JEE Advanced 2019 Paper 1 · Mathematics · Sets, Relations and Functions · medium

Let f:mathbb R→mathbb R be differentiable with f(x+y)=f(x)+f(y)+2xy for all real x,y, and f'(0)=-2. The area of the region bounded by the curve…

Single correct option

Let

f:\mathbb R\to\mathbb R

be differentiable with

f(x+y)=f(x)+f(y)+2xy

for all real

x,y

, and

f'(0)=-2

. The area of the region bounded by the curve

y=f(x)

and the line

y=x

Show answer and solution

Answer: (A) $\dfrac92$

Putting

x=y=0

gives

f(0)=0

. Then

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\left[\frac{f(h)}h+2x\right]=f'(0)+2x=2x-2

. Integrating with

f(0)=0

gives

f(x)=x^2-2x

. The curve meets the line

y=x

where

x^2-2x=x

, i.e. at

x=0

and

x=3

, and between them the line lies above the parabola. The area is

\int_0^3\left[x-(x^2-2x)\right]dx=\int_0^3(3x-x^2)\,dx=\frac{27}2-9=\frac92

. The value

\frac43

is the area between the parabola and the

x

-axis, and

9

is just

\int_0^3x^2\,dx

, the parabola term alone.

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