The slope of the normal to the curve \(y=x^3-4\sin x\) at x = 0 is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{4}\)
Slope of the tangent:
\( \displaystyle \frac{dy}{dx} = 3x^2-4\cos x \)
\( \displaystyle \left.\frac{dy}{dx}\right|_{x=0} = 0-4 = -4 \)
The normal is perpendicular to the tangent, so its slope is the negative reciprocal:
\( \displaystyle \text{Slope of normal} = -\frac{1}{-4} = \frac{1}{4} \)
Why the other options are incorrect
Option (1) −4 — Incorrect. This is the slope of the tangent, not of the normal.
Option (3) 4 — Incorrect. This only changes the sign of the tangent slope; the normal slope also needs the reciprocal.
Option (4) \(-\frac{1}{4}\) — Incorrect. This is the reciprocal of −4 without the sign change; then the product of the slopes is \(+1\), not −1.