The slope of the normal to the curve y = 2x2 + 3sin x at x = 0, is:

The correct answer is Option (4) → \(-\displaystyle\frac{1}{3}\)

The slope of the tangent is the value of \(\frac{dy}{dx}\):

    \( \displaystyle \frac{dy}{dx} = 4x + 3\cos x \)

    \(\text{At } x = 0:\ 4(0) + 3\cos 0 = 3\)

The normal is perpendicular to the tangent, so its slope is the negative reciprocal:

    \( \displaystyle \text{slope of normal} = -\frac{1}{3} \)

Why the other options are incorrect

• Option (1) 3 — Incorrect. This is the slope of the tangent at x = 0, not of the normal.

• Option (2) −3 — Incorrect. Only the sign of the tangent slope is changed; the reciprocal is not taken.

• Option (3) \(\displaystyle\frac{1}{3}\) — Incorrect. Only the reciprocal is taken; the minus sign is missing, and the product of the two slopes would be 1, not −1.