JEE Main 2022 — Definite Integration Question with Solution

Given,

$$f(x)=x+{\int }_0^1(x-t)f(t)dt$$

$$=x+x{\int }_0^{1}f(t)dt-{\int }_0^{1}tf(t)dt$$

$$=x\left(1+{\int }_0^{1}f(t)dt\right)-{\int }_0^{1}tf(t)dt$$

Now,

let $$A=1+{\int }_0^{1}f(t)dt$$

and $$B={\int }_0^{1}tf(t)dt$$

$$\therefore$$ $$f(x)=Ax-B$$

$$\Rightarrow f(t)=At-B$$

So, $$A=1+{\int }_0^{1}f(t)dt$$

$$=1+{\int }_0^1(At-B)dt$$

$$=1+{\left[\frac{A{t}^2}{2}-Bt\right]}_0^1$$

$$=1+\frac{A}{2}-B$$

$$\Rightarrow \frac{A}{2}=1-B$$ ...... (1)

$$B={\int }_0^{1}tf(t)dt$$

$$={\int }_0^{1}t(At-B)dt$$

$$={\int }_0^1(A{t}^2-Bt)dt$$

$$={\left[\frac{A{t}^3}{3}-\frac{B{t}^2}{2}\right]}_0^1$$

$$=\frac{A}{3}-\frac{B}{2}$$

$$\Rightarrow \frac{3B}{2}=\frac{A}{3}$$ ....... (2)

Solving equation (1) and (2) we get,

$$A=\frac{18}{13}$$ and $$B=\frac{4}{13}$$

$$\therefore$$ $$f(x)=\frac{18}{13}x-\frac{4}{13}$$

By checking all the options you can see when x = 6 we get

$$y=f(x)=\frac{18}{13}\times 6-\frac{4}{13}$$

$$=\frac{108-4}{13}=8$$

$$\therefore$$ Point (6, 8) lies on the curve.