Question

The function $f(x)=x^3-6x^2+ax+b$ is such that $f(2)=f(4)=0.$ Consider two statements.
(S1) there exists $x_1,x_2\in (2,4),x_1<x_2$, such that $f^{\prime }\left(x_1\right)=-1$ and $f^{\prime }\left(x_2\right)=0$
(S2) there exists $x_3,x_4\in (2,4),x_3<x_4$, such that $f$ is decreasing in $\left(2,x_4\right)$, increasing in $\left(x_4,4\right)$ and $2f^{\prime }\left(x_3\right)=\sqrt{3}f\left(x_4\right)$
Then

• A. both (S1) and (S2) are true
• B. (S1) is false and (S2) is true
• C. both (S1) and (S2) are false
• D. (S1) is true and (S2) is false

Answer 1

Answer: (A)

$f(x)=x^3-6x^2+ax+b$
$f(2)=8-24+2a+b=0$
$2a+b=16\dots (1)$
$f(4)=64-96+4a+b=0$
$4a+b=32\dots (2)$
Solving (1) and (2)
$a=8,b=0$
$f(x)=x^3-6x^2+8x$
$f(x)=x^3-6x^2+8x$
$f^{\prime }(x)=3x^2-12x+8$
$f^{\prime }(x)=6x-12$
$\Rightarrow f^{\prime }(x)$ is $\uparrow$ for $x>2$, and $f^{\prime }(x)$ is $\downarrow$ for $x<2$
$f^{\prime }(2)=12-24+8=-4$
$f^{\prime }(4)=48-48+8=8$
$f^{\prime }(x)=3x^2-12x+8$
vertex $(2,-4)$
$f^{\prime }(2)=-4,f^{\prime }(4)=8,f^{\prime }(3)=27-36+8$

$f^{\prime }\left(x_1\right)=-1$, then $x_1=3$
$f^{\prime }\left(x_2\right)=0$
Again
$f^{\prime }(x)<0$ for $x\in \left(2,x_4\right)$
$f^{\prime }(x)>0$ for $x\in \left(x_4,4\right)$
$x_4\in (3,4)$
$f(x)=x^3-6x^2+8x$
$f(3)=27-54+24=-3$
$f(4)=64-96+32=0$
For $x_4(3,4)$
$f\left(x_4\right)<-3\sqrt{3}$
and $f^{\prime }\left(x_3\right)>-4$
$2f^{\prime }\left(x_3\right)>-8$
So, $2f^{\prime }\left(x_3\right)=\sqrt{3}f\left(x_4\right)$