Clearly p′(x)=λ(x−1)(x−2) where λ>0
Integrating both sides
p(x)=λ[3x3−23x2+2x]+C
Given
p(1)=8⇒λ(31−23+2)+C=8⇒65λ+C=8...(1)
p(2)=4⇒λ(38−6+4)+C=4⇒32λ+C=4...(2)
By (1) and (2) ⇒C=−12
⇒p(0)=−12
JEE Main 2020 — Mathematics Calculus
If p(x) be a polynomial of degree three that has a local maximum value 8 at x=1 and a local minimum value 4 at x=2 then p(0) is equal to
Held on 2 Sept 2020 · Verified 6 Jul 2026.
6
−12
24
12
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in \mathbf{R}$. Consider the following two statements : (I) $f(x)$ is discontinous at $x=1$. (II) $f(x)$ is continous at $x=-1$. Then,
The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$ is:
The value of ∫₀¹ x·eˣ dx is:
Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
Work through every JEE Main Calculus PYQ, year by year.