If f(x) = xe^-≤ft( 1/|x| + 1/x ), & x ≠ 0 \\ 0, & x = 0 then f(x) is: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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If f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0 then f (x) is:

Solution & Step-by-step Explanation

Check for x = 0 :LHL: x \to 0^{-}, ∣ x ∣ = - x . Power = - (- 1/ x + 1/ x) = 0 . f (x) \to 0 \cdot e^{0} = 0 .RHL: x \to 0^{+}, ∣ x ∣ = x . Power = - (1/ x + 1/ x) = - 2/ x . f (x) = x e^{- 2/ x} \to 0 \cdot 0 = 0 .So f (x) is continuous at x = 0 .Differentiability:LHD: lim_{h \to 0} \frac{f ( 0 - h ) - f ( 0 )}{- h} = \frac{- h e ^{0}}{- h} = 1 .RHD: lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h} = \frac{h e ^{- 2/ h}}{h} = e^{- \infty} = 0 .LHD \neq = RHD.Thus, continuous but not differentiable at x = 0 .

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If f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0 then f (x) is:

continuous as well as differentiable for all x

continuous for all x but not differentiable at x = 0

neither differentiable nor continuous at x = 0

discontinuous everywhere

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