Question

$f(x)$ is a polynomial of degree $6$ which decreases in the interval $(0, \infty)$ and increases in the interval $(-\infty, 0) .$ If $f'(2)=0, f'(0)=0, f''(0)=0, f(0)=1$ and $f(1)-f(-1)=\frac{8}{5}$, then $-3(f(1)+f(-1))$ equals

Answer 1

Since, $f'(x)=k x^{3}(x-2)^{2}=k x^{3}\left(x^{2}-4 x+4\right)$
$f'(x)=k\left(x^{5}-4 x^{4}+4 x^{3}\right)$
$\therefore f(x)=k\left(\frac{x^{6}}{6}-\frac{4 x^{5}}{5}+x^{4}\right)+c$
Also $f(0)=1 \Rightarrow c=1$.
and $f(1)-f(-1)=-\frac{8}{5} k=\frac{8}{5}$
$\Rightarrow k=-1 .$
So $f(1)+f(-1)=\frac{7}{3} k+2 c=-\frac{7}{3}+2=-\frac{1}{3}$