JEE Main 2023MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2023 — Continuity and Differentiability Question with Solution

JEE Main 2023 (11 Apr Shift 1)

Question

Let fx=x2-x+-x+x, where x∈ℝ and t denotes the greatest integer less than or equal to t. Then, f is

Choose an option

Show full solutionCorrect option: B
Correct answer
Bcontinuous at x=1, but not continuous at x=0

Step-by-step explanation

Given,

fx=x2-x+-x+x

⇒fx=x2-x+x-x,  as -A=A

⇒fx=x2-x+x, as x+x=x 

⇒fx=x2-x+x  (∵x≥0)

Now at x=0, f0=0

And f0+=-1, as x2-x<0 for x→0+

So, function is discontinuous at x=0

Now at x=1,  f1=0, 

f1+=0+0=0 and f1-=-1+1=0

Hence, the function is continuous at x=1.

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About this question

This is a previous-year question from JEE Main 2023, covering the Continuity and Differentiability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.