Question

Let $f\left(x\right)$ be a polynomial of degree 3 having extremum at $x=\frac{-1}{3},1$ and $f\left(2\right)=0.$

Which of the following statement is correct?

• A. $f\left(x\right)$ is increasing in $(\frac{-1}{3},1)$
• B. $f\left(x\right)$ has a local maximum at $x=\frac{1}{3}$
  
• C. $f\left(x\right)$ is negative in $(-\infty ,2)$ and positive in $(2,\infty )$
• D. $f^{\prime }\left(x\right)$ is decreasing in $(\frac{1}{3},\infty )$

Answer 1

The correct option is C $f\left(x\right)$ is negative in $(-\infty ,2)$ and positive in $(2,\infty )$

Let $f^{\prime }\left(x\right)=3(x+\frac{1}{3})(x-1)=(3x+1)(x-1)=3x^2-2x-1$

$\therefore f\left(x\right)=x^3-x^2-x+\lambda$

As $f\left(2\right)=0\Rightarrow 8-4-2+\lambda =0$
$\Rightarrow \lambda =-2$

Hence, $f\left(x\right)=x^3-x^2-x-2$
Clearly, $f\left(x\right)$ is negative in $(-\infty ,2)$ and positive in $(2,\infty )$

$f^{\prime }\left(x\right)=3x^2-2x-1$

and $f^{\prime \prime }\left(x\right)=6x-2=2(3x-1)=6(x-\frac{1}{3})$

$\Rightarrow f^{\prime }\left(x\right)$ has a local minimum at $x=\frac{1}{3}$. Also $f^{\prime }\left(x\right)$ is increasing on $(\frac{1}{3},\infty )$