Question

Let $f (x) =x^{3}+ax^{2}+bx+c$, where $a, b, c$ are real numbers If $f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-\frac{1}{3}$ and $f (2)=0$, then $\int\limits_{-1}^{1}f \left(x\right) dx $ equals

• A. $\frac{14}{3}$
• B. $\frac{-14}{3}$
• C. $\frac{7}{3}$
• D. $\frac{-7}{3}$

Answer 1

Answer: (B)

We have,
$f (x) =x^{3} +ax^{2}+bx+c$
$f (2) =8+4a+2b +c=0 \, \dots(i)$
$[\because f (2)=0]$
$f'(x) =3x^{2}+2ax+b $
$f'(1) =3+2a+b=0 \, \dots(ii)$
$[\because f' (1)=0]$
$f'\left(\frac{-1}{3}\right)=\frac{1}{3}-\frac{2a}{3}+b=0 \,\ldots\left(iii\right)$
$\left[\because f' \left(\frac{-1}{3}\right)=0\right]$
From Eqs. $\left(ii\right) $ and $\left(iii\right)$, we get
$a=-1, b=-1$
Putting the values of a and b in Eq. $\left(i\right)$, we get $c = - 2$
$\therefore f \left(x\right)=x^{3}-x^{2}-x-2$
$\int\limits_{-1}^{1}f\left(x\right)dx=\int\limits_{-1}^{1}\left(x^{3}-x^{2}-x-2\right)dx$
$=-2 \int\limits_{0}^{1}\left(x^{2}+2\right)dx $
$[\because x^{3}$ and -$x$ are odd functions
$=-2\left[\frac{x^{3}}{3}+2x\right]_{0}^{1}$
$=-2\left[\frac{1}{3}+2\right]=-\frac{14}{3}$