JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
The function $f(x)=x^{3}-6x^{2}+ax+b$ is such that $f(2)=f(4)=0$. Consider two statements.
$S_1$: there exists $x_1,x_2\in(2,4)$, $x_1<x_2$, such that $f'(x_1)=-1$ and $f'(x_2)=0$.
$S_2$: there exists $x_3,x_4\in(2,4)$, $x_3<x_4$, such that $f$ is decreasing in $(2,x_4)$, increasing in $(x_4,4)$ and $2f'(x_3)=\sqrt3f(x_4)$. Then
Differentiation and its Application
Solution
Concept. Find $f$; then $f'$ increases on $(2,4)$ from $-4$ to $8$, which settles both statements. Formulas used. (1) Intermediate value theorem for the continuous $f'$Step 1 - $f$. $f(2)=0$: $-16+2a+b=0$. $f(4)=0$: $-32+4a+b=0$. So $a=8$, $b=0$ and $f=x(x-2)(x-4)$, $f'=3x^2-12x+8$. Step 2 - $S_1$. $f''=6x-12>0$ on $(2,4)$, so $f'$ increases from $f'(2)=-4$ to $f'(4)=8$. By (1), $f'$ takes the values $-1$ and then $0$ at points $x_1<x_2$. True. Step 3 - $x_4$. $f'=0$ at $x_4=2+\frac{2}{\sqrt3}$; $f$ decreases on $(2,x_4)$ and increases on $(x_4,4)$. Here $$f(x_4)=\left(2+\tfrac{2}{\sqrt3}\right)\cdot\tfrac{2}{\sqrt3}\cdot\left(\tfrac{2}{\sqrt3}-2\right)=-\frac{16}{3\sqrt3},\qquad\sqrt3f(x_4)=-\frac{16}{3}.$$ Step 4 - $S_2$. We need $f'(x_3)=-\frac83$ with $x_3<x_4$. On $(2,x_4)$, $f'$ runs from $-4$ to $0$, so by (1) such $x_3$ exists. True. Answer: (C)