Question

Let $f(x)=x^3+a{x}^2+bx+c$, 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 $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$ equals

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

Answer 1

Answer: (B)

We have,
$f(x)=x^3+a{x}^2+bx+c$
$f(2)=8+4a+2b+c=0\dots (i)$
$[\because f(2)=0]$
$f^{\prime }(x)=3x^2+2ax+b$
$f^{\prime }(1)=3+2a+b=0\dots (ii)$
$[\because f^{\prime }(1)=0]$
$f^{\prime }\left(\frac{-1}{3}\right)=\frac{1}{3}-\frac{2a}{3}+b=0\dots \left(iii\right)$
$\left[\because f^{\prime }\left(\frac{-1}{3}\right)=0\right]$
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 $c=-2$
$\therefore f\left(x\right)=x^3-x^2-x-2$
$\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\underset{-1}{\overset{1}{\int }}\left(x^3-x^2-x-2\right)dx$
$=-2\underset{0}{\overset{1}{\int }}\left(x^2+2\right)dx$
$[\because x^3$ and -$x$ are odd functions
$=-2{\left[\frac{x^3}{3}+2x\right]}_0^1$
$=-2\left[\frac{1}{3}+2\right]=-\frac{14}{3}$