JEE Main 2019MathematicsDifferentiationMediumMCQ

JEE Main 2019 — Differentiation Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

Let, f:R→R be a function such that fx=x3+x2f'1+xf''2+f'''3, ∀x∈R. Then f2 equals

Choose an option

Show full solutionCorrect option: D
Correct answer
D-2

Step-by-step explanation

Let, fx=x3+ax2+bx+c

⇒f'x=3x2+2ax+b

⇒f''x=6x+2a

⇒f'''x=6

According to the question, 

a=f'1=3+2a+b⇒a+b=-3 ...i

b=f''2=12+2a⇒2a-b=-12 ...ii

c=f'''3⇒c=6

Solving equations i & ii, we get, a= -5 & b=2

⇒fx=x3-5x2+2x+6

∴ f2=8-20+4+6=-2.

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About this question

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