Let f(x) = begin{cases} x^alpha ln x, & x 0 | vedclass

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

Question

(D) For Rolle's theorem to be applicable to $f(x)$ on $[0, 1]$,the following conditions must be satisfied:$1$. $f(x)$ must be continuous on $[0, 1]$.$

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) For Rolle's theorem to be applicable to $f(x)$ on $[0, 1]$,the following conditions must be satisfied:$1$. $f(x)$ must be continuous on $[0, 1]$.$2$. $f(x)$ must be differentiable on $(0, 1)$.$3$. $f(0) = f(1)$.Checking condition $3$: $f(1) = 1^\alpha \ln(1) = 0$ and $f(0) = 0$. Thus,$f(0) = f(1) = 0$ holds for any $\alpha$.Checking condition $1$ (Continuity at $x=0$): We require $\lim_{x 	o 0^+} x^\alpha \ln x = 0$. This limit exists and equals $0$ if and only if $\alpha > 0$.Checking condition $2$ (Differentiability on $(0, 1)$): $f'(x) = \alpha x^{\alpha-1} \ln x + x^{\alpha-1} = x^{\alpha-1}(\alpha \ln x + 1)$. This is defined for all $x \in (0, 1)$ for any $\alpha$.Among the given options,only $\alpha = 0.5$ satisfies the condition $\alpha > 0$.

Let $f(x) = \begin{cases} x^\alpha \ln x, & x > 0 \\ 0, & x = 0 \end{cases}$. Rolle's theorem is applicable to $f$ for $x \in [0, 1]$ if $\alpha = $

Similar Questions

The value $c$ of Lagrange's mean value theorem for $f(x)=e^{x}+24$ in $[0,1]$ is

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

If $f(x)$ is a twice differentiable polynomial function such that $f(1) = 1, f(2) = 4, f(3) = 9$,then:

Let $a > 0$ and $f$ be continuous in $[-a, a]$. Suppose that $f'(x)$ exists and $f'(x) \le 1$ for all $x \in (-a, a)$. If $f(a) = a$ and $f(-a) = -a$,then $f(0)$ is:

Let $f(x)$ and $g(x)$ be two differentiable functions in $R$ such that $f(2) = 8, g(2) = 0, f(4) = 10$,and $g(4) = 8$. Then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module