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Maths Functions, Limits, Continuity and Differentiability JEE Main 2023 - ( Continuity and Differentiability ) Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: August 13, 2026

Let f(x)= [x 2 – x] +|- x +[x]|, where x R and [t] denotes the greatest integer less than or equal to t. Then, f is

continuous at x = 0, but not continuous at x = 1

continuous at x = 1, but not continuous at x = 0

continuous at x = 0 and x = 1

not continuous at x = 0 and x = 1

Text Solution

Verified by Experts

The correct answer is:

Given,

f(x)=[x 2 -x]+ |-x+[x]|

f(x)=[x 2 -x]+|x-[x]|,{as|-A| = |A|}

f(x)=[x 2 - x]+|{x}|, {as [x]+{x}= x}

f(x)= [x 2 -x] +{x} ( {x} 0)

Now at x = 0, f(0)= 0

And f(0 + )= -1, {as x 2 - x < 0 for x 0 + }

So, function is discontinuous at x = 0

Now at x = 1, f(a)=0,

f(1 + )= 0 + 0 = 0 and f(l - )= -1 + 1 = 0

Hence, the function is continuous at x = 1.

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