JEE Main 2026 — 4 April, Morning Shift — Question 40

Let degree of polynomial be ' n ' $\mathrm{n}=\mathrm{n}-1+\mathrm{n}-2 \Rightarrow \mathrm{n}=3$ $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}$ $\left(a x^{3}+b x^{2}+c x\right)=\left(3 a x^{2}+2 b x+c\right)(6 a x+2 b)$ $x^{3}: a=18 a^{2} \Rightarrow a=\frac{1}{18}$ $x^{2}: b=6 a b+12 a b$ $x: c=4 b^{2}+6 a c \Rightarrow \frac{2 C}{3}=4 b^{2} \Rightarrow b^{2}=\frac{c}{6}$ const : $\mathrm{bc}=0 \Rightarrow \mathrm{~b}=\mathrm{c}=0$

$$\begin{aligned} & \int_{0}^{2} f(x) d x+f^{\prime}(2)+f^{\prime \prime}(2) \\& \int_{0}^{2} a x^{3} d x+12 a+12 a=4 a+24 a=28 a=\frac{28}{18}=\frac{14}{9} \\& 36\left(\frac{14}{9}\right)=56 \end{aligned}$$