text{x}^2-2}}$
- C
$\sqrt{\frac{\text{x}-1}{\text{x}-2}}$
- D
$\sqrt{\frac{\text{x}+1}{\text{x}+2}}$
AnswerCorrect option: A. $\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
View full question & answer→MCQ 1401 Mark
The number of solutions for the equation $ 2\sin ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x}+1 } } +\cos ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x} } } =\frac { 3\pi }{ 2 }$ is:
Answer$ 2\sin ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x}+1 } } +\cos ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x} } } =\frac { 3\pi }{ 2 }$
For existence of domain of
$\sin^{-1}\sqrt{\text{x} ^2-\text{x}+1}-1\leq \sqrt{\text{x}^2-\text{x}}+1\leq 1$
$ 0 \leq \text{x}^2-\text{x}+1\leq 1$
$\text{x}\in [0,1]$
For $\cos^{-1}\sqrt{\text{x}^2-\text{x}0}\leq \text{x}^2-\text{x}\leq 1$
$ \Rightarrow \text{x}^2−\text{x}\geq 0$
$ \Rightarrow \text{x}−\text{x}\geq 0$
$\text{x}\in [−\infty ,0] \cup [1,\infty ]$ Only two points are common in their domains
i.e.$0$ and $1$ which also satisfies the given equation.
So option $B$ is correct.
View full question & answer→MCQ 1411 Mark
$\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha,$ then $\frac{\text{x}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha+\frac{\text{y}^2}{\text{b}^2}=$
- ✓
$\sin^2\alpha$
- B
$\cos^2\alpha$
- C
$\tan^2\alpha$
- D
$\cot^2\alpha$
AnswerCorrect option: A. $\sin^2\alpha$
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
Consider, $\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha$
$\Rightarrow\cos^{-1}\bigg(\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}=\cos\alpha$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\cos\alpha=\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}$
Squaring on both sides,
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\Big(1-\frac{\text{x}^2}{\text{a}^2}\Big)\Big(1-\frac{\text{y}^2}{\text{a}^2}\Big)$
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{a}^2}+\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\cos^2\alpha$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\sin^2\alpha$
View full question & answer→MCQ 1421 Mark
$\sin−10 $ is equal to:
- ✓
$0$
- B
$ \dfrac{\pi }{6}$
- C
$ \dfrac{\pi}{2}$
- D
$ \dfrac{\pi}{3}$
AnswerAs we know that $\sin{0} = 0\sin0=0$
$\Rightarrow 0 = \sin^{-1}{\left( 0 \right)}$
Hence the value of $ \sin^{-1}{\left( 0 \right)}$ is $0.$
View full question & answer→MCQ 1431 Mark
If $ \alpha \leq 2\sin^{-1} \text{x} + \cos^{-1} \text{x}\leq \betaα$ then:
- A
$ \alpha = 0, \beta = \pi(2)$
- B
$ \alpha = 0, \beta = 2\pi$
- ✓
$ \alpha = 0, \beta = \pi$
- D
$ \alpha = 0, \beta = 5\pi$
AnswerCorrect option: C. $ \alpha = 0, \beta = \pi$
We know that, $ -\frac{\pi}{2}\le \sin^{-1}\text{x}\le \frac{\pi}{2}$
Now add $ \frac{\pi}{2}$
each sides $ \frac{\pi}{2}-\frac{\pi}{2}\le \sin^{-1}\text{x}+\frac{\pi}{2}\le \frac{\pi}{2}+\frac{\pi}{2}$
$ \Rightarrow 0\le \sin^{-1}+\dfrac{\pi}{2}\le \pi$
$ \Rightarrow 0\le \sin^{-1}+(\sin^{-1}\text{x}+\cos^{-1}\text{x})\le \pi$ use the identity
$ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$ \therefore 0\le 2\sin^{-1}\text{x}+\cos^{-1}\text{x}\le \pi$
Hence $ \alpha=0, \beta=\pi$
View full question & answer→MCQ 1441 Mark
If $x > 1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:
- A
$4\tan^{-1}\text{x}$
- B
$0$
- ✓
$\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: C. $\frac{\pi}{2}$
$2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=2\tan^{-1}\text{x}+2\tan^{-1}\text{x}$
$\Big[\because \sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$=4\tan^{-1}\text{x}$
View full question & answer→MCQ 1451 Mark
Choose the correct answer from the given four options.The number of real solutions of the equation $\sqrt{1+\cos2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x})$ in $\Big[\frac{\pi}{2},\pi\Big]$ is:
AnswerWe have $\sqrt{1+\cos2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x}),$
$\text{x}\in\Big[\frac{\pi}{2},\pi\Big]$
$\Rightarrow\ \sqrt{2\cos^2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \sqrt{2}\cos\text{x}=\sqrt{2}\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \cos\text{x}=\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \cos\text{x}=\text{x}$
$[\because\ \cos^{-1}(\cos\text{x})=\text{x}]$
For $\text{x}\in\Big[\frac{\pi}{2},\pi\Big],\ \cos\text{x}\leq0$
$\therefore \cos x = x$ is not possible for any value of $x.$
View full question & answer→MCQ 1461 Mark
$\cos^{-1}\Big(\frac{1}{2}\Big)$
- A
$-\frac{\pi}{3}$
- ✓
$\frac{\pi}{3}$
- C
$\frac{\pi}{2}$
- D
$\frac{2\pi}{3}$
AnswerCorrect option: B. $\frac{\pi}{3}$
View full question & answer→MCQ 1471 Mark
If $4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of $x$ is:
- A
$\frac{3}{2}$
- B
$\frac{1}{\sqrt2}$
- ✓
$\frac{\sqrt3}{2}$
- D
$\frac{2}{\sqrt3}$
AnswerCorrect option: C. $\frac{\sqrt3}{2}$
We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi$
$\Rightarrow4\cos^{-1}\text{x}+\frac{\pi}{2}-\cos^{-1}\text{x}=\pi $
$\Rightarrow3\cos^{-1}\text{x}=\pi-\frac{\pi}{2}$
$\Rightarrow3\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$
View full question & answer→MCQ 1481 Mark
If $\sin\Big(\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}\Big)=1,$ then the value of $x$ is:
- A
$-1$
- B
$\frac{2}{5}$
- C
$\frac{1}{3}$
- ✓
$\frac{1}{5}$
AnswerCorrect option: D. $\frac{1}{5}$
View full question & answer→MCQ 1491 Mark
Find the value of $x$ if $ \sin (\text{arc} \sin \text{x}) = \frac {\sqrt {2}}{4}:$
- ✓
$ \frac {\sqrt {2}}{4}$
- B
$ \frac {\sqrt {2}}{6}$
- C
$ \frac {\sqrt {6}}{4}$
- D
$ \frac {\sqrt {2}}{3}$
AnswerCorrect option: A. $ \frac {\sqrt {2}}{4}$
Given, $ \sin \text{arc} \sin { \text{x} } =\frac { \sqrt { 2 } }{ 4 } =\frac { 1 }{ 2\sqrt { 2 } }$
$ \therefore \text{arc}\sin { \text{x} } =\text{arc}\sin \left (\frac {1}{2\sqrt2}\right)=0.36136$
$ \therefore \text{x}=\sin(0.36136)=\frac { 1 }{ 2\sqrt { 2 } }$
View full question & answer→MCQ 1501 Mark
ind the value of $\text{cot}\text{(tan}^1\text{a}+\text{cot}^1\text{a}).$
AnswerWe know,
$\tan^1\text{a}+\cot^{-1}\text{a}=\frac{\pi}{2}$
Therefore,
$\cot(\tan^{−1}\text{a}+\cot^{−1}a)=\cot\frac{\pi}{2}=0$
View full question & answer→