The slope of the normal to the curve $y=2x^2-3sinx$ at $x=0$ is

The correct answer is $\dfrac{1}{3}$

Given equation of curve $y=2x^2-3$ $sinx$

The slope of tangent to the curve = $m_1$ = $\dfrac{dy}{dx}$ = $4x - 3$ $cosx$


At $x=0$, $\left( \dfrac{dy}{dx} \right)_{x=0}$ = $4(0) - 3$ $cos0$ = $-3$


The slope of normal to the curve = $m_2$ = $-\dfrac{1}{m1}$ = $-\dfrac{1}{-3}$ = $\dfrac{1}{3}$