JEE Main 2026 · 4 Apr · Shift 1 · Mathematics Q17

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MathematicsSingle correct+4 / -1Question 17 of 75

Let $f$ be a real polynomial of degree $n$ such that $f(x)=f'(x)f''(x)$ , for all $x\in\mathbb R$ . If $f(0)=0$ , then $36\left(f'(2)+f''(2)+\displaystyle\int_0^2f(x)\,dx\right)$ is equal to:

Answer and worked solution

Official answer: C

Degree. $n=(n-1)+(n-2)$ gives $n=3$ .

Form. Write $f=A(x-p)^3+C(x-p)+D$ . Then $f'f''=18A^2(x-p)^3+6AC(x-p)$ . Matching gives $A=\frac1{18}$ , $C=0$ , $D=0$ , and $f(0)=0$ gives $p=0$ : $f=\dfrac{x^3}{18}$ .

Values. $f'(2)=\frac23$ , $f''(2)=\frac23$ , $\displaystyle\int_0^2f=\frac29$ . So $36\cdot\frac{14}9=56$ .

(C)

Topics:Higher-order derivatives Fundamental theorem of calculus

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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 17