Question

Let $f\left(x\right)=4x^2-4ax+a^2-2a+2$ be a quadratic polynomial in $x$, $a\in R$. If $y=f\left(x\right)$ takes minimum value of $3$ on $[0,2]$ and x-coordinate of vertex is greater than $2$, then value of $a$ is

• A. $5-\sqrt{10}$
• B. $10-\sqrt{10}$
• C. $5+\sqrt{10}$
• D. $10-\sqrt{5}$

Answer 1

Answer: (C)

$5+\sqrt{10}$

$f\left(x\right)=4x^2-4ax+a^2-2a+2,x,aϵR$

x- co ordinate of vertex of $y=f\left(x\right)$ is greater than $2$.

$\therefore f\left(x\right)$ is monotonically decreasing in $xϵ[0,2]$

$\therefore f\left(2\right)$ is minimum for $kϵ[0,2]$

$\therefore 4\left(4\right)-8a+a^2-2a+2=3$

$a^2-10a+18=3$

$a^2-10a+15=0$

$a=5\pm \sqrt{10}$

$\therefore a=5+\sqrt{10}(\because a>4)$

($\because$Vertex at $x=\frac{a}{2}>2$)