The function

`
Question:

The function $f(x)=x^{3}-6 x^{2}+a x+b$ is such that $f(2)=f(4)=0 .$ Consider two statements.

$(\mathrm{S} 1)$ there exists $\mathrm{x}_{1}, \mathrm{x}_{2} \in(2,4), \mathrm{x}_{1}<\mathrm{x}_{2}$, such that $f^{\prime}\left(x_{1}\right)=-1$ and $f^{\prime}\left(x_{2}\right)=0 .$

$(\mathrm{S} 2)$ there exists $\mathrm{x}_{3}, \mathrm{x}_{4} \in(2,4), \mathrm{x}_{3}<\mathrm{x}_{4}$, such that

$f$ is decreasing in $\left(2, x_{4}\right)$, increasing in $\left(x_{4}, 4\right)$

and $2 f^{\prime}\left(\mathrm{x}_{3}\right)=\sqrt{3} f\left(\mathrm{x}_{4}\right)$.

Then

  1. both (S1) and (S2) are true

  2. $(\mathrm{S} 1)$ is false and $(\mathrm{S} 2)$ is true

  3. both (S1) and (S2) are false

  4. (S1) is true and (S2) is false


Correct Option: 1

Solution:

$f(x)=x^{3}-6 x^{2}+a x+b$

$f(2)=8-24+2 a+b=0$

$2 a+b=16 \ldots(1)$...(1)

$f(4)=64-96+4 a+b=0$

$4 a+b=32 \ldots(2)$...(2)

Solving (1) and (2)

$a=8, b=0$

$f(x)=x^{3}-6 x^{2}+8 x$

$f(x)=x^{3}-6 x^{2}+8 x$

$f^{\prime}(x)=3 x^{2}-12 x+8$

$f^{\prime \prime}(x)=6 x-12$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$ is $\uparrow$ for $\mathrm{x}>2$, and $\mathrm{f}^{\prime}(\mathrm{x})$ is $\downarrow$ for $\mathrm{x}<2$

$f^{\prime}(2)=12-24+8=-4$

$f^{\prime}(4)=48-48+8=8$

$f^{\prime}(x)=3 x^{2}-12 x+8$

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 $x \in\left(2, x_{4}\right)$

$\mathrm{f}^{\prime}(\mathrm{x})>0$ for $\mathrm{x} \in\left(\mathrm{x}_{4}, 4\right)$

$x_{4} \in(3,4)$

$f(x)=x^{3}-6 x^{2}+8 x$

$f(3)=27-54+24=-3$

$f(4)=64-96+32=0$

For $x_{4}(3,4)$

$f\left(x_{4}\right)<-3 \sqrt{3}$

and $f^{\prime}\left(x_{3}\right)>-4$

$2 \mathrm{f}^{\prime}\left(\mathrm{x}_{3}\right)>-8$

So, $2 \mathrm{f}^{\prime}\left(\mathrm{x}_{2}\right)=\sqrt{3} \mathrm{f}\left(\mathrm{x}_{4}\right)$

Correct Ans. (1)

Leave a comment