Let f(x) = |x|^a where a is a non-zero real number. For what values of a is f(x) differentiable at x = 0? A. For all non-zero a. B. For all a > 1. C. For no values of a. D. For all a different from 1 and 0. Please answer fast..

							
if a is negative, then f(x) = |x|^a is not continuous at x=0 (since it is not defined at x=0). hence, it cant be differentiable at x=0 for negative a.
now, if a is positive, then
f’(0)= Lt x tends to 0 [f(x) – f(0)]/(x – 0)
also, f(0)= |0|^a= 0
so, for f to be differentiable at x = 0, the following must exist finitely:
Lt x tends to 0 f(x)/x= Lt x tends to 0 |x|^a/x.
now, RHL= Lt x tends to 0+ x^a/x = Lt x tends to 0+ x^(a-1), clearly this lim is infinity when a – 1 is less than 0 or a is less than 1. so a must be greater than or equal to 1.
when a=1, y=|x| is clearly not diff at x=0 as can be seen by graph or by noting that RHL= 1 while LHL= – 1
For a greater than 1, RHL= Lt x tends to 0+ x^(a-1)= 0.
for LHL, we ll use the fact that |x|= – x when x is less than 0.
so LHL= Lt x tends to 0- |x|^a/x= Lt x tends to 0- |x|^a/( – |x|)
= – Lt x tends to 0- |x|^(a-1)= – 0= 0
hence, we observe that f’(0) exists for all a greater than 1 since RHL= LHL.
option B is correct.
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