Mathematics · Definite Integration

JEE Main 2026 — 2 April, Morning Shift — Question 40

Let $P(x)$ be a real polynomial of degree 3 which vanishes at $x=-3$ . Let $P(x)$ have local minima at $x=1$ ,

local maxima at $x=-1$ and $\int_{-1}^{1} P(x) d x=18$ , then the sum of all the coefficients of the polynomial $P(x)$ is equal to $\_\_\_\_\_\_$ .

Answer: 8

Numerical answer — enter this value.

$P^{\prime}(x)$ is second degree polynomial and vanishes at $x= \pm 1$ let $P^{\prime}(x)=k(x-1)(x+1)=k x^{2}-k$

$\Rightarrow P(x)=\frac{k}{3} \cdot x^{3}-k x+\lambda$

$\int_{-1}^{1} p(x) d x=18 \Rightarrow \int_{-1}^{1}\left(\frac{k}{3} \cdot x^{3}-k x+\lambda\right) d x=2 \lambda=18$

$\Rightarrow \lambda=9$

Also $P(-3)=0$

$\Rightarrow \frac{\mathrm{k}}{3}(-3)^{3}-\mathrm{k}(-3)+9=0 \Rightarrow \mathrm{k}=\frac{3}{2}$

$P(x)=\frac{1}{2} x^{3}-\frac{3}{2} x+9 \Rightarrow$ sum of all the coefficients $=8$

Answer key and solution verified before publishing.

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