Maths M.C.Q (1 Marks)

1. $\frac{1}{\sqrt3}$

Solution:

Let $\text{x}=\tan\text{y}$

Then,

$3\sin^{-1}\Big(\frac{\tan2\text{y}}{1+\tan^2\text{y}}\Big)-4\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)+2\tan^{-1}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)=\frac{\pi}{3}$

$\Rightarrow3\sin^{-1}(\sin2\text{y})-4\cos^{-1}(\cos2\text{y})+2\tan^{-1}(\tan2\text{y})=\frac{\pi}{3}$

$\Big[\because\ \sin2\text{y}=\Big(\frac{2\tan\text{y}}{1+\tan^2\text{y}}\Big),\cos2\text{y}=\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)\text{ and }\tan2\text{y}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)\Big]$

$\Rightarrow3\times2\text{y}-4\times2\text{y}+2\times2\text{y}=\frac{\pi}{3}$

$\Rightarrow6\text{y}-8\text{y}+4\text{y}=\frac{\pi}{3}$

$\Rightarrow2\text{y}=\frac{\pi}{3}$

$\Rightarrow\text{y}=\frac{\pi}{6}$

$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$ $\big[\because\ \tan^{-1}\text{x}=\text{y}\big]$

$\Rightarrow\text{x}=\tan\frac{\pi}{6}$

$\Rightarrow\text{x}=\frac{1}{\sqrt3}$

1. $ \frac{π}{2}$

solution:

$ \sin-1\text{⁡x}+\cos-1\text{⁡x}=π2; \text{x} ∈ [-1,1] $

4. $\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$

Solution:

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 } } }$

3. x + 2y = 2

Solution:

Given, $\text{x}=\cos^{-1}(\cos 4)$

$ = 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)$

3. $\frac { 1 }{ \sqrt { 2 } }$

Solution:

$ \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}​$

$= \cos^{-1}\text{x}=\frac{\pi}{4}\text{x}$

$=\cos (\frac{\pi}{4})=\frac{1}{\sqrt 2}$

3. $\frac{\text{x}-\text{y}}{1+\text{xy}}$

4. 0

Solution:

We have

$\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$

$=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{\sin\Big(2\pi-\frac{\pi}{3}\Big)\Big\}$

$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{-\sin\Big(\frac{\pi}{3}\Big)\Big\}$

$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}-\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$

$=\frac{\pi}{3}-\frac{\pi}{3}$

$=0$

1. $4\tan^{-1}\text{x}$

Solution:

We have, $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$

Let $\text{x}=\tan\theta$

$\therefore\ 2\tan^{-1}\tan\theta+\sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta}$

$[\because\ \tan^{-1}(\tan\text{x})=\text{x}]$

$=2\theta+\sin^{-1}\sin2\theta$

$\Big[\because\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$

$=2\theta+2\theta$

$[\because\ \sin^{-1}=(\sin\text{x})=\text{x}]$

$=4\theta\ [\because\ \theta=\tan^{-1}\text{x}]$

$=4\tan^{-1}\text{x}$

1. $\sqrt{\text{ab}}$

1. $\tan^2\Big(\frac{\alpha}{2}\Big)$

4. $-\frac{24}{25}$

4. $[1.(1+\pi)^2]$

Solution:

$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$

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 $\text{f(x)}=[1,(1+\pi)^2]$

2. $\sqrt{5}-2$

2. $0$

4. 5

Solution:

Given, $\cos \left ( 2\sin^{-1}\text{x} \right )=\frac{1}{9}$

Let, $\sin^{-1}\text{x}=θ.$

Then, $\cos 2\theta =\frac{1}{9}​$

$ 1-2\sin ^{2}\theta =\frac{1}{9}$

or $1-2\text{x}^{2}=\frac{1}{9}$

$\text{x}^{2}=\frac{4}{9}\text{x}$

$ ∴ \text{a}+\text{b}=2+3=5$

4. $-\frac{\pi}{6}$

1. $\sin2\alpha$

Solution:

$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha$

$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}=\tan\alpha$

$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}=\tan\alpha$

$\frac{1+\text{x}^2-2\sqrt{1-\text{x}^2}\sqrt{1+\text{x}^2}+1-\text{x}^2}{1+\text{x}^2-1+\text{x}^2}=\tan\alpha$

$\frac{1-\sqrt{1-\text{x}^4}}{\text{x}^2}=\tan\alpha$

$1-\sqrt{1-\text{x}^4}=\text{x}^2\tan\alpha$

$\big(1-\text{x}^2\tan\alpha\big)^2=1-\text{x}^4$

$1-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=1-\text{x}^4$

$\text{x}^4-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=0$

$\text{x}^2\big(\text{x}^2-2\tan\alpha+\text{x}^2\tan^2\alpha\big)=0$

$\text{x}^2=\frac{2\tan\alpha}{1+\tan^2\alpha}$

$\text{x}^2=\frac{2\tan\alpha}{\sec^2\alpha}$

$\text{x}^2=2\tan\alpha\cos^2\alpha$

$\text{x}^2=2\sin\alpha\cos\alpha=2\sin\alpha$

1. $ \dfrac {2\pi}{3}$

Solution:

$ \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}$

2. $-\frac{\pi}{2}$

Solution:

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.

$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$

$=\tan^{-1}\Big(\frac{-\text{a}-\text{a}}{1-1}\Big)$

$=\tan^{-1}(-\infty)$

$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{2}\Big)\Big\}$

$=-\frac{\pi}{2}$

2. 15

Solution:

$ \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$

2. $\sqrt{5}-2$

Solution:

We have, $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$

Let $\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}=\theta$

$\Rightarrow\ \cos^{-1}\frac{2}{\sqrt{5}}=2\theta$

$\Rightarrow\ \cos2\theta=\frac{2}{\sqrt{5}}$

$\therefore\ 2\cos^{2}\theta-1=\frac{2}{\sqrt{5}}$

$\Rightarrow\ \cos^2\theta=\frac{1}{2}+\frac{1}{\sqrt{5}}$

$\Rightarrow\ \cos\theta=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}$

$\therefore\ \tan\theta=\frac{\sin\theta}{\cos\theta}$

$=\sqrt{\frac{\frac{1}{2}-\frac{1}{\sqrt{5}}}{\frac{1}{2}+\frac{1}{\sqrt{5}}}}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$

$=\sqrt{\frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)}}=\sqrt{5}-2$

1. $\frac{6}{17}$

4. {-1, 1}

Solution:

$ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +co\sec ^{ -1 }{ \text{x}}$

Domain of $\cot^{−1}\text{x}=(−∞,∞)$

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 $\text{f}_1​(\text{x}) ∩$

domain of $\text{f}_2(\text{x}) ∩$

domain of $\text{f}_\text{n}(\text{x})$

3. 0.96

Solution:

We have, $\sin\big[2\tan^{-1}(0.75)\big]$

$=\sin\Big(2\tan^{-1}\frac{3}{4}\Big)$

$=\sin\Bigg(\sin^{-1}\frac{2.\frac{3}{4}}{1+\frac{9}{16}}\Bigg)$

$\Big(\because\ 2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big)$

$=\sin\Bigg(\sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}}\Bigg)$

$=\sin\Big(\sin^{-1}\frac{24}{25}\Big)=\frac{24}{25}=0.96$

3. $\pm\sqrt{\frac{5}{2}}$

2. $\frac{\pi}{4}$

4. -3

Solution:

The equation is $ \cos-1 \text{x} +\cos-1 \text{y} + \cos-1 \text{z} = 3π$

This means $ \cos-1 \text{x} = π, \cos-1 \text{y} = π$ and $ \cos-1 \text{z} = π$

This will be only possible when it is in maxima.

As, $\cos-1 \text{x} = π$ so,$ \text{x} = \cos-1 π = -1$ similarly, y = z = -1

Therefore, x + y + z = -1 -1 -1

So, x + y + z = -3.

4. $0,\pm\frac{1}{2}$

1. $\sin^{-1}\text{⁡x}$

Solution:

[-1, 1] is the domain for $\sin^{-1}\text{⁡x}$

The domain for $\cot^{-1}\text{⁡x}$ is (-∞, ∞).

The domain for $\tan^{-1}\text{⁡x}$ is (-∞, ∞).

The domain for $\sec^{-1}\text{⁡x}$ is (-∞, -1) ∪ (1, ∞).

3. 8π - 24

Solution:

12 rad lies in 4th quadrant

$ \frac{7\pi}{2}<12<4\pi$

Let θ be an acute angle such that

$ 12+\theta=4\pi$

$∴12=4π−θ or \theta=4\pi-12θ=4π−12$

$ \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(4π−24)$

$ =8π−24​$

$ ∴\cos−1(\cos12)−\sin−1(\sin12)=8π−24$

3. $\text{x}+\text{y}=\frac{\pi}{2}$

Solution:

$\because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$

$\therefore \text{x}+\text{y}=\sin ^{ -1 }{ \text{K} } +\cos ^{ -1 }{ \text{K} } =\frac { \pi }{ 2 }$

2. $\frac{\sqrt{3}}{2}$

3. $\frac{24}{25}$

2. $\frac{1}{5\sqrt{2}}$

3. $36\sin^{2}\theta$

Solution:

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)

$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^{2}\theta-12\text{xy}\cos\theta$

$\Rightarrow36-4\text{y}^2-9\text{x}^2=36\cos^2\theta-12\text{xy}\cos\theta$

$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36-36\cos^2\theta$

$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36\sin^2\theta$

3. 6

Solution:

The domain of cos^-1x is $[0,\pi]$

We are given that, $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi$

Which is possible only when $\alpha=\beta=\gamma=\cos\pi\ \text{or }-1$

Now, $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$

$=-1(-1-1)-1(-1-1)-1(-1-1)$

$=2+2+2$

$=6$

4. $\frac{3\pi}{4}$

3. either -2π + 6 or 2π + 6

Solution:

We 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).$

2. $ \tan^{-1}\left (\frac {120}{119}\right )$

Solution:

$\text{x}=4{ \tan }^{ -1 }\left(\dfrac { 1 }{ 5 } \right)\\$

$ =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)\\$

4. $\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$

2. 2 only

Solution:

We know that $\tan^{-1} { \text{x} } + \cot^{-1} { \text{x} } =\frac { \pi }{ 2 }$$ \text{x} ∈\text{Rand}\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.

2. $[\frac{\pi}{4},\frac{3\pi}{4}]$

4. $\frac{2\pi}{3}$

2. 8π - 26

Solution:

The 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$

2. $\frac{\pi}{4}$

Solution:

We know,

$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$

$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$

$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$

$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$

$=\tan^{-1}(1)$

$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$

$=\frac{\pi}{4}$

3. 2

Solution:

$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$

$\Rightarrow\sqrt{2\cos^2\text{x}}=\sqrt2(-\pi-\text{x})$

$\Rightarrow|\cos\text{x}|=\text{x}$

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 $[-\pi,\pi]$

2. $\frac{5\pi}{6}$

2. $\sqrt{\frac{3}{76}}$