Skip to content
dronacharya.ai
← JEE Advanced 2015 · Paper 2
MathematicsMultiple correct+4 / -2Question 60 of 60

JEE Advanced 2015 · Paper 2, Mathematics question 60

Passage

PARAGRAPH 2

Let F:R→RF:\mathbb R\to\mathbb R be a thrice differentiable function. Suppose that F(1)=0F(1)=0, F(3)=−4F(3)=-4 and F′(x)<0F'(x)<0 for all x∈(1/2,3)x\in(1/2,3). Let f(x)=xF(x)f(x)=xF(x) for all x∈Rx\in\mathbb R.

If ∫13x2F′(x) dx=−12\displaystyle\int_1^3x^2F'(x)\,dx=-12 and ∫13x3F′′(x) dx=40\displaystyle\int_1^3x^3F''(x)\,dx=40, then the correct expression(s) is(are)

Choose one or more options
Answer and worked solution

Official answer: C, D

(D). Integrate by parts:
∫13xF dx=[x22F]13−12∫13x2F′ dx=−18+6=−12.\int_1^3xF\,dx=\Big[\tfrac{x^2}2F\Big]_1^3-\tfrac12\int_1^3x^2F'\,dx=-18+6=-12.
So ∫13f=−12\int_1^3f=-12: (D) is true and (B) is false.

Relation for F′F'. Integrate by parts again:
∫13x3F′′ dx=[x3F′]13−3∫13x2F′ dx=27F′(3)−F′(1)+36=40,\int_1^3x^3F''\,dx=\Big[x^3F'\Big]_1^3-3\int_1^3x^2F'\,dx=27F'(3)-F'(1)+36=40,
so 27F′(3)−F′(1)=427F'(3)-F'(1)=4.

(C). f′(3)=F(3)+3F′(3)=−4+3F′(3)f'(3)=F(3)+3F'(3)=-4+3F'(3) and f′(1)=F′(1)f'(1)=F'(1). Then
9f′(3)−f′(1)=−36+4=−32.9f'(3)-f'(1)=-36+4=-32.
So 9f′(3)−f′(1)+32=09f'(3)-f'(1)+32=0: (C) is true and (A) is false.

(C), (D)

Topics:Integration by partsProperties of definite integrals
← Previous question