LimitsHard
Question
If f(x) = a0 + 
, where ai ′s are real constants, then f(x) is
, where ai ′s are real constants, then f(x) is Options
A.continuous at x = 0 for all ai
B.differentiable at x = 0 for all ai ∈ R
C.differentiable at x = 0 for all a2k - 1 = 0
D.none of these
Solution
f(x) = 
= a0 + a1 |x| + a2|x|2 + a3|x|3 + ............+ an|x|n
f(0) = a0 we know that
|x| = 0
f(x) = a0
f(x) is continous for x = 0
f(x) is not differentiable at x = 0, due to presence of |x|
If all a2k+1 = 0, f(x) does not contains |x|
⇒ f(x) is differentiable at x = 0
= a0 + a1 |x| + a2|x|2 + a3|x|3 + ............+ an|x|n
f(0) = a0 we know that
|x| = 0f(x) = a0f(x) is continous for x = 0
f(x) is not differentiable at x = 0, due to presence of |x|
If all a2k+1 = 0, f(x) does not contains |x|
⇒ f(x) is differentiable at x = 0
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