Question

If a cubic equation $f\left(x\right)$ vanishes at $x=-2$ and has relative maximum/minimum at $x=-1$ and $x=\frac{1}{3}$ and $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{14}{3}.$ Then which of the following is/are correct ?

• A. $f\left(0\right)=2$
• B. $f\left(1\right)=3$
• C. $f\left(2\right)=4$
• D. $f(-1)=3$

Answer 1

Answer: (A), (B), (D)

$f(-1)=3$

As, $f\left(x\right)$ has relative extrema at $x=-1,\frac{1}{3}$
Let $f^{\prime }\left(x\right)=a(x+1)(x-\frac{1}{3})$
Integrating both sides, we get
$\Rightarrow f\left(x\right)=a(\frac{x^3}{3}+\frac{x^2}{3}-\frac{x}{3})+C$
$\because f(-2)=0;C=\frac{2a}{3}$
$\Rightarrow f\left(x\right)=\frac{a}{3}(x^3+x^2-x+2)$
Again,
$\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{14}{3}\Rightarrow \frac{2a}{3}{(0+\frac{x^3}{3}-0+2x)}_0^1=\frac{14}{3}$
$[\because x^3,-x\text{is odd and}{x}^2,2\text{is even}]$
$\Rightarrow a=3$
So, $f\left(x\right)=x^3+x^2-x+2$
$\therefore f\left(1\right)=3,f\left(2\right)=12,f\left(0\right)=2$ and $f(-1)=3.$