Question

A cubic function $f\left(x\right)$ vanishes at $x=-2$ and has a relative minima/maxima at $x=1$ and $x=1/3$ if ${\int }_{-1}^{1}f\left(x\right)dx=\frac{14}{3}$ then $f\left(x\right)$ equals

• A. $(x+2)(x^2-x+1)$
• B. $(x+2)(x^2+x-1)$
• C. $(x+3)(x+2)(x-1)$
• D. None of these

Answer 1

The correct option is A $(x+2)(x^2-x+1)$

Let the cubic function $f\left(x\right)=A{x}^3+B{x}^2+Cx+D$
$f^{{}^{\prime }}\left(x\right)=3A{x}^2+2Bx+c$
$f^{{}^{\prime }}\left(x\right)=3A{x}^2+2Bx+C$
From the given data ,
$f(-2)=0\Rightarrow -8A+4B-2C+D=0.....\left\{1\right\}$
$f^{{}^{\prime }}\left(1\right)=3A+2B+C=0......\left\{2\right\}$
$f^{{}^{\prime }}\left(\frac{1}{3}\right)=\frac{A}{3}+\frac{2b}{3}+C=0......\left\{3\right\}$
${\int }_{-1}^{1}f\left(x\right)dx=2(\frac{B}{3}+D)=\frac{14}{3}.....\left\{4\right\}$
By solving {1} {2} {3} and {4} we get ,
A = 1 , B = 1 , C = - 1 , D = 2

ie , $f\left(x\right)=x^3+x^2-x+2=(x+2)(x^2-x+1).....Ans$