Let $f: R \rightarrow R$ be a continuous function such that $f\left(x^2\right)=f\left(x^3\right)$ for all $x \in R$. Consider the following statements.

$I.$ $f$ is an odd function.

$II.$ $f$ is an even function.

$III$. $f$ is differentiable everywhere. Then,

  • [KVPY 2019]
  • A

    $I$ is true and $III$ is false

  • B

    $II$ is true and $III$ is false

  • C

    Both $I$ and $III$ are true

  • D

    Both $II$ and $III$ are true

Similar Questions

Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.

(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )

Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$

 $LIST I$ $LIST II$
$P$ The range of $f$ is $1$ $\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$
$Q$ The range of $g$ contains $2$ $(0,1)$
$R$ The domain of $f$ contains $3$ $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$S$ The domain of $g$ is $4$ $(-\infty, 0) \cup(0, \infty)$
  $5$ $\left(-\infty, \frac{ e }{ e -1}\right]$
  $6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$

The correct option is:

  • [IIT 2018]

The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is

  • [JEE MAIN 2023]

Let $S=\{1,2,3,4\}$. Then the number of elements in the set $\{f: S \times S \rightarrow S: f$ is onto and $f(a, b)=f(b, a)$ $\geq a; \forall(a, b) \in S \times S\}$ is

  • [JEE MAIN 2022]

Consider a function $f:\left[ { - 1,1} \right] \to R$ where $f(x) = {\alpha _1}{\sin ^{ - 1}}x + {\alpha _3}\left( {{{\sin }^{ - 1}}{x^3}} \right) + ..... + {\alpha _{(2n + 1)}}{({\sin ^{ - 1}}x)^{(2n + 1)}} - {\cot ^{ - 1}}x$ Where $\alpha _i\ 's$ are positive constants and $n \in N < 100$ , then $f(x)$ is

Range of the function

$f(x) = \sqrt {\left| {{{\sin }^{ - 1}}\left| {\sin x} \right|} \right| - {{\cos }^{ - 1}}\left| {\cos x} \right|} $ is