JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q7

← JEE Main 2022 · 29 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 7 of 90

Let $f$ be a real valued continuous function on $[0,1]$ and $f(x)=x+\displaystyle\int_0^1(x-t)f(t)\,dt$ . Then, which of the following points $(x,y)$ lies on the curve $y=f(x)$ ?

(2, 4)

(1, 2)

(4, 17)

(6, 8)

Answer and worked solution

Official answer: D

Form. $f(x)=x(1+a)-b$ with $a=\int_0^1f$ , $b=\int_0^1tf$ .

Solve. $a=\frac{1+a}2-b$ and $b=\frac{1+a}3-\frac b2$ give $a=\frac5{13}$ , $b=\frac4{13}$ , so $f(x)=\frac{18x-4}{13}$ .

Check. $f(6)=8$ .

(D)

Topics:Properties of definite integrals

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JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 7