JEE Main 2020 (Online) 2nd September Morning ShiftQuestion 1
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 :
Differentiation and its Application
Solution
Concept. Extrema at $1$ and $2$ give $p'=k(x-1)(x-2)$; the two values fix $k$ and the constant. Formulas used. (1) $p=k\left(\dfrac{x^3}{3}-\dfrac{3x^2}{2}+2x\right)+C$Step 1 - Values. By (1): $p(1)=\frac{5k}{6}+C=8$ and $p(2)=\frac{2k}{3}+C=4$. Step 2 - Solve. Subtracting: $\frac k6=4$, so $k=24$ and $C=8-20=-12$. ($k>0$ makes $x=1$ the maximum.) Step 3 - Value. $p(0)=C=-12$. Answer: (A)