Question

Page No
Date :
If f(x)= x ^Alpha.logx & f(0)=0 then the value
of alpha
for which Rolle's Theorem can be
applied in [0,1]..​

Answer 1

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Answer 2

Given:

$f(x) = \left \{ {{x^{\alpha} logx \ \ \ x>0 } \atop {0}\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \ x=0} \right.$

[0 , 1]

To find:

Find the value of the α.

Solution:

If f(x) is continuous in [a , b] and differentiable in (a , b) there exists a 'c' such that

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Since Rolle's theorem is applicable

Checking continuity at x = 0

$\lim_{x \to 0} f(x) = f(0) = 0\\\\ \lim_{x \to 0} x^{\alpha} logx = (x^{- \alpha} )^{-1}logx = \frac{logx}{x^{- \alpha} } = \frac{ \infty }{\infty} \\\\L ' Hospital \ Rule\\\\ \lim_{x \to 0} = \frac{1/x}{(- \alpha)x^{- \alpha -1} } = \frac{1}{-\alpha}.\frac{1}{x}.\frac{1}{x^{-\alpha -1} } = \frac{1}{-\alpha} .\frac{1}{x^{-\alpha} } = \frac{x^{\alpha} }{-\alpha}$

For the limit to exist α > 0

For α > 0

$\lim_{x \to x^{+} } f(x) = 0$

So, the answer is 0.

Value of alpha is 0.