Question

If the number of integral value(s) of $a$ for which $f\left(x\right)=\frac{2}{3}{x}^3+a{x}^2-(a^2-6)x+7$ has a negative point of minima is $k.$ Then the value of $f\left(k\right)$ is equal to

Answer 1

Given : $f\left(x\right)=\frac{2}{3}{x}^3+a{x}^2-(a^2-6)x+7$
$\Rightarrow f^{\prime }\left(x\right)=2x^2+2ax-(a^2-6)$
As, the given function is defined as $f:R\to R$
Graph of $f\left(x\right)$ can be :
from graph, it is clear that for minima to be at negative side, both roots of $f^{\prime }\left(x\right)=0$ will be negative.
$\Rightarrow D>0$ and $-\frac{b}{2a}<0$ and $f^{\prime }\left(0\right)>0$
$\Rightarrow 4a^2+8(a^2-6)>0$
$\Rightarrow 3a^2-12>0$
$\Rightarrow a\in (-\infty ,-2)\cup (2,\infty )\cdots \left(i\right)$
and $\frac{-b}{2a}=-a<0$
$\Rightarrow a>0\cdots \left(ii\right)$
and $f^{\prime }\left(0\right)=-(a^2-6)>0$
$\Rightarrow a^2-6<0$
$\Rightarrow a\in (-\sqrt{6},\sqrt{6})\cdots \left(iii\right)$
From $\left(i\right),\left(iii\right)$ and $\left(iii\right)$, we get :
$a\in (2,\sqrt{6})$
So, there is no integral value of $a,$ is possible.
$\therefore k=0,f\left(0\right)=7$