Question

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

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