Question

Consider the function $f\left(x\right)=\sin \left({\sin }^{-1}2x\right)+\sec \left({\sec }^{-1}3x\right)+\tan \left({\tan }^{-1}4x\right)$ and $g\left(x\right)=9x$ , then which of the following is correct?

• A. $f\left(x\right)$ and $g\left(x\right)$ are equal functions
• B. $f\left(x\right)$ is an odd function
• C. Number of integers in range of $f\left(x\right)$ are $4$
• D. Number of integers in domain of $f\left(x\right)$ are $2.$

Answer 1

Answer: (B), (C)

Number of integers in range of $f\left(x\right)$ are $4$

For domain :
$\sin \left({\sin }^{-1}2x\right)=2x$ $\Rightarrow -\frac{1}{2}\le x\le \frac{1}{2}\cdots \left(1\right)$ and
$\sec \left({\sec }^{-1}3x\right)=3x$ $\Rightarrow x\le -\frac{1}{3}\text{or}x\ge \frac{1}{3}\cdots \left(2\right)$
and $\tan \left({\tan }^{-1}4x\right)=4x\Rightarrow x\in R\cdots \left(3\right)$
$\therefore$Domain of $f\left(x\right)$ : $\left(1\right)\cap \left(2\right)\cap \left(3\right)$
$\Rightarrow x\in [-\frac{1}{2},-\frac{1}{3}]\cup [\frac{1}{3},\frac{1}{2}]$
In the given domain, we have $f\left(x\right)=2x+3x+4x=9x$

Domain of $f\left(x\right)$ and $g\left(x\right)$ is different, thus they are not equal functions.
$f(-x)=9(-x)=-9x=-f\left(x\right)$
Clearly, $f\left(x\right)$ is an odd function.

Range of $f\left(x\right)=[-\frac{9}{2},-3]\cup [3,\frac{9}{2}]$
Integers $:-4,-3,3,4$
$\Rightarrow$ Number of integers $=4$