Question

If the global maximum value of $f\left(x\right)=-x^2+ax-a^2-2a-2$ for $xϵ[0,1]$ be '-5', then 'a' can take the value

• A. $\frac{-4+2\sqrt{13}}{3}$
• B. $\frac{-4-2\sqrt{13}}{3}$
• C. $1$
• D. $-3$

Answer 1

Answer: (A), (D)

$-3$
$\frac{-4+2\sqrt{13}}{3}$

Vertex of $f\left(x\right)$ will be at x = $\frac{a}{2}$
Case-| : if $a<0$, then maximum value of $f\left(x\right)$ is at $x=0$
$\therefore f\left(0\right)=-5\Rightarrow a^2+2a-3=0$
$\Rightarrow a=-3or1\Rightarrow a=-3$
Case-|| : if $a>2$, then maximum value of $f\left(x\right)$ is at $x=1$
$\therefore f\left(1\right)=-5\Rightarrow a^2+a-2=0$
$\Rightarrow a=-2or1$ which is not possible
Case-||| : if $0<a<2$, then maximum value is at $x=\frac{a}{2}$
$\Rightarrow f\left(\frac{a}{2}\right)=-5\Rightarrow 3a^2+8a-12=0\Rightarrow a=\frac{-4\pm 2\sqrt{13}}{3}$
$\Rightarrow a=\frac{-4+2\sqrt{13}}{3}$ is possible
Ans: A,D