JEE Main 2019MathematicsApplication of DerivativesHardMCQ

JEE Main 2019 — Application of Derivatives Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Let f:0, 2→R be a twice differentiable function such that f''x>0, for all  x∈0, 2. If ϕx= fx+ f2–x, then ϕ is

Choose an option

Show full solutionCorrect option: D
Correct answer
Ddecreasing on 0,1  and increasing on (1,2)

Step-by-step explanation

fx:0,2→R and f''x> 0  for x∈ [0,2]
    ⇒f'(x) is increasing for x∈ [0,2]
Now, ϕ(x)=f(x)+f(2-x)
    ⇒ϕ'x=f'x-f'(2-x)
For  x∈ [0,1) , x < 2 – x  

    ⇒f'(x)<f'(2-x)⇒ϕ'(x)<0
For x∈ (1, 2] , x > 2 – x

    ⇒f'x>f'2-x⇒ϕ'x>0
Hence, ϕ is decreasing on 0,1 and increasing on 1,2.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Application of Derivatives chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, 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.