JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
If the function $f:\mathbf{R}\to\mathbf{R}$ defined by $f(x)=\frac{x^{3}}{3}-(2a+3b)x^{2}+24abx+5$ has a local maxima at $-4$ and a local minima at $4$, then a value of $4a+3b$ is
Differentiation and its Application
Solution
Concept. $-4$ and $4$ are the roots of $f'$; use their sum and product. Formulas used. (1) $f'(x)=x^2-2(2a+3b)x+24ab$Step 1 - Sum and product. By (1): $2(2a+3b)=0$ and $24ab=-16$. So $2a=-3b$ and $ab=-\frac23$. Step 2 - Solve. $a=-\frac{3b}{2}$ gives $-\frac{3b^2}{2}=-\frac23$, so $b=\pm\frac23$, with $a=\mp1$. Step 3 - Check the max/min. $f'=x^2-16$ in both cases, which goes $+\to-$ at $-4$ and $-\to+$ at $4$. Fine. Step 4 - Value. $4a+3b=-4+2=-2$ or $4-2=2$. Only $2$ is an option. Answer: (B)