Let f: R rightarrow R be a continuous functio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x^2) = f(x^3)$ for all $x \in R$.For any $x > 0$,let $x = t^6$. Then $f(t^{12}) = f(t^{18})$.More generally,for any $x > 0$,we can write

Vedclass · Accepted Answer

Both $II$ and $III$ are true

Vedclass · Answer

(D) Given $f(x^2) = f(x^3)$ for all $x \in R$.For any $x > 0$,let $x = t^6$. Then $f(t^{12}) = f(t^{18})$.More generally,for any $x > 0$,we can write $x = t^{6^n}$.By repeating the substitution,$f(x) = f(x^{2/3}) = f(x^{(2/3)^2}) = \dots = f(x^{(2/3)^n})$.As $n 	o \infty$,$(2/3)^n 	o 0$,so $f(x) = f(x^0) = f(1)$ for all $x > 0$.Since $f$ is continuous at $x = 0$,$f(0) = \lim_{x 	o 0^+} f(x) = f(1)$.For $x  0$. Then $f((-t)^2) = f((-t)^3) \implies f(t^2) = f(-t^3)$.Since $f(t^2) = f(1)$,we have $f(-t^3) = f(1)$.As $t^3$ covers all positive values,$f(x) = f(1)$ for all $x Thus,$f(x) = c$ (a constant) for all $x \in R$.$A$ constant function is an even function because $f(-x) = c = f(x)$.$A$ constant function is also differentiable everywhere with $f'(x) = 0$.Therefore,both $II$ and $III$ are true.

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ 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,

Similar Questions

Show that the function $f$ given by $f(x) = \begin{cases} x^3 + 3, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases}$ is not continuous at $x = 0$.

Let $f: R \rightarrow R$ be the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$. Then $f$ is:

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is:

Let $f(x) = [2x^2 + 1]$ and $g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}$,where $[t]$ denotes the greatest integer function $\leq t$. Then,in the open interval $(-1, 1)$,the number of points where $f(g(x))$ is discontinuous is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ 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,