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,
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
The domain of the definition of the function $f\left( x \right) = \frac{1}{{4 - {x^2}}} + \log \,\left( {{x^3} - x} \right)$ is
The domain of the definition of the function $f\left( x \right) = \frac{1}{{4 - {x^2}}} + \log \,\left( {{x^3} - x} \right)$ is
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Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{a+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{\prime}(6)+f(7)$ is equal to
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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:
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 domain of definition of the function $f (x) = {\log _{\left[ {x + \frac{1}{x}} \right]}}|{x^2} - x - 6|+ ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$ is
Where $[x]$ denotes greatest integer function.
The domain of definition of the function $f (x) = {\log _{\left[ {x + \frac{1}{x}} \right]}}|{x^2} - x - 6|+ ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$ is
Where $[x]$ denotes greatest integer function.
Let $f : R -\{0,1\} \rightarrow R$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :
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- [JEE MAIN 2023]