MCQ 11 Mark
The principal value of $\tan^{-1}\Big(\tan\frac{3\pi}{5}\Big)$ is:
- A
$\frac{2\pi}{5}$
- ✓
$\frac{-2\pi}{5}$
- C
$\frac{3\pi}{5}$
- D
$\frac{-3\pi}{5}$
AnswerCorrect option: B. $\frac{-2\pi}{5}$
View full question & answer→MCQ 21 Mark
$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:
- A
$\frac{6}{25}$
- B
$\frac{24}{25}$
- C
$\frac{4}{5}$
- ✓
$-\frac{24}{25}$
AnswerCorrect option: D. $-\frac{24}{25}$
View full question & answer→MCQ 31 Mark
$\tan^{-1}(\sqrt{3})$
- A
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- C
$\frac{2\pi}{3}$
- D
$\frac{5\pi}{6}$
AnswerCorrect option: B. $\frac{\pi}{3}$
View full question & answer→MCQ 41 Mark
If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:
- ✓
$\frac{1}{\sqrt2}<\text{x}\leq1$
- B
$0\leq\text{x}\leq\frac{1}{\sqrt2}$
- C
$-1\leq\text{x}<\frac{1}{\sqrt2}$
- D
$\text{x}>0$
AnswerCorrect option: A. $\frac{1}{\sqrt2}<\text{x}\leq1$
View full question & answer→MCQ 51 Mark
Choose the correct answer from the given four options.The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$ is:
AnswerCorrect option: B. $\sqrt{5}-2$
We have, $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$
Let $\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}=\theta$
$\Rightarrow\ \cos^{-1}\frac{2}{\sqrt{5}}=2\theta$
$\Rightarrow\ \cos2\theta=\frac{2}{\sqrt{5}}$
$\therefore\ 2\cos^{2}\theta-1=\frac{2}{\sqrt{5}}$
$\Rightarrow\ \cos^2\theta=\frac{1}{2}+\frac{1}{\sqrt{5}}$
$\Rightarrow\ \cos\theta=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}$
$\therefore\ \tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\sqrt{\frac{\frac{1}{2}-\frac{1}{\sqrt{5}}}{\frac{1}{2}+\frac{1}{\sqrt{5}}}}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$
$=\sqrt{\frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)}}=\sqrt{5}-2$
View full question & answer→MCQ 61 Mark
If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of $x$ is:
Answer$\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}+1}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}}{1-\frac{\text{x}+1}{\text{x}-1}\times\frac{\text{x}-1}{\text{x}}}\Bigg)=\tan^{-1}(-7)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}^2+\text{x}+\text{x}^2-2\text{x}+1}{\text{x}^2-\text{x}-(\text{x}^2-1)}\Big)\tan^{-1}(-7)$
$\Rightarrow\frac{2\text{x}^2-\text{x}+1}{-\text{x}+1}=-7$
$\Rightarrow2\text{x}^2-\text{x}+1=7\text{x}-7$
$\Rightarrow2\text{x}^2-8\text{x}+8=0$
$\Rightarrow\text{x}^2-4\text{x}+4=0$
$\Rightarrow(\text{x}-2)^2=0$
$\Rightarrow\text{x}=2$
View full question & answer→MCQ 71 Mark
$\sin^{-1}\Big(\frac{-1}{2}\Big)$
- A
$\frac{\pi}{3}$
- B
$-\frac{\pi}{3}$
- C
$\frac{\pi}{6}$
- ✓
$-\frac{\pi}{6}$
AnswerCorrect option: D. $-\frac{\pi}{6}$
View full question & answer→MCQ 81 Mark
The number of solutions of the equation $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$ is:
AnswerWe know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big).$
$\therefore \tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}\Big)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}=1$
$\Rightarrow5\text{x}=1-6\text{x}^2$
$\Rightarrow6\text{x}^2+5\text{x}-1=0$
View full question & answer→MCQ 91 Mark
If $ \frac{3\text{x}+1}{(\text{x}-1)(\text{x}+3)} = \frac{\text{A}}{\text{x}-1}+\frac{B}{\text{x}+3} $ then $ {\sin}^{-1} \frac{\text{A}}{\text{B}} :$
- A
$ \frac{\pi}{2}$
- B
$ \frac{\pi}{3}$
- ✓
$ \frac{\pi}{6}$
- D
$ \frac{\pi}{8}$
AnswerCorrect option: C. $ \frac{\pi}{6}$
We have,$ \frac{3\text{x}+1}{(\text{x}-1)(\text{x}+3)} = \frac{\text{A}}{\text{x}-1}+\frac{B}{\text{x}+3} $
$\Rightarrow 3x + 1 = A (x + 3) + B(x - 1)$
Substitute $x = 1$ both sides
$\Rightarrow 3(1) + 1 = A(1 + 3) + 0$
$ \Rightarrow A = 1$
Substitute $x = - 3x$ both sides
$\Rightarrow 3( -3) + 1 = 0 + B(-3 -1)$
$\Rightarrow -8 - 4B$
$ \Rightarrow B = 2$
Hence $ \sin^{-1}\frac{\text{A}}{\text{B}}=\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$
View full question & answer→MCQ 101 Mark
If $\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\text{x}\in(0,1),$ then the value of $x$ is:
AnswerCorrect option: D. $\frac{2\text{a}}{1-\text{a}^2}$
$\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Let, $\text{a}=\tan\theta\Rightarrow\theta=\tan^{-1}\text{a}$
$\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)=2\tan^{-1}(\text{x})$
$2\theta+2\theta=2\tan^{-1}(\text{x})$
$4\theta=2\tan^{-1}(\text{x})$
$2\tan^{-1}\text{a}=\tan^{-1}(\text{x})$
$\tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)=\tan^{-1}(\text{x})$
$\text{x}=\frac{2\text{a}}{1-\text{a}^2}$
View full question & answer→MCQ 111 Mark
Choose the correct answer from the given four options.The value of $\cos^{-1}\Big(\cos\frac{3\pi}{2}\Big)$ is equal to:
- ✓
$\frac{\pi}{2}$
- B
$\frac{3\pi}{2}$
- C
$\frac{5\pi}{2}$
- D
$\frac{7\pi}{2}$
AnswerCorrect option: A. $\frac{\pi}{2}$
We have, $\cos^{-1}\Big(\cos\frac{3\pi}{2}\Big)$
$=\cos^{-1}\cos\Big(2\pi-\frac{\pi}{2}\Big)$
$=\cos^{-1}\cos\Big(\frac{\pi}{2}\Big)$
$[\because\ \cos(2\pi-\theta)=\cos\theta]$
$=\frac{\pi}{2}\ \Big[\because\ \cos^{-1}(\cos\text{x})=\text{x},\ \text{x}\in[0,\pi]\Big]$
View full question & answer→MCQ 121 Mark
$2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$ is equal to:
- A
$\cot^{-1}\text{x}$
- B
$\cot^{-1}\text{x}$
- ✓
$\tan^{-1}\text{x}$
- D
AnswerCorrect option: C. $\tan^{-1}\text{x}$
$\therefore\ 2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$
$=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\Big(\tan^{-1}\frac{1}{\text{x}}\Big)\Big\}$
$=\frac{3}{29}=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\frac{1}{\text{x}}\Big\}$
$=2\tan^{-1}\Big\{\text{cosec y}-\frac{1}{\tan\text{y}}\Big\}$
$=2\tan^{-1}\Big\{\frac{1-\cos\text{y}}{\sin\text{y}}\Big\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{\sin\text{y}}\bigg\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{2\sin\frac{\text{y}}{2}\cos\frac{\text{y}}{2}}\bigg\}$
$=2\tan^{-1}\Big\{\tan\frac{\text{y}}{2}\Big\}$
$=\text{y}$
$=\tan^{-1}\text{x}$
View full question & answer→MCQ 131 Mark
Choose the correct answer from the given four options.Which of the following is the principal value branch of $\operatorname{cosec}^{-1} x ?$
- A
$\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
- B
$[0,\pi]-\Big\{\frac{\pi}{2}\Big\}$
- C
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
- ✓
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$
AnswerCorrect option: D. $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$
We know that, the principal value branch of $\operatorname{cosec}^{-1} x ?$ is $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$
View full question & answer→MCQ 141 Mark
Choose the correct answer from the given four options.If $\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\ \text{x}\in[0,1]$ then the value of $x$ is:
AnswerCorrect option: D. $\frac{2\text{a}}{1-\text{a}^2}$
We have, $\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\ 2\tan^{-1}\text{a}+2\tan^{-1}\text{a}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$\bigg[\because\ 2\tan^{-1}\text{a}=\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)=\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)\bigg]$
$\Rightarrow\ 4\tan^{-1}\text{a}=2\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\ 2\tan^{-1}\text{a}=\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\ \tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)=\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{a}=\tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)\Big]$
$\Rightarrow\ \text{x}=\frac{2\text{a}}{1-\text{a}^2}$
View full question & answer→MCQ 151 Mark
If $\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=\text{x},$ then $\sin\text{x}$ is equal to:
- ✓
$\tan^2\Big(\frac{\alpha}{2}\Big)$
- B
$\cot^2\Big(\frac{\alpha}{2}\Big)$
- C
$\tan\alpha$
- D
$\cot\Big(\frac{\alpha}{2}\Big)$
AnswerCorrect option: A. $\tan^2\Big(\frac{\alpha}{2}\Big)$
View full question & answer→MCQ 161 Mark
If ${ \sin }^{ -1 }\frac { \text{x} }{ 5 } +{ \text{cosec} }^{ -1 }\frac { 5 }{ 4 } $ then x is equal to:
Answer${ \sin }^{ -1 }\frac { x }{ 5 } +{ \text{cosec} }^{ -1 }\frac { 5 }{ 4 } =\frac{ \pi }{ 2 }$
$ \Rightarrow { \sin }^{ -1 }\frac { \text{x} }{ 5 } +{ \sin }^{ -1 }\frac { 4 }{ 5 } =\frac { \pi }{ 2 }$
$ \Rightarrow \sin^{-1}\frac{\text{x}}{5}=\frac{\pi}{2}-\sin^{-1}\frac{4}{5}=\cos^{-1}\frac{4}{5}$
$ \Rightarrow \text{x}=5\sin\cos^{-1}\frac{4}{5}=5\sin\sin^{-1}\frac{3}{5}=3$
View full question & answer→MCQ 171 Mark
What is $ \tan ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +\tan ^{ -1 }{ \left( \frac { 1 }{ 3 } \right) }$equal to?
- A
$ \frac { \pi }{ 3 }$
- ✓
$ \frac { \pi }{ 4 }$
- C
$ \frac { \pi }{ 6 }$
- D
$ \frac { \pi }{ 9 }$
AnswerCorrect option: B. $ \frac { \pi }{ 4 }$
We know the formula $ \tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\left(\frac { \text{a}+\text{b} }{ 1-\text{ab} } \right)$
So $\tan^{-1}\big(\frac{1}{2}\big)+\tan^{-1}\big(\frac{1}{3}\big)=\tan^{-1}\Bigg(\frac{\big(\frac{1}{2}\big)+\big(\frac{1}{3}\big)}{1-\big(\frac{1}{2}\big)\big(\frac{1}{2}\big)}\Bigg)$
$=\tan^{-1}\Bigg(\frac{\big(\frac{5}{6}\big)}{\big(\frac{5}{6}\big)}\Bigg)=\tan^{-1}(1)=\frac{\pi}{4}$
View full question & answer→MCQ 181 Mark
If $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4},$ then $x$ is:
- ✓
$\frac{1}{6}$
- B
$1$
- C
$(\frac{1}{6},-1)$
- D
AnswerCorrect option: A. $\frac{1}{6}$
View full question & answer→MCQ 191 Mark
The value of $\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$ is:
- A
$\frac{33}{5}$
- ✓
$-\frac{\pi}{10}$
- C
$\frac{\pi}{10}$
- D
$\frac{7\pi}{5}$
AnswerCorrect option: B. $-\frac{\pi}{10}$
$\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$
$=\sin^{-1}\Big(\cos\Big(6\pi+\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\cos\Big(\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(\frac{\pi}{2}-\frac{3\pi}{5}\Big)\Big)$
$=\frac{\pi}{2}-\frac{3\pi}{5}$
$=\frac{-\pi}{10}$
View full question & answer→MCQ 201 Mark
If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$
- ✓
$\frac{\pi}{6}$
- B
$\frac{\pi}{3}$
- C
$\frac{\pi}{2}$
- D
$-\frac{\pi}{3}$
AnswerCorrect option: A. $\frac{\pi}{6}$
We have
$\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
Now, $\alpha-\beta=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-1}\Big)-\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}-\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}{1+{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\times\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}}\end{pmatrix}$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{\sqrt{3}\text{y}(2\text{y}-\text{x})+\sqrt{3}\text{x}(2\text{z}-\text{y})}}{\sqrt{3}\text{y}(2\text{y}-\text{x})}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{2\sqrt3\text{y}^2-\sqrt3\text{xy}+2\sqrt3\text{x}^2-\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{2\text{y}^2+2\text{x}^2-2\text{xy}}{2\sqrt3\text{y}^2+2\sqrt3\text{x}^2-2\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=\frac{\pi}{6}$
View full question & answer→MCQ 211 Mark
If $\text{x}=\sin ^{ -1 }{ \text{K} },\text{y}=\cos ^{ -1 }\text{K}, -1\le \text{K}\le 1$, then the correct relationship is:
- A
$\text{x}+\text{y}=\frac{\pi}{8}$
- B
$\text{x}+\text{y}={2}$
- ✓
$\text{x}+\text{y}=\frac{\pi}{2}$
- D
$\text{x}+\text{y}=\frac{\pi}{8}$
AnswerCorrect option: C. $\text{x}+\text{y}=\frac{\pi}{2}$
$\because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$
$\therefore \text{x}+\text{y}=\sin ^{ -1 }{ \text{K} } +\cos ^{ -1 }{ \text{K} } =\frac { \pi }{ 2 }$
View full question & answer→MCQ 221 Mark
If $\sin^{-1}(\text{x}^2-7\text{x}+12)=\text{n}\pi,\forall\text{ n }\in\text{ I},$ then $x =$
View full question & answer→MCQ 231 Mark
Choose the correct answer from the given four options.The value of $\sin\big[2\tan^{-1}(0.75)\big]$ is equal to:
- A
$0.75$
- B
$1.5$
- ✓
$0.96$
- D
$\sin1.5$
AnswerCorrect option: C. $0.96$
We have, $\sin\big[2\tan^{-1}(0.75)\big]$
$=\sin\Big(2\tan^{-1}\frac{3}{4}\Big)$
$=\sin\Bigg(\sin^{-1}\frac{2.\frac{3}{4}}{1+\frac{9}{16}}\Bigg)$
$\Big(\because\ 2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\sin\Bigg(\sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}}\Bigg)$
$=\sin\Big(\sin^{-1}\frac{24}{25}\Big)=\frac{24}{25}=0.96$
View full question & answer→MCQ 241 Mark
Choose the correct answer from the given four options.The value of $\sin^{-1}\bigg[\cos\Big(\frac{33\pi}{5}\Big)\bigg]$ is:
- A
$\frac{3\pi}{5}$
- B
$\frac{-7\pi}{5}$
- C
$\frac{\pi}{10}$
- ✓
$\frac{-\pi}{10}$
AnswerCorrect option: D. $\frac{-\pi}{10}$
We have, $\sin^{-1}\bigg[\cos\Big(\frac{33\pi}{5}\Big)\bigg]=\sin^{-1}\bigg[\cos\Big(6\pi+\frac{33\pi}{5}\Big)\bigg]$
$=\sin^{-1}\bigg[\cos\Big(\frac{3\pi}{5}\Big)\bigg]$
$\Big[\because\ \cos(2\text{n}\pi+\theta)=\cos\theta\Big]$
$=\sin^{-1}\Big[\cos\Big(\frac{\pi}{2}+\frac{\pi}{10}\Big)\Big]$
$=\sin^{-1}\Big(-\sin\frac{\pi}{10}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{10}\Big)$
$[\because\ \sin^{-1}(-\text{x})=-\sin^{-1}\text{x}]$
$=-\frac{\pi}{10}\ \Big[\because\ \sin^{-1}(\sin\text{x})=\text{x},\ \text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)\Big]$
View full question & answer→MCQ 251 Mark
$\sin\begin{Bmatrix}2\cos^{-1}\Big(\frac{-3}{5}\Big)\end{Bmatrix}$ is equal to:
- A
$\frac{6}{25}$
- B
$\frac{24}{25}$
- C
$\frac{4}{5}$
- ✓
$-\frac{24}{25}$
AnswerCorrect option: D. $-\frac{24}{25}$
View full question & answer→MCQ 261 Mark
The range of the function, $\text{f(x)}=(1+\sec^{-1}\text{x})(1+\cos^{-1}\text{x})$ is:
- A
$(-\infty,\infty)$
- B
$(-\infty,0]\cup[4.\infty)$
- C
$\big\{0,(1+\pi^2)\big\}$
- ✓
$[1.(1+\pi)^2]$
AnswerCorrect option: D. $[1.(1+\pi)^2]$
$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$
Here the limiting component is $\cos−1(\text{x}),$ since the domain of $\cos−1(\text{x}),$ is $[−1, 1].$
Therefore,
$\text{f}(1)=(1+0)(1+0)$
$=1$
$\text{f}(−1)=(1+\pi(1+\pi)$
$=(1+\pi)^2 $
Hence range of $\text{f(x)}=[1,(1+\pi)^2]$
View full question & answer→MCQ 271 Mark
The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{3}}\Big)$
- A
$2+\sqrt{5}$
- ✓
$\sqrt{5}-2$
- C
$\frac{\sqrt{5}+2}{2}$
- D
$5+\sqrt{2}$
AnswerCorrect option: B. $\sqrt{5}-2$
View full question & answer→MCQ 281 Mark
$\cos\Big(2\tan^{-1}\frac{1}{7}\Big)-\sin\Big(4\sin^{-1}\frac{1}{3}\Big)=$
- A
$1$
- ✓
$0$
- C
$\frac{1}{2}$
- D
$-\frac{1}{2}$
View full question & answer→MCQ 291 Mark
Choose the correct answer from the given four options.The domain of the function $\cos ^{-1}(2 x-1)$ is:
- ✓
$[0,1]$
- B
$[-1,1]$
- C
$(-1,1)$
- D
$[0,\pi]$
AnswerCorrect option: A. $[0,1]$
We have,$\cos ^{-1}(2 x-1)$
Now, we know that the domain of $\cos ^{-1}(x)$ is $-1\leq\text{x}\leq1$
$\therefore\ -1\leq2\text{x}-1\leq1$
Adding $1$ to all terms, we get
$\Rightarrow\ 0\leq2\text{x}\leq2$
Dividing all terms by $2,$ we get
$\Rightarrow\ 0\leq\text{x}\leq1$
$\therefore\ \text{x}\in[0,1]$
View full question & answer→MCQ 301 Mark
If $\cos \left ( 2\sin^{-1}\text{x} \right )=\frac{1}{9}$, the value of $x$ which satify equation is $ \pm \frac{a}{b}$. Find the value of $a + b:$
AnswerGiven, $\cos \left ( 2\sin^{-1}\text{x} \right )=\frac{1}{9}$
Let, $\sin^{-1}\text{x}=\theta .$
Then, $\cos 2\theta =\frac{1}{9}$
$ 1-2\sin ^{2}\theta =\frac{1}{9}$
or $1-2\text{x}^{2}=\frac{1}{9}$
$\text{x}^{2}=\frac{4}{9}\text{x}$
$ \therefore \text{a}+\text{b}=2+3=5$
View full question & answer→MCQ 311 Mark
Choose the correct answer from the given four options.If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:
- A
$\frac{1}{\sqrt{2}}<\text{x}\leq1$
- B
$0\leq\text{x}<\frac{1}{\sqrt{2}}$
- ✓
$-1\leq\text{x}<\frac{1}{\sqrt{2}}$
- D
$\text{x}>0$
AnswerCorrect option: C. $-1\leq\text{x}<\frac{1}{\sqrt{2}}$
We have, $\cos^{-1}\text{x}>\sin^{-1}\text{x}$
$\Rightarrow\ \frac{\pi}{2}-\sin^{-1}\text{x}>\sin^{-1}\text{x}$
$\Rightarrow\ \frac{\pi}{2}>2\sin^{-1}\text{x}$
$\Rightarrow\ \sin^{-1}\text{x}<\frac{\pi}{4}\ ....(\text{i})$
But $-\frac{\pi}{2}\leq\sin^{-1}\text{x}\leq\frac{\pi}{2}\ ....(\text{ii})$
From $(i)$ and $(ii), -\frac{\pi}{2}\leq\sin^{-1}\text{x}<\frac{\pi}{4}$
$\Rightarrow\ \sin\Big(-\frac{\pi}{2}\Big)\leq\text{x}<\sin\frac{\pi}{4}$
$\Rightarrow\ -1\leq\text{x}<\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 321 Mark
If $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+2}\Big)+\tan^{-1}\Big(\frac{\text{x}+1}{\text{x}+2}\Big)=\frac{\pi}{4},$ then $x$ is equal to:
- A
$\frac{1}{\sqrt{2}}$
- B
$-\frac{1}{\sqrt2}$
- ✓
$\pm\sqrt{\frac{5}{2}}$
- D
$\pm\frac{1}{2}$
AnswerCorrect option: C. $\pm\sqrt{\frac{5}{2}}$
View full question & answer→MCQ 331 Mark
If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$
- ✓
$\sin2\alpha$
- B
$\sin\alpha$
- C
$\cos2\alpha$
- D
$\cos\alpha$
AnswerCorrect option: A. $\sin2\alpha$
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{1+\text{x}^2-2\sqrt{1-\text{x}^2}\sqrt{1+\text{x}^2}+1-\text{x}^2}{1+\text{x}^2-1+\text{x}^2}=\tan\alpha$
$\frac{1-\sqrt{1-\text{x}^4}}{\text{x}^2}=\tan\alpha$
$1-\sqrt{1-\text{x}^4}=\text{x}^2\tan\alpha$
$\big(1-\text{x}^2\tan\alpha\big)^2=1-\text{x}^4$
$1-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=1-\text{x}^4$
$\text{x}^4-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=0$
$\text{x}^2\big(\text{x}^2-2\tan\alpha+\text{x}^2\tan^2\alpha\big)=0$
$\text{x}^2=\frac{2\tan\alpha}{1+\tan^2\alpha}$
$\text{x}^2=\frac{2\tan\alpha}{\sec^2\alpha}$
$\text{x}^2=2\tan\alpha\cos^2\alpha$
$\text{x}^2=2\sin\alpha\cos\alpha=2\sin\alpha$
View full question & answer→MCQ 341 Mark
The value of $ \cos^{-1}\left (\cot \left (\dfrac {\pi}{2}\right )\right ) + \cos^{-1} \left (\sin \left (\dfrac {2\pi}{3}\right )\right )$ is:
- ✓
$ \dfrac {2\pi}{3}$
- B
$2$
- C
$3$
- D
$\pi $
AnswerCorrect option: A. $ \dfrac {2\pi}{3}$
$ \cos^{-1}\left (\cot \dfrac {\pi}{2}\right ) + \cos^{-1} \left (\sin \dfrac {2\pi}{3}\right ) = \cos^{-1} (0) + \cos^{-1} \left (\dfrac {\sqrt {3}}{2}\right )$
$=\frac{\pi}{2}+\cos^{-1}\bigg(\cos\frac{\pi}{6}\bigg)$
$ = \frac {\pi}{2} + \frac {\pi}{6}$
$ = \frac {4\pi}{6}$
$ = \frac {2\pi}{3}$
View full question & answer→MCQ 351 Mark
If $x < 0, y < 0$ such that $xy = 1,$ then $\tan^{-1}\text{x}+\tan^{-1}\text{y}$ equals:
- A
$\frac{\pi}{2}$
- ✓
$-\frac{\pi}{2}$
- C
$-\pi$
- D
AnswerCorrect option: B. $-\frac{\pi}{2}$
We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$x < 0, y < 0$ such that
$xy = 1$
Let $x = -a$ and $y = -b,$ where $a$ and $b$ both are positive.
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{-\text{a}-\text{a}}{1-1}\Big)$
$=\tan^{-1}(-\infty)$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{2}\Big)\Big\}$
$=-\frac{\pi}{2}$
View full question & answer→MCQ 361 Mark
Find the value of $:\ \sec^2 (\tan^{-1} 2) +\csc^2 (\cot^{-1} 3)$
Answer$ \sec^2 (\tan^{-1} 2) +\csc^2 (\cot^{-1} 3)$
$ =1+\tan^2 (\tan^{-1} 2) +1+\cot^2 (\cot^{-1} 3)$
$ =1+[\tan (\tan^{-1} 2)]^2 +1+[\cot (\cot^{-1} 3)]^2$
$ =1+2^2+1+3^2=15$
View full question & answer→MCQ 371 Mark
$\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)=$
- ✓
$\frac{6}{17}$
- B
$\frac{3}{17}$
- C
$\frac{4}{17}$
- D
$\frac{5}{17}$
AnswerCorrect option: A. $\frac{6}{17}$
View full question & answer→MCQ 381 Mark
The value of $\sin(2\tan^{-1}(0.75))$ is equal to:
- A
$0.75$
- B
$1.5$
- ✓
$0.96$
- D
$\sin1.5$
AnswerCorrect option: C. $0.96$
View full question & answer→MCQ 391 Mark
Domain of $ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +\text{cosec} ^{ -1 }{ \text{x}}$ is:
AnswerCorrect option: D. $\{-1, 1\}$
$ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +\text{cosec} ^{ -1 }{ \text{x}}$
Domain of $\cot^{−1}\text{x}=(−\infty ,\infty )$
Domain of $\cos^{−1}\text{x}=(−1,1)$
Domain of $ \text{cosec}^{-1}\text{x} = (-\infty, -1]\cup [1, \infty)c$
These function are in addition.So,
we have to take the intersection of all domains.So,
answer is $\{-1, 1\}$
concept: $ \text{f}(\text{x}) = \text{f}_1(\text{x}) +\text{f}_2(\text{x}) + ...+\text{f}_\text{n}(\text{x})$
domain of $ \text{f}(\text{x})$
Domain of $\text{f}_1(\text{x}) \cap$
domain of $\text{f}_2(\text{x}) \cap$
domain of $\text{f}_\text{n}(\text{x})$
View full question & answer→MCQ 401 Mark
The value of $\cot^{-1}9+\text{cosec}^{-1}(\frac{\sqrt{41}}{4})$ is given by:
- A
$0$
- ✓
$\frac{\pi}{4}$
- C
$\tan^{-1}2$
- D
$\frac{\pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{4}$
View full question & answer→MCQ 411 Mark
$\cot(\frac{\pi}{4}-2\cot^{-1}3)=$
View full question & answer→MCQ 421 Mark
The value of $\tan\Big\{\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{4}{\sqrt{17}}\Big\}$ is:
- A
$\frac{\sqrt{29}}{3}$
- B
$\frac{29}{3}$
- C
$\frac{\sqrt3}{29}$
- ✓
$\frac{3}{29}$
AnswerCorrect option: D. $\frac{3}{29}$
Let, $\cos^{-1}\frac{1}{5\sqrt2}=\text{y}$ and $\sin^{-1}\frac{4}{\sqrt{17}}=\text{z}$
$\therefore\ \cos\text{y}=\frac{1}{5\sqrt2}$
$\Rightarrow\sin\text{y}=\frac{7}{5\sqrt2}$
$\Rightarrow\tan\text{y}=7$
$\sin\text{z}=\frac{4}{\sqrt{17}}$
$\Rightarrow\cos\text{z}=\frac{1}{\sqrt{17}}$
$\Rightarrow\tan\text{z}=4$
$\therefore\ \tan\Big(\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{1}{\sqrt{17}}\Big)=\tan(\text{y}-\text{z})$
$=\frac{\tan\text{y}-\tan\text{z}}{1+\tan\text{y}\tan\text{z}}$
$=\frac{7-4}{1+7\times4}$
$=\frac{3}{29}$
View full question & answer→MCQ 431 Mark
What will be the value of $ \text{x} + \text{y} + \text{z }$ if $\cos-1 \text{x} + \cos-1 \text{y} + \cos-1 \text{z} = 3\pi ?$
- A
$ \frac{-1}{3}$
- B
$1$
- C
$3$
- ✓
$-3$
AnswerThe equation is $ \cos-1 \text{x} +\cos-1 \text{y} + \cos-1 \text{z} = 3\pi $
This means $ \cos-1 \text{x} = \pi , \cos-1 \text{y} = \pi $ and $ \cos-1 \text{z} = \pi $
This will be only possible when it is in maxima.
As, $\cos-1 \text{x} = \pi $ so,$ \text{x} = \cos-1 \pi = -1$ similarly$, y = z = -1$
Therefore$, x + y + z = -1 -1 -1$
So$, x + y + z = -3.$
View full question & answer→MCQ 441 Mark
If $\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}+\tan^{-1}(\text{x}+1)=\tan^{-1}3\text{x},$ then the values of $x$ are:
- A
$\pm\frac{1}{2}$
- B
$0,\frac{1}{2}$
- C
$0,-\frac{1}{2}$
- ✓
$0,\pm\frac{1}{2}$
AnswerCorrect option: D. $0,\pm\frac{1}{2}$
View full question & answer→MCQ 451 Mark
$\cos[\tan^{-1}\{\sin(\cot^{-1}\text{x})\}]$ is equal to:
- A
$\sqrt{\frac{\text{x}^2+2}{\text{x}^3+3}}$
- B
$\sqrt{\frac{\text{x}^2+2}{\text{x}^2+1}}$
- ✓
$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
- D
AnswerCorrect option: C. $\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
View full question & answer→MCQ 461 Mark
$[-1, 1]$ is the domain for which of the following inverse trigonometric functions?
- ✓
$\sin^{-1}\text{x}$
- B
$\cot^{-1}\text{x}$
- C
$\tan^{-1}\text{x}$
- D
$\sec^{-1}\text{x}$
AnswerCorrect option: A. $\sin^{-1}\text{x}$
$[-1, 1]$ is the domain for $\sin^{-1}\text{x}$
The domain for $\cot^{-1}\text{x}$ is $(-\infty , \infty ).$
The domain for $\tan^{-1}\text{x}$ is $(-\infty , \infty ).$
The domain for $\sec^{-1}\text{x}$ is $(-\infty , -1) \cup (1, \infty ).$
View full question & answer→MCQ 471 Mark
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 12)$ is:
- A
$0$
- B
$ \pi $
- ✓
$ 8\pi - 24$
- D
AnswerCorrect option: C. $ 8\pi - 24$
$12$ rad lies in $4^{th}$ quadrant
$ \frac{7\pi}{2}<12<4\pi$
Let $\theta$ be an acute angle such that
$ 12+\theta=4\pi$
$\therefore 12=4\pi −\theta$ or $\theta=4\pi-12\theta =4\pi −12$
$ \cos^{-1}(\cos12)-\sin^{-1}(\sin12)$
$ =\cos^{-1}(\cos(4\pi-\theta))-\sin^{-1}(\sin(4\pi-\theta))$
$ =\cos^{-1}(\cos\theta)-\sin^{-1}(-\sin\theta)$
$=\cos^{-1}(\cos\theta)-\sin^{-1}(\sin(-\theta))$
$ =\theta-(-\theta)$
$ =2\theta$
$ =2(4\pi −24)$
$ =8\pi −24$
$ \therefore \cos−1(\cos12)−\sin−1(\sin12)=8\pi −24$
View full question & answer→MCQ 481 Mark
If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then $x =$
- A
$\frac{1}{2}$
- ✓
$\frac{\sqrt{3}}{2}$
- C
$-\frac{1}{2}$
- D
$-\frac{\sqrt{3}}{2}$
AnswerCorrect option: B. $\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 491 Mark
If $\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2},$ then, $4\text{x}^2-12\text{xy}\cos^2\frac{\theta}{2}+9\text{y}^2=$
- A
$36$
- B
$36-36\cos\theta$
- ✓
$18-18\cos\theta$
- D
$18+18\cos\theta$
AnswerCorrect option: C. $18-18\cos\theta$
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
$\Rightarrow\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2}$
$\Rightarrow\cos^{-1}\Bigg(\frac{\text{x}}{3}\times\frac{\text{y}}{2}-\sqrt{1-\Big(\frac{\text{x}}{3}\Big)^2}\sqrt{1-\Big(\frac{\text{y}}{2}\Big)^2}\Bigg)=\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}}{6}-\sqrt{1-\Big(\frac{\text{x}^2}{9}\Big)}\sqrt{1-\Big(\frac{\text{y}^2}{4}\Big)}=\cos\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}-6\cos\frac{\theta}{2}}{6}=\frac{\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}}{6}$
$\Rightarrow\text{xy}-6\cos\frac{\theta}{2}=\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}$
Taking square on both sides,
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=\big(9-\text{x}^2\big)\big(4-\text{y}^2\big)$
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=36-9\text{y}^2-4\text{x}^2+\text{x}^2\text{y}^2$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\cos^2\frac{\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\frac{1+\cos\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\cos^2\frac{\theta}{2}=18-18\cos\theta$
View full question & answer→MCQ 501 Mark
The value of $\sin\bigg[\cos^{-1}\Big(\frac{7}{25}\Big)\bigg]$ is:
- A
$\frac{25}{24}$
- B
$\frac{25}{7}$
- ✓
$\frac{24}{25}$
- D
$\frac{7}{24}$
AnswerCorrect option: C. $\frac{24}{25}$
View full question & answer→