The values of b for which y=2x^2-bx+2 touches the x-axis are :

The correct answer is Option (4) → ±4

$y=2x^2-bx+2$  ...(1)

y touches x axis when $\frac{dy}{dx}=0$ and $y=0$

from (1) at touch point

$2x^2-bx+2=0$

$2(\frac{b}{4})^2-b(\frac{b}{4})+2=0$

$⇒\frac{b^2}{8}-\frac{b^2}{4}+2=0⇒2=\frac{b^2}{8}$

$⇒b=±4$