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Question Id : 11667 | Context :NIMCET 2024

Question

Let f:RR be a function such that f(0)=1π and f(x)=xeπx1 for x0, then
🎥 Video solution / Text Solution of this question is given below:

Analysis of Continuity and Differentiability

Function:

f(x)={xeπx1,x01π,x=0

✅ Continuity at x=0:

lim

✏️ Differentiability at x = 0 :

f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2}

✅ Final Result:

  • Function is continuous at x = 0
  • Function is differentiable at x = 0
  • f'(0) = \boxed{-\frac{1}{2}}

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