Question

The function $f\left(x\right)=x^3-6x^2+ax+b$ is such that $f\left(2\right)=f\left(4\right)=0$. Consider two statements.

$\left(S1\right)$ There exists $x_1,x_2\in (2,4),x_1<x_2$, such that $f^{\prime }\left(x_1\right)=-1$ and $f^{\prime }\left(x_2\right)=0$

$\left(S2\right)$ There exists $x_3,x_4\in (2,4),x_1<x_4$, such that $f$ is decreasing in $(2,x_4)$, increasing in $(x_4,4)$ and $2f^{\prime }\left(x_3\right)=\sqrt{3}f\left(x_4\right)$

• A. Both $\left(S1\right)$ and $\left(S2\right)$ are true
• B. Both $\left(S1\right)$ and $\left(S2\right)$ are false
• C. $\left(S1\right)$ is true and $\left(S2\right)$ is false
• D. $\left(S1\right)$ is false and $\left(S2\right)$ is true

Answer 1

The correct option is A Both $\left(S1\right)$ and $\left(S2\right)$ are true

$\because f\left(2\right)=f\left(4\right)=0\Rightarrow a=8\text{and}b=0$
$f\left(x\right)=x^3-6x^2+8x;$
$f^{\prime }\left(x\right)=3x^2-12x+8=0\Rightarrow x=2+\frac{2}{\sqrt{3}}$
For statement $S1,x_2=2+\frac{2}{\sqrt{3}}$
$\because f^{\prime }\left(2\right)=-4\text{and}{f}^{\prime }\left(x_2\right)=0$ hence there exist $x_1$
such that $x_1\in (2,x_2)\text{and}{f}^{\prime }\left(x_1\right)=-1$
$\Rightarrow$ Statement $S1$ is true.

For statement $S2$; $x_4=2+\frac{2}{\sqrt{3}}$
$\text{So,}{f}^{\prime }\left(x_3\right)=\frac{\sqrt{3}}{2}f\left(x_4\right)=-\frac{8}{3}$
$f^{\prime }\left(2\right)<f^{\prime }\left(x_3\right)<f^{\prime }\left(x_4\right)$ so statement $S2$ is also true.