Let P′(x)=a(x−1)(x+1)=a(x2−1)
P(x)=a∫(x2−1)dx+c
=a(3x3−x)+c
Now P(−3)=0
⇒a(−327+3)+c=0
⇒−6a+c=0…(1)
Now ∫−11(a(3x3−x)+c)dx=18
=2c=18⇒c=9...(2)
From (1)&(2) ⇒−6a+9=0⇒a=23
⇒P(x)=23(3x3−x)+9
Then, the Sum of coefficient=21−23+9
=8
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 ∫−11P(x)dx=18, then the sum of all the coefficients of the polynomial P(x) is equal to ___ .
Held on 18 Mar 2021 · Verified 6 Jul 2026.
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 $y = y(x)$ be the solution of the differential equation $x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$, $y(1) = \dfrac{\pi}{2}$ and let $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Then the number of integral values of $p$, for which the equation $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ represents a circle of radius $r \leq 6$, is __________.
Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$. Then $\lim _{x \rightarrow 1}\left(\log _{\mathrm{e}}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$ is equal to :
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 :
If the function $f(x)=\frac{e^{x}\left(e^{\tan x-x}-1\right)+\log _{e}(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to
Work through every JEE Main Calculus PYQ, year by year.