The slope of the normal to the curve $y = 2{x^2} + 3\sin x$ at $x = 0$ is

(A) $3$

(B) $\cfrac{1}{3}$

(C) $- 3$

(D) $- \cfrac{1}{3}$

Option D is correct

We have,
$y = 2{x^2} + 3\sin x$ …(i)
$= \cfrac{{dy}}{{dx}} = 4x + 3\cos x$

Slope of the tangent to (i) at $x = 0$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 0}} = 4 \cdot 0 + 3\cos 0 = 3$

So, slope of the normal to (i) at $x = 0$ is $\cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{tangent}}}} = - \cfrac{1}{3}$