JEE Main 2021 (Online) 26th February Morning ShiftQuestion 1
The maximum slope of the curve $y=\frac{1}{2}x^{4}-5x^{3}+18x^{2}-19x$ occurs at the point :
Differentiation and its Application
Solution
Concept. The slope is $m=y'$; its local maximum is where $m'=0$ and $m''<0$. Formulas used. (1) $m=2x^3-15x^2+36x-19$, $m'=6(x-2)(x-3)$, $m''=12x-30$Step 1 - Critical points of $m$. By (1), $x=2$ or $3$. Step 2 - Max. $m''(2)=-6<0$, so the slope is (locally) largest at $x=2$; $m(2)=9>m(3)=8$. Step 3 - Point. $y(2)=8-40+72-38=2$, so the point is $(2,2)$. Answer: (B)