JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
Let $f(x)$ be a polynomial of degree $6$ in $x$, in which the coefficient of $x^{6}$ is unity and it has extrema at $x=-1$ and $x=1$. If $\lim_{x\to0}\frac{f(x)}{x^{3}}=1$, then $5\cdot f(2)$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$144$
Solution
Concept. The limit fixes the low powers; the two extrema fix the remaining coefficients. Formulas used. (1) $\lim_{x\to0}\frac{f(x)}{x^3}=1$ for a polynomial means $f=x^3+(\text{higher powers})$Step 1 - Form. By (1), $f=x^6+ax^5+bx^4+x^3$, and $$f'=6x^5+5ax^4+4bx^3+3x^2.$$ Step 2 - Extrema. $f'(1)=9+5a+4b=0$ and $f'(-1)=-3+5a-4b=0$. Adding: $10a+6=0$, $a=-\frac35$; then $b=-\frac32$. Step 3 - Value. $f(2)=64-\frac{96}{5}-24+8=\frac{144}{5}$, so $5f(2)=144$. Answer: 144