Question

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

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

Answer 1

Answer: (C)

$1-\sqrt{2}$

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

$f\left(x\right)={(2x-a)}^2-2a+2$

Vertex of $y=f\left(x\right)$ is at $x=\frac{a}{2}<0$

$\therefore a<0$

Since $f\left(x\right)$ has vertex at less than $0$

$\therefore f\left(x\right)$ is monotonically increasing from $0$ to $2$

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

$\therefore a^2-2a+2=3$

$a^2-2a-1=0$

$\therefore a=1-\sqrt{2},a<0,\therefore a\ne 1+\sqrt{2}$