JEE Advanced 2018 Paper 1 · Mathematics · Indefinite and Definite Integrals · hard

A continuous function f:[0,1]→mathbb R satisfies f(x)=x+∫0^1 (xt+t^2f(t) dt for all x∈[0,1].

One or more options correct

A continuous function

f:[0,1]\to\mathbb R

satisfies

f(x)=x+\displaystyle\int_0^1 (xt+t^2)f(t)\,dt

for all

x\in[0,1]

. Which of the following is (are) correct?

Show answer and solution

Answer: A

The integral is linear in

x

, so

f(x)=x+ax+b

with

a=\int_0^1 tf(t)\,dt

and

b=\int_0^1 t^2f(t)\,dt

. Substituting

f(t)=(1+a)t+b

a=\frac{1+a}{3}+\frac b2

and

b=\frac{1+a}{4}+\frac b3

. The second equation gives

\frac{2b}{3}=\frac{1+a}{4}

, so

b=\frac{3(1+a)}{8}

; the first then gives

a=\frac{1+a}{3}+\frac{3(1+a)}{16}=\frac{25(1+a)}{48}

, i.e.

23a=25

, so

a=\frac{25}{23}

and

b=\frac{3}{8}\cdot\frac{48}{23}=\frac{18}{23}

. Both constants reproduce the original integrals, which was also confirmed symbolically. Hence

f(x)=\frac{48x+18}{23}

. Then

f(1)=\frac{66}{23}

(option A true),

f(0)=\frac{18}{23}\neq1

(C false),

\int_0^1f=\frac{24+18}{23}=\frac{42}{23}\neq2

(B false) and

f'(x)=\frac{48}{23}\neq2

(D false).

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