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Maths Functions, Limits, Continuity and Differentiability NTA Abhiyas Question - Continuity and Differentiability Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: August 13, 2026

Consider the function f(x) = (x 3 - x) |x 2 - 6x + 5|, x R, then f(x) is

discontinuous at x = 1

discontinuous at x = 5

non differentiable at x = 1

non differentiable at x = 5

Text Solution

Verified by Experts

The correct answer is:

non differentiable at x = 5

f(x) = (x 3 -x)|(x-l)(x-5)|

= x(x + l)(x-l)|(x-l)(x-5)|

(x 3 - x), | x 2 - 6x + 5| both are continuous

f(x) is also continuous

Now,

Now, check differentiability At x = l

f' (1 - ) = (1 3 - 1) (2 - 6) + (1 - 6 + 5) (3 - 1) = 0

f' (1 + ) = - (1 - 1) (2 - 6) - (1 - 6 + 5) (3 - 1) = 0

f' (1 - ) = f(1 + ) at x = 1 differentiable

At x = 5

f'(5 - ) = (125-5)(10-6) + 0

f'(5 + ) = -(125-5)(10-6)-0

Clearly, f'(5 - ) f '(5 + )

non-differentiable at x = 5

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