Inverse Trigonometric Functions

$ \begin{aligned} & \text { } \tan ^{-1}(x)+\tan ^{-1}(2 x)=\tan ^{-1}(l) \\ & \tan ^{-1}\left(\frac{3 x}{1-2 x^2}\right)=\tan ^{-1}(l) \\ & 3 x=1-2 x^2 \\ & 2 x^2+3 x-1=0 \\ & \quad x=\frac{-3 \pm \sqrt{9+8}}{4} \\ & \quad x=\frac{-3 \pm \sqrt{17}}{4} \end{aligned} $

So, $\quad x=\frac{-3+\sqrt{17}}{4}$ will satisfy Number of solution is 1 .

$ \begin{aligned} &\text { A is false since }\\ &\begin{aligned} \sin ^{-1}(\log x) & +\cos ^{-1}(\log x) \\ & =\frac{\pi}{2} \text { only when }-1 \leq \log x \leq 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { So, } \int \sqrt{x-3}\left(\sin ^{-1}(\log x)+\cos ^{-1}(\log x) d x\right. \\ & \qquad \neq \frac{\pi}{2} \times(x-3)^{\frac{3}{2}} \times \frac{2}{3}+C \\ & \sin ^{-1}(f(x))+\cos ^{-1}\left(f(x)=\frac{\pi}{2}\right. \text { when } \\ & |f(x)| \leq 1 \end{aligned}\\ &\text { So } A \text { is false and } B \text { is true. } \end{aligned} $

$ \text { } \begin{aligned} & \sin ^{-1}(-\cos 2)=-\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2\right)\right) \\ & \Rightarrow \quad \cos ^{-1}(\sin 3)=\cos ^{-1}\left(\cos \left(3-\frac{\pi}{2}\right)\right) \\ & \Rightarrow \quad \tan ^{-1}(\cot 5)=\tan ^{-1}\left(\tan \left(\frac{3 \pi}{2}-5\right)\right) \\ & \text { So, } \sin ^{-1}(-\cos 2)+\cos ^{-1}(\sin 3) \\ &+ \tan ^{-1}(\cot 5) \\ &=2-\frac{\pi}{2}+3-\frac{\pi}{2}+\frac{3 \pi}{2}-5 \\ &= \frac{3 \pi}{2}-\pi=\frac{\pi}{2} \end{aligned} $

$f(x)=\cos ^{-1}(2 x-5)-\sin ^{-1}(x-2)$

Domain of $f(x)$

$ \begin{aligned} & =|2 x-5| \leq 1 \cap|x-2| \leq 1 \\ & =-1 \leq 2 x-5 \leq 1 \cap-1 \leq x-2 \leq 1 \\ & =2 \leq x \leq 3 \cap 1 \leq x \leq 3, x \in[2,3] \\ f^{\prime}(x) & =\frac{-1 \times \hat{2}}{\sqrt{1-(2 x-5)^2}}-\frac{1 \times 1}{\sqrt{1-(x-2)^2}} \end{aligned} $

Now, $f^{\prime}(x)$ exist only

$ \begin{aligned} & \quad 1-(2 x-5)^2>0 \text { and } 1-(x-2)^2>0 \\ & \Rightarrow \quad x \in(2,3) \text {. } \\ & \text { Domain }=(2,3) \end{aligned} $

We have,

$ \begin{aligned} & \tan ^{-1}\left(x+\frac{\sqrt{2}}{x}\right)+\tan ^{-1}\left(x-\frac{\sqrt{2}}{x}\right) \\ & =\tan ^{-1} x \\ & x \neq 0 \\ & \text { Now, } \tan ^{-1}\left(\frac{x+\frac{\sqrt{2}}{x}+x-\frac{\sqrt{2}}{x}}{1-\left(x+\frac{\sqrt{2}}{x}\right)\left(x-\frac{\sqrt{2}}{x}\right)}\right) \\ & =\tan ^{-1} x \\ & \Rightarrow \quad \frac{2 x}{1-\left(x^2-\frac{2}{x^2}\right)}=x \\ & \Rightarrow \quad 2=1-x^2+\frac{2}{x^2} \\ & \quad x^2-\frac{2}{x^2}+1=0 \\ & \Rightarrow \quad x^4+x^2-2=0 \\ & \quad\left(x^2+2\right)\left(x^2-1\right)=0 \\ & \quad x^2=2 \Rightarrow x= \pm 1 \end{aligned} $

∴ Number of values of $x=2$

$ \begin{aligned} &\text { } y=\sec ^{-1} x\\ &\begin{aligned} & x=\sec y, \cos y=\frac{1}{x}, \sin y=\frac{\sqrt{x^2-1}}{x} \\ & 1=\sec y \tan y \frac{d y}{d x} \\ \Rightarrow & \frac{d y}{d x}=\cot y \cdot \cos y \\ \Rightarrow & \frac{d^2 y}{d x^2}=\left[\cot y \cdot(-\sin y)+\cos y\left(-\operatorname{cosec}^2 y\right)\right] \frac{d y}{d x} \\ = & -\cos y\left(1+\frac{1}{\sin ^2 y}\right) \cdot \cot y \cdot \cos y \\ = & -\cos ^3 y \frac{\left(1+\sin ^2 y\right)}{\sin ^3 y}=\frac{1}{x|x|} \frac{\left(1-2 x^2\right)}{\left(x^2-1\right)^{\frac{3}{2}}} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan ^{-1}(2 x-1) & +\tan ^{-1} 2 x \\ = & \tan ^{-1} 4 x-\tan ^{-1}(2 x+1) \\ \Rightarrow \tan ^{-1}(2 x-1) & +\tan ^{-1}(2 x+1) \\ = & \tan ^{-1} 4 x-\tan ^{-1} 2 x \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \tan ^{-1}\left(\frac{2 x-1+2 x+1}{1-(2 x-1)(2 x+1)}\right) \\ & =\tan ^{-1}\left(\frac{4 x-2 x}{1+(4 x)(2 x)}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{4 x}{1-4 x^2+1}\right)=\tan ^{-1}\left(\frac{2 x}{1+8 x^2}\right) \\ & \therefore \frac{4 x}{2-4 x^2}=\frac{2 x}{1+8 x^2} \\ & \quad x=0 \\ & \quad 1-x^2=1+8 x^2 \\ & \quad x^2=0 \Rightarrow x=0 \end{aligned}\\ &\text { ∴ Only one solution. } \end{aligned} $

$ \begin{aligned} & \sinh ^{-1} x=\cosh ^{-1} y=\log (1+\sqrt{2}) \\ & \sinh ^{-1} x=\log (1+\sqrt{2}) \\ & \log \left(x+\sqrt{x^2}+1\right)=\log (1+\sqrt{2}) \\ & x=1 \\ & \text { And } \cosh ^{-1} y=\log (1+\sqrt{2}) \\ & \Rightarrow \log \left(y+\sqrt{y^2-1}\right)=\log (1+\sqrt{2}) \\ & \therefore \quad y=\sqrt{2} \end{aligned} $

Now, $\tan ^{-1}(x+y)$

$ \begin{aligned} & =\tan ^{-1}(1+\sqrt{2}) \\ & =67.5^{\circ} \end{aligned} $

By LHS, $\tan ^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\tan ^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)$$ +\tan ^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) $

Let $a=\sqrt{\frac{x(x+y+z)}{y z}}, b=\sqrt{\frac{y(x+y+z)}{x z}}$ and $z=\sqrt{\frac{z(x+y+z)}{x y}}$

Here, $a+b+c=\frac{(x+y+z)^{1 / 2}(x+y+z)}{\sqrt{x y z}}=\frac{(x+y+z)^{3 / 2}}{\sqrt{x y z}}$

And $a b c=\frac{(x+y+z)^{3 / 2}}{\sqrt{x y z}}$

Since, $a+b+c=a b c$

$ \therefore \quad \text { LHS }=\pi $

$\Rightarrow$ Assertion is true and since, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$.

if $x y<1$

$\Rightarrow$ Reason is false.

$ \begin{aligned} &\text { We know that }\\ &\begin{aligned} & \sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ &\therefore \sinh ^{-1} 2=\log (2+\sqrt{4+1})=\log (2+\sqrt{5}) \\ & \cosh ^{-1} x=\log \left(x+\sqrt{x^2-1}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore \cosh ^{-1} \sqrt{6}=\log (\sqrt{6}+\sqrt{6-1})=\log (\sqrt{6}+\sqrt{5}) \\ & \therefore \sinh ^{-1} 2+\cosh ^{-1} \sqrt{6} \\ & =\log (2+\sqrt{5})+\log (\sqrt{6}+\sqrt{5}) \\ & =\log [(2+\sqrt{5})(\sqrt{6}+\sqrt{5})] \\ & e^{\left(\sinh ^{-1} \sqrt{2}+\cosh ^{-1} \sqrt{6}\right)}=e^{\log ((2+\sqrt{5})(\sqrt{6}+\sqrt{5})]} \\ & =(2+\sqrt{5})(\sqrt{6}+\sqrt{5})=2 \sqrt{6}+2 \sqrt{5}+\sqrt{30}+5 \\ & =5+2 \sqrt{5}+2 \sqrt{6}+\sqrt{30}=5+2 \sqrt{5}+2 \sqrt{6}+\sqrt{5} \sqrt{6} \\ & =5+(2+\sqrt{6}) \sqrt{5}+2 \sqrt{6}=a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c} \\ & \therefore a=5, b=2, c=6 \\ & \therefore \quad a+b+c=5+2+6=13 \end{aligned} $

Given,

$ \begin{aligned} & \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right) \\ & \text { By LHS, } \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right) \\ & =\frac{1}{1+\left(\frac{1+x}{1-x}\right)^2} \frac{d}{d x}\left(\frac{1+x}{1-x}\right) \\ & =\frac{(1-x)^2}{(1-x)^2+(1+x)^2}\left[\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}\right] \\ & =\frac{(1-x)^2}{2\left(1+x^2\right)}\left[\frac{1-x+1+x}{(1-x)^2}\right]=\frac{1}{(1+x)^2} \end{aligned} $

And by RHS, $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$

$\therefore$ Assertion is true.

For reason ( R )

when $x<1$, let $x=\tan \theta$

$ \begin{array}{ll} \Rightarrow & \theta=\tan ^{-1} x \\ \therefore & \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right)=\frac{\pi}{4}+\theta \\ & =\frac{\pi}{4}+\tan ^{-1} x \end{array} $

When $x>1$, then $\tan ^{-1} x>1$

$ \therefore \quad \frac{1+x}{1-x}<0 $

so, $\tan ^{-1}\left(\frac{1+x}{1-x}\right)$ lies in $\left(-\frac{\pi}{2}, 0\right)$

$ \therefore \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{-3 \pi}{4}+\tan ^{-1} x $

$\therefore$ Reason is also true.

We have, $y=\left(\sin ^{-1} x\right)^2$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=2\left(\sin ^{-1} x\right) \frac{1}{\sqrt{1-x^2}} \\ & \Rightarrow \quad \sqrt{1-x^2} \frac{d y}{d x}=2 \sin ^{-1} x \\ & \Rightarrow \quad\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=4\left(\sin ^{-1} x\right)^2 \\ & \Rightarrow \quad\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=4 y \text { (Using Eq. } \end{aligned} $

Again, differentiating w.r.t $x$, we get

$ \begin{aligned} & (-2 x)\left(\frac{d y}{d x}\right)^2+\left(1-x^2\right)\left(2 \frac{d y}{d x}\right) \frac{d^2 y}{d x^2}=4 \frac{d y}{d x} \\ \Rightarrow & (-2 x) \frac{d y}{d x}+2\left(1-x^2\right) \frac{d^2 y}{d x^2}=4 \\ \Rightarrow & \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=2 \end{aligned} $

Given, $f(x)=\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$

The domain of $\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$ is $[-1,1]$

$ \therefore \quad-1 \leq \sqrt{x^2+x+1} \leq 1 $

Since $\sqrt{x^2+x+1}$ is always,

Let $g(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$

The minimum value of $g(x)$ is $3 / 4$, which occurs at $x=-\frac{1}{2}$

$ \begin{array}{ll} \therefore & x^2+x+1 \geq \frac{3}{4} \\ \Rightarrow & \sqrt{x^2+x+1} \geq \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \\ \Rightarrow & \frac{\sqrt{3}}{2} \leq \sqrt{x^2+x+1} \leq 1 \\ \Rightarrow & \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) \leq \sin ^{-1}\left(\sqrt{x^2+x+1}\right) \\ \Rightarrow & \frac{\pi}{3} \leq f(x) \leq \frac{\pi}{2} \quad \leq \sin ^{-1} \quad \end{array} $

Thus, the range of $f(x)=\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$ is $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$

$ \begin{aligned} & \text { } \begin{aligned} \tan & \left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{6}{41}\right) \\ & =\tan ^{-1}\left(\frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5} \cdot \frac{6}{41}}\right)=\tan ^{-1}\left(\frac{\frac{153}{205}}{\frac{187}{205}}\right) \\ & =\tan ^{-1}\left(\frac{153}{187}\right) \\ \text { Now, } & \tan { }^{-1}\left(\frac{153}{187}\right)+\tan ^{-1}\left(\frac{9}{191}\right) \\ & =\tan ^{-1}\left(\frac{\frac{153}{187}+\frac{9}{191}}{1-\frac{153}{187} \cdot \frac{9}{191}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{29223+1683}{35717}}{1-\frac{1377}{35717}}\right) \\ & =\tan ^{-1}\left(\frac{30906}{34340}\right)=\tan ^{-1}\left(\frac{9}{10}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } 2 \tanh ^{-1} x=\sinh ^{-1}\left(\frac{4}{3}\right) \\ & \begin{array}{l} \tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \\ \Rightarrow \quad 2 \tanh ^{-1}(x)=\ln \left(\frac{1+x}{1-x}\right) \\ \text { So, } \ln \left(\frac{1+x}{1-x}\right)=\sinh ^{-1}\left(\frac{4}{3}\right) \\ =\ln \left(\frac{4}{3}+\sqrt{\left(\frac{4}{3}\right)^2+1}\right) \\ {\left[\because \sinh ^{-1}(y)=\ln \left(y+\sqrt{y^2+1}\right)\right]} \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\ln \left(\frac{4}{3}+\sqrt{\frac{16}{9}+1}\right) \\ & =\ln \left(\frac{4}{3}+\frac{5}{3}\right) \Rightarrow \ln \frac{9}{3}=\ln (3) \\ & =\frac{1+x}{1-x}=3 \Rightarrow 1+x=3-3 x \\ & =4 x=2 \Rightarrow x=\frac{1}{2} \end{aligned}\\ &\text { Now, } \cosh ^{-1}\left(\frac{1}{x}\right)=\cosh ^{-1}(4\\ &\begin{aligned} & =\ln (2+\sqrt{4-1}) \\ & \quad\left[\because \cosh ^{-1}(z)=\ln \left(z+\sqrt{z^2-1}\right)\right] \\ & =\ln (2+\sqrt{3}) \end{aligned} \end{aligned} $

Given, $f(x)=\sqrt{\cos ^{-1} \sqrt{1-x^2}}$

Let $x=\sin \theta$.

Then, $\sqrt{1-x^2}=\sqrt{1-\sin ^2 \theta}$

$ \Rightarrow \quad \sqrt{\cos ^2 \theta}=|\cos \theta| $

Since, $\theta \in\left[0, \frac{\pi}{2}\right]$, so $\cos \theta \geq 0$

$ \therefore \quad \sqrt{1-x^2}=\cos \theta $

So, $f(x)=\sqrt{\cos ^{-1}(\cos \theta)}$

$ \begin{aligned} \Rightarrow \quad \sqrt{\theta} & =\sqrt{\sin ^{-1}(x)} \\ f^{\prime}(x) & =\frac{d}{d x} \sqrt{\sin ^{-1}(x)} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad & \frac{1}{\sqrt{1-x^2}} \cdot \frac{1}{2 \sqrt{\sin ^{-1}(x)}} \\ & f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}} \cdot \frac{1}{2 \cdot \sqrt{\sin ^{-1} \frac{1}{2}}} \\ & =\frac{1}{\frac{\sqrt{3}}{2}} \cdot \frac{1}{2 \cdot \sqrt{\frac{\pi}{6}}} \end{aligned} $

$ \Rightarrow \quad \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{6}}{2 \sqrt{\pi}} \Rightarrow \frac{\sqrt{2}}{\sqrt{\pi}}=\sqrt{\frac{2}{\pi}} $

$ \therefore \quad f^{\prime}(x)=\sqrt{\frac{2}{\pi}} $

$ \sin ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} $

$ \Rightarrow \frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{6} $

$ \Rightarrow \cos ^{-1} x+\cos ^{-1} 2 x=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3} $

$ \Rightarrow \cos ^{-1}\left(x \cdot 2 x-\sqrt{1-x^{2}} \sqrt{1-4 x^{2}}\right)=\frac{\pi}{3} $

$ \Rightarrow 2 x^{2}-\sqrt{1-x^{2}} \sqrt{1-4 x^{2}}=\cos \frac{\pi}{3}=\frac{1}{2} $

$ \Rightarrow \sqrt{1-x^{2}} \sqrt{1-4 x^{2}}=\left(2 x^{2}-\frac{1}{2}\right)^{2} $

On solving, we get

$ x=\frac{1}{2} $

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

$ \begin{array}{l} =\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right) \\\\ =\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4} \end{array} $

$ \sec h^{-1} \frac{3}{5}-\tan ^{-1} \frac{3}{5} $

Let $ \operatorname{sech}^{-1} \frac{3}{5}=x $ and $ \tan h^{-1} \frac{3}{5}=y $

$ \begin{array}{l} \sec h x=\frac{3}{5} \text { and } \tan h y=\frac{3}{5} \\\\ x=\ln 3 \quad e^{2 y}=4 \\\\ e y=2 \\\\ y=\ln 2 \end{array} $

Now, $ x-y=\ln 3-\ln 2=\ln \frac{3}{2} $

To find the domain of the function $ f(x) = \sin^{-1}\left(\log_{2}\left(\frac{x^{2}}{2}\right)\right) $, we need to determine where the expression $\log_{2}\left(\frac{x^{2}}{2}\right)$ falls within the domain of the inverse sine function, which is $[-1, 1]$.

First, ensure that:

$ -1 \leq \log_{2}\left(\frac{x^{2}}{2}\right) \leq 1 $

This inequality translates to:

$ 2^{-1} \leq \frac{x^{2}}{2} \leq 2 $

Solving the inequality:

$ \frac{1}{2} \leq \frac{x^{2}}{2} \leq 2 $

Multiply through by 2 to clear the fraction:

$ 1 \leq x^{2} \leq 4 $

This implies that $ x $ must satisfy:

$ x \in [-2, -1] \cup [1, 2] $

We also need to ensure that $\frac{x^{2}}{2} \geq 0$, which is always true since $x^2 \geq 0$.

Therefore, considering the intersection of the feasible solutions:

$ x \in [-2, -1] \cup [1, 2] $

Thus, the domain of the function is:

$ x \in [-2, -1] \cup [1, 2] $

To solve the trigonometric equation $\sin^{-1} x = 2 \sin^{-1} a$, we must consider the following:

Given that $\sin^{-1} x = 2 \sin^{-1} a$, it implies:

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

This corresponds to:

$ 2 \sin^{-1} a \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

From this, we can deduce:

$ \sin^{-1} a \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] $

Consequently, the values for $ a $ must satisfy:

$ a \in \left[-\sin \frac{\pi}{4}, \sin \frac{\pi}{4}\right] $

This simplifies to:

$ a \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] $

Therefore, the condition for $ a $ is:

$ |a| \leq \frac{1}{\sqrt{2}} $

Given,

$ \begin{array}{c} f(x)=\sin ^{-1}\left(x^{2}-1\right)-3 \log _{3}\left(3^{x}-2\right) \\\\ -1 \leq x^{2}-1 \leq 1 \\\\ 0 \leq x^{2} \leq 2 \\\\ -\sqrt{2} < x < \sqrt{2} \end{array} $

$\therefore \sin ^{-1}\left(x^2-1\right)$ is not defined for $(-\infty,-\sqrt{2})$ and $(\sqrt{2}, 0)$

For $3 \log _3\left(3^x-2\right) \quad \Rightarrow \quad 3^x-2>0$

$ \begin{aligned} & 3^x>2 \\\\ & x>\log _3 2 \end{aligned} $

$\therefore 3 \log _3\left(3^x-2\right)$ is not defined for $x \leq \log _3(2)$

So, intervel $[-\sqrt{2}, \sqrt{2}]$ combine with $x>\log _3(2)$. The domain of $f(x)$ is $\log _3(2)

So, $f(x)$ is not defined for

$ \begin{aligned} & \left(-\infty, \log _3 2\right) \cup(\sqrt{2}, 0) \\\\ & a=\log _3 2, b=\sqrt{2} \\\\ & 3^a+b^2=2+2=4 \end{aligned} $

Given, $ \sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6} $

$ \Rightarrow \frac{\pi}{2}-\cos ^{-1} 4 x-\cos ^{-1} 3 x=\frac{\pi}{6} $

$ \Rightarrow \cos ^{-1} 4 x+\cos ^{-1} 3 x=\frac{\pi}{3} $

$ \Rightarrow 4 x \times 3 x-\sqrt{1-16 x^{2}} \sqrt{1-9 x^{2}}=\cos \frac{\pi}{3} $

$ \Rightarrow\left(12 x^{2}-\frac{1}{2}\right)^{2}=\left(1-16 x^{2}\right)\left(1-9 x^{2}\right) $

$ \Rightarrow 144 x^{4}+\frac{1}{4}-12 x^{2}=1-25 x^{2}+144 x^{4} $

$ 13 x^{2}=\frac{3}{4} \Rightarrow x^{2}=\frac{3}{4 \times 13} \Rightarrow x=\frac{\sqrt{3}}{2 \sqrt{13}} $

Given,

$ \sinh ^{-1}(-\sqrt{3})+\cosh ^{-1}(2)=K $

$ \log (-\sqrt{3}+\sqrt{3+1})+\log (2+\sqrt{3})=K $

$ \Rightarrow K=\log [(2-\sqrt{3})(2+\sqrt{3})] $

$ \Rightarrow K=\log 1=0 $

$ \cosh K=\cosh 0=\frac{e^{0}+e^{-0}}{2}=1 $

$ y=\cos ^{-1}\left(\frac{-2 x^{2}+6 x-4}{2 x^{2}-6 x+5}\right) $

$ \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{6 x-2 x^{2}-4}{2 x^{2}-6 x+5}\right)^{2}}} $

$ \begin{array}{l} \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{\frac{\left(2 x^{2}-6 x+5\right)^{2}-\left(-\left(2 x^{2}-6 x+4\right)^{2}\right.}{\left(2 x^{2}-6 x+5\right)}}} \\\\ \frac{\left(12 x^{2}-36 x+30-8 x^{3}+24 x^{2}-20 x\right)}{-\left(24 x^{2}-8 x^{3}-16 x-36 x+12 x^{2}+24\right.} \frac{\left(2 x^{2}-6 x+5\right)^{2}}{} \\\\ \Rightarrow \frac{d y}{d x}=\frac{-1\left(2 x^{2}-6 x+5\right)}{\sqrt{4 x^{2}-12 x+9}} . \\\\ 12 x^{2}-36 x+30-8 x^{3}+24 x^{2}-20 x \\\\ -24 x^{2}+8 x^{3}+16 x+36 x-12 x-24 \\\\ \Rightarrow \frac{d y}{d x}=\frac{-1\left(2 x^{2}-6 x+5\right)^{2}}{\sqrt{4 x^{2}-12 x+9}} \cdot \frac{(-4 x+6}{\left(2 x^{2}-6 x+5\right)^{2}} \\\\ \Rightarrow \frac{d y}{d x}=\frac{2(2 x-3)}{(2 x-3)\left(2 x^{2}-6 x+5\right)} \\\\ \Rightarrow \frac{d y}{d x}=\frac{2}{2 x^{2}-6 x+5} \end{array} $

Given, $ 2 \tan ^{-1} x=3 \sin ^{-1} x $ and $ x \neq 0 $

$ \begin{array}{l} \therefore \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ \Rightarrow \quad \frac{2 x}{1+x^{2}}=x\left(3-4 x^{2}\right) \\\\ \Rightarrow \quad 2=\left(1+x^{2}\right)\left(3-4 x^{2}\right) \quad[\because x \neq 0] \\\\ \Rightarrow \quad 4 x^{4}+x^{2}-1=0 \\\\ \Rightarrow \quad x^{2}=\frac{-1+\sqrt{(1)^{2}-4(4)(-1)}}{2(4)}\left[\because 0 < x^{2}\right] \\\\ \Rightarrow \quad 8 x^{2}=-1+\sqrt{17} \Rightarrow 8 x^{2}+1=\sqrt{17} \end{array} $

(a) $\sinh x$ is an odd function and its range is $R$ and domain is $R$.

(b) $\operatorname{sech} x$ is an even function and its range is ( 0,1 ] and domain is $R$.

(c) $\tanh x$ is an odd function and its range is $(-1,1)$ and domain is $R$.

(d) $\operatorname{cosec}^{-1} x$ is neither even nor odd function.

Hence, A-IV, B-III, C-V, D-II

We have,

$ \begin{aligned} & \cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\tan ^{-1} \frac{16}{63} \\\\ & =\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{16}{63} \\\\ & {\left[\because \cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{4}{3} \text { and } \sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{5}{12}\right]} \\\\ & =\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{16}{63} \\\\ & =\cot ^{-1} \frac{16}{63}+\tan ^{-1} \frac{16}{63}=\frac{\pi}{2} \end{aligned} $

We have,

$ \begin{aligned} & \cosh ^{-1}\left(\frac{5}{3}\right)+\sinh ^{-1}\left(\frac{3}{4}\right)=k \\\\ & \ln \left(\frac{5}{3}+\sqrt{\left(\frac{5}{3}\right)^2-1}\right)+\ln \left[\left(\frac{3}{4}\right)+\sqrt{\left(\frac{3}{4}\right)^2+1}\right]=k \end{aligned} $

$\begin{aligned} & {\left[\begin{array}{r}\because \cos h^{-1} x=\ln \left(x+\sqrt{x^2-1}\right) \\\\ \text { and } \sin h^{-1} x=\ln \left(x+\sqrt{x^2+1}\right)\end{array}\right]} \\\\ & \ln \left(\frac{5}{3}+\frac{4}{3}\right)+\ln \left(\frac{3}{4}+\frac{5}{4}\right)=k \\\\ & \ln 3+\ln 2=k \\\\ & \ln 6=k \\\\ & e^k=6\end{aligned}$

$\left(\sin ^{-1} y\right)^2+\cos ^{-1} x=\frac{n \pi^2}{4}$ and $\left(\cos ^{-1} x\right)\left(\sin ^{-1} y\right)^2=\frac{\pi^4}{16}$

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad a+b^2=\frac{n \pi^2}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } a \cdot b^2=\frac{\pi^4}{16} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \begin{aligned} \therefore a-b & =\sqrt{\left(a+b^2\right)^2-4 a b} \\ & =\sqrt{\frac{n^2 \pi^4}{16}-\frac{4 \pi^4}{16}} \end{aligned} \end{aligned}\\ &\text { Now, least value of } n \text { can be } 2\\ &\begin{aligned} a-b^2 & =\sqrt{\frac{4 \pi^4}{16}-\frac{4 \pi^4}{16}} \\ \Rightarrow \quad a-b^2 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} \end{aligned} $

Adding Eqs. (i) and (iii), we get

$ \begin{aligned} & \quad 2 a=\frac{n \pi^2}{4} \quad \text { or } \quad 2 a=\frac{2 \pi^2}{4}(\because n=2) \\ & \text { or } \quad a=\frac{\pi^2}{4} \Rightarrow \cos ^{-1} x=\frac{\pi^2}{4} \\ & \Rightarrow \quad x=\cos \left(\frac{\pi^2}{4}\right) \end{aligned} $

Now, from Eq. (iii), $a=b$

$ \begin{aligned} & \left(\sin ^{-1} y\right)^2=\frac{\pi^2}{4} \text { or } \sin ^{-1} y= \pm \frac{\pi}{2} \\or\,\,\,\,\,\,\, & y=\sin \left( \pm \frac{\pi}{2}\right) \text { or } y= \pm 1 \end{aligned} $

Therefore, ordered pair is $\left(\cos \frac{\pi^2}{4}, \pm 1\right)$.

Given that,

$ x=\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right) $

As we know,

$ \begin{aligned} \tan ^{-1} x & +\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ \therefore \quad x & =\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{8+5}{40}}{\frac{40-1}{40}}\right) \\ x & =\tan ^{-1} \frac{13}{39}=\tan ^{-1}\left(\frac{1}{3}\right) \end{aligned} $

$ \therefore \tan x=\frac{1}{3} $

And $\sin x=\frac{1}{\sqrt{10}}$ and $\cos x=\frac{3}{\sqrt{10}}$

Putting values of $\sin x, \cos x, \tan x$

$ \frac{\sin x+\cos x}{\tan x}=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{3}}=\frac{\frac{4}{\sqrt{10}}}{\frac{1}{3}}=\frac{12}{\sqrt{10}} $

Given that, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x=\log _e(f(x))$ for $|x|>1$

$ \begin{align*} & \because \quad \tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right) \\ & \left(\because \tanh ^{-1} x=\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right)\right) \\ & \tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \quad \operatorname{cot} \mathrm{h}^{-1} x=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

Adding the both Eqs. (i) and (ii), we get

$ \begin{aligned} \tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth} \mathrm{h}^{-1} x & =\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \\ & +\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) \\ & =\log _e\left(\frac{x+1}{x-1}\right) \\ \tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x & =\log _e(f(x)) \end{aligned} $

On comparing, we get

$ f(x)=\frac{x+1}{x-1} $

At $x=-5, f(-5)=\frac{-5+1}{-5-1}=\frac{-4}{-6}$

$ \Rightarrow f(-5)=2 / 3 $

We have,

$ f(x)=\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right] $

$f(x)$ is defined of

$ \begin{array}{ll} & \log _5\left(x^2+7 x+15\right) \in[-1,1] \\ & -1 \leq \log _5\left(x^2+7 x+15\right) \leq 1 \\ & \frac{1}{5} \leq x^2+7 x+15 \leq 5 \\ \because & x^2+7 x+15 \leq 5 \\ \Rightarrow \quad & x^2+7 x+10 \leq 0 \\ & (x+5)(x+2) \leq 0 \\ \Rightarrow \quad & x \in[-5,-2] \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) & =\tan ^{-1} \alpha \\ \sum_{n=1}^k \tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right) & =\tan ^{-1} \alpha \\ \Rightarrow \sum_{n=1}^k\left(\tan ^{-1}(n+2)-\tan ^{-1}(n+1)\right) & =\tan ^{-1} \alpha \\ \Rightarrow\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4\right. & \left.-\tan ^{-1} 2\right) \\ \ldots . . \tan ^{-1}(k+2)-\tan ^{-1}(k+1) & =\tan ^{-1} \alpha \\ \tan ^{-1}(k+2)-\tan ^{-1} 2 & =\tan ^{-1} \alpha \end{aligned} \end{aligned} $

$ \begin{aligned} & \tan ^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right)=\tan ^{-1} \alpha \\ \Rightarrow & \frac{k}{1+2 k+4}=\alpha \Rightarrow \alpha=\frac{k}{2 k+5} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right) =\tan ^{-1}\left(\frac{2}{3}\right) \\ & \tan ^{-1} \frac{\left(\frac{x}{x-2}-\frac{x}{2 x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2 x-1}\right)}=\tan ^{-1} \frac{2}{3} \\ \Rightarrow \quad & \frac{2 x^2-x-x^2+2 x}{2 x^2-4 x-x+2+x^2}=\frac{2}{3} \\ \Rightarrow \quad & \frac{x^2+x}{3 x^2-5 x+2}=\frac{2}{3} \\ \Rightarrow \quad & \quad 3 x^2+3 x=6 x^2-10 x+4 \\ \Rightarrow \quad & \quad 3 x^2-13 x+4=0 \\ \Rightarrow \quad & \quad(3 x-1)(x-4)=0 \\ \Rightarrow \quad & x=\frac{1}{3}, 4, \quad x \in\left\{\frac{1}{3}, 4\right\} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2} \\ & \Rightarrow \\ & \sin ^{-1}\left(\frac{12}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \sin ^{-1}\left(\frac{12}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \cos ^{-1}\left(\frac{\sqrt{x^2-144}}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right) \\ & \Rightarrow \quad \frac{\sqrt{x^2-144}}{x}=\frac{5}{x} \\ & \Rightarrow \quad \sqrt{x^2-144}=5 \Rightarrow x^2-144=25 \\ & \Rightarrow \quad x^2=144+25 \\ & \Rightarrow \quad x^2=169 \\ & \Rightarrow \quad x=13 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \begin{array}{c} \operatorname{cosec}^{-1}\left[\begin{array}{l} \left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}\right. \left.\left.\quad+\cos \frac{\alpha}{2}\right) \cot 5 \alpha \sec \frac{11 \alpha}{2}\right] \end{array}\right. \\ =\operatorname{cosec}^{-1}\left[\left(-\cos 2\left(\frac{\alpha-\pi}{4}\right)\right.\right. \left.\left.\quad+\cos \frac{\alpha}{2} \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right] \end{array} \\ & =\operatorname{cosec}{ }^{-1}\left[\left(-\sin \frac{\alpha}{2}\right.\right. \left.\left.\quad+\cos \frac{\alpha}{2} \frac{\cos 5 \alpha}{\sin 5 \alpha}\right) \sec \frac{11 \alpha}{2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\operatorname{cosec}^{-1}\left[\left(\frac{-\sin 5 \alpha \sin \frac{\alpha}{2}+\cos 5 \alpha \cos \frac{\alpha}{2}}{\sin 5 \alpha}\right)\right. \left.\sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}\left[\frac{\cos \left(5 \alpha+\frac{\alpha}{2}\right)}{\sin 5 \alpha} \cdot \sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}\left[\operatorname{cosec} 5 \alpha \cdot \cos \frac{11 \alpha}{2} \sec \frac{11 \alpha}{2}\right] \\ & =\operatorname{cosec}^{-1}(\operatorname{cosec} 5 \alpha)=5 \alpha \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8} \\ \Rightarrow & \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & \tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \times \frac{1}{8}}\right]+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & \tan ^{-1} \frac{13}{39}+\frac{1}{2} \sec ^{-1} x=\frac{\pi}{8} \\ \Rightarrow & 2 \tan ^{-1} \frac{13}{39}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & 2 \tan ^{-1} \frac{1}{3}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{3}{4}+\sec ^{-1} x=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1} \frac{3}{4}+\tan ^{-1} \sqrt{x^2-1}=\frac{\pi}{4} \\ \Rightarrow & \tan ^{-1}\left[\frac{\frac{3}{4}+\sqrt{x^2-1}}{1-\frac{3}{4} \sqrt{x^2-1}}\right]=\frac{\pi}{4} \end{aligned} \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \frac{3+4 \sqrt{x^2-1}}{4-3 \sqrt{x^2-1}}=1 \\ \Rightarrow & 3+4 \sqrt{x^2-1}=4-3 \sqrt{x^2-1} \\ \Rightarrow & 7 \sqrt{x^2-1}=1 \\ \Rightarrow & \sqrt{x^2-1}=\frac{1}{7} \\ \Rightarrow & x^2-1=\frac{1}{49} \\ \Rightarrow & x^2=\frac{50}{49} \end{array} $

Reason (R) $e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2-2 p-x=0$

We have, $x p^2-2 p-x=0$

$ \begin{aligned} \Rightarrow \quad p & =\frac{2 \pm \sqrt{4+4 x^2}}{2 x}=\frac{1 \pm \sqrt{1+x^2}}{x} \\ & =\frac{1+\sqrt{1+x^2}}{x}, \frac{1-\sqrt{1+x^2}}{x} \\ & =e^{\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)}, e^{\ln \left(\frac{1-\sqrt{1+x^2}}{x}\right)} \end{aligned} $

Since, $\operatorname{cosech}^{-1} x=\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)$

$\therefore e^{\operatorname{cosech}^{-1} x}$ is a solution of given equation.

$ \text { } \begin{aligned} Now,\,\, \operatorname{cosech}^{-1}(3) & =\ln \left(\frac{1+\sqrt{1+3^2}}{3}\right) \\ & =\ln \left(\frac{1+\sqrt{10}}{3}\right) \end{aligned} $