JEE Main 2023MathematicsApplication of DerivativesMediumMCQ

JEE Main 2023 — Application of Derivatives Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

The absolute minimum value, of the function fx=x2-x+1+x2-x+1, where t denotes the greatest integer function, in the interval -1,2, is

Choose an option

Show full solutionCorrect option: D
Correct answer
D34

Step-by-step explanation

fx=x2-x+1+x2-x+1  x∈-1,2

∵x2-x+1→x2-x+1>0

∴fx=x2-x+1+x2-x+1

Now,

Consider g(x)=x2-x+1

For the minimum value of g(x),

g'(x)=0⇒2x-1=0

⇒x=12

x2-x+1 attains its minimum value at x=12

And minx2-x+1=0 as x2-x+1>0

⇒fx attains its minimum at x=12

So, f12=34+0=34

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About this question

This is a previous-year question from JEE Main 2023, covering the Application of Derivatives 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.