Question

Consider two functions $f$ and $g$ defined by $f\left(x\right)={\sin }^{-1}x+{\tan }^{-1}x$ and $g\left(x\right)={\cos }^{-1}x+{\cot }^{-1}x.$ Let $A$ and $B$ be sets of non-negative integers in $R_f$ and $R_g$ respectively, where $R_f$ is the range of $f$ and $R_g$ is the range of $g.$ Then

• A. number of real solutions of $f\left(x\right)=g\left(x\right)$ is one.
• B. if $R_f$ is $[a\pi ,b\pi ]$ and $R_g$ is $[c\pi ,d\pi ],$ then $\frac{c+d}{b-a}=\frac{4}{3}.$
• C. number of one-one functions from $A$ to $B$ is $60.$
• D. number of values of $x$ satisfying $|f\left(x\right)|+|g\left(x\right)|=\pi$ is infinitely many.

Answer 1

Answer: (A), (B), (C), (D)

number of values of $x$ satisfying $|f\left(x\right)|+|g\left(x\right)|=\pi$ is infinitely many.

$f\left(x\right)={\sin }^{-1}x+{\tan }^{-1}x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}+\frac{1}{1+x^2}>0$
$f\left(x\right)$ is a strictly increasing function.

$g\left(x\right)={\cos }^{-1}x+{\cot }^{-1}x$
$\Rightarrow g^{\prime }\left(x\right)=\frac{-1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}<0$
$g\left(x\right)$ is a strictly decreasing function.

Also, domain of $f$ is $[-1,1]$
$\therefore$ Range of $f,$ $R_f=[-\frac{3\pi }{4},\frac{3\pi }{4}]$
Domain of $g$ is $[-1,1]$
$\therefore$ Range of $g,$ $R_g=[\frac{\pi }{4},\frac{7\pi }{4}]$

Now, ​$A=\{0,1,2\}$ and $B=\{1,2,3,4,5\}$

As $f$ is strictly increasing and $g$ is strictly decreasing for $x\in [-1,1],$ so $f\left(x\right)=g\left(x\right)$ has only one solution.

$\frac{c+d}{b-a}=\frac{\frac{1}{4}+\frac{7}{4}}{\frac{3}{4}+\frac{3}{4}}=\frac{8}{6}=\frac{4}{3}$​​​

Number of one-one functions from $A$ to $B$ is ${}^5{C}_3\cdot 3!=60$

$|f\left(x\right)|+|g\left(x\right)|=\pi$
As $g\left(x\right)$ is always positive,
$\Rightarrow |f\left(x\right)|+g\left(x\right)=\pi$
$\Rightarrow |f\left(x\right)|=\pi -{\cos }^{-1}x-{\cot }^{-1}x\Rightarrow |f\left(x\right)|={\sin }^{-1}x+{\tan }^{-1}x\Rightarrow |f\left(x\right)|=f\left(x\right)$
Clearly, above equation is true for all $x\in [0,1]$