JEE Main 2020 (Online) 8th January Evening ShiftQuestion 1
Let $f(x)$ be a polynomial of degree $3$ such that $f(-1)=10$, $f(1)=-6$, $f(x)$ has a critical point at $x=-1$ and $f'(x)$ has a critical point at $x=1$. Then $f(x)$ has a local minima at $x=$ ________.
Differentiation and its Application
Answer$3$
Solution
Concept. $f''(1)=0$ and $f'(-1)=0$, with the two values, fix the cubic. Formulas used. (1) $f=ax^3+bx^2+cx+d$, $f'=3ax^2+2bx+c$, $f''=6ax+2b$Step 1 - Conditions. By (1): $f''(1)=6a+2b=0$, so $b=-3a$. $f'(-1)=3a-2b+c=0$, so $c=-9a$. Step 2 - Values. $f(-1)=-a-3a+9a+d=5a+d=10$ and $f(1)=a-3a-9a+d=-11a+d=-6$. So $a=1$, $d=5$. Step 3 - Minimum. $f'=3x^2-6x-9=3(x-3)(x+1)$; it goes $-\to+$ at $x=3$. Answer: 3