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

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(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

$0.5$

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(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 = $

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