Question

A cubic polynomial $f\left(x\right)$ vanishes at $x=-2$ and has a relative minimum/maximum at $x=-1$ and $x=\frac{1}{3}$. Then

• A. If $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{13}{3}$, then $f\left(x\right)=\frac{1}{4}(5x^3+5x^2-5x+5)$
• B. If $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{14}{3}$, then $f\left(x\right)=x^3+x^2-x+2$
• C. If $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{5}{3}$, then $f\left(x\right)=\frac{1}{6}(5x^3+5x^2-5x+15)$
• D. If ${\int }_{-1}^{1}f\left(x\right)dx=\frac{10}{3}$, then $f\left(x\right)=\frac{1}{7}(5x^3+5x^2-5x+10)$

Answer 1

Answer: (B), (D)

If ${\int }_{-1}^{1}f\left(x\right)dx=\frac{10}{3}$, then $f\left(x\right)=\frac{1}{7}(5x^3+5x^2-5x+10)$

As the function has relative minimum/maximum at $x=-1$ and $x=\frac{1}{3}$
So, $f^{\prime }\left(x\right)=a(x+1)(x-\frac{1}{3})$
where $a$ is constant.
$\Rightarrow f^{\prime }\left(x\right)=a(x^2+\frac{2x}{3}-\frac{1}{3})$
$\Rightarrow f\left(x\right)=a(\frac{x^3}{3}+\frac{x^2}{3}-\frac{x}{3})+b$
where $b$ is constant of integration.

Now $f(-2)=0$
$\Rightarrow \frac{-8a}{3}+\frac{4a}{3}+\frac{2a}{3}+b=0$
$\Rightarrow b=\frac{2a}{3}$
$\Rightarrow f\left(x\right)=\frac{a}{3}(x^3+x^2-x+2)$

When $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{14}{3}$
$\Rightarrow \frac{a}{3}\underset{-1}{\overset{1}{\int }}(x^3+x^2-x+2)dx=\frac{14}{3}$
$\Rightarrow a\underset{-1}{\overset{1}{\int }}(x^2+2)=14$
$\Rightarrow 2a\underset{0}{\overset{1}{\int }}(x^2+2)=14$
$\Rightarrow 2a(\frac{1}{3}+2)=14$
$\Rightarrow \frac{14}{3}a=14\Rightarrow a=3$

So the function will be
$f\left(x\right)=x^3+x^2-x+2$

Similarly, when $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{10}{3}$
$\Rightarrow \frac{14}{3}a=10$
$\Rightarrow a=\frac{15}{7}$

So the function will be
$f\left(x\right)=\frac{1}{7}(5x^3+5x^2-5x+10)$
Similarly, remaining two options can be found to be incorrect.