Question

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

• A. $\dfrac{1}{3}$
• B. $3$
• C. $-\dfrac{1}{3}$
• D. $-3$

Hint:

The slope of the normal to a curve at a point is the negative reciprocal of the slope of the tangent at that point.

Answer 1

The Correct Option is D

Step 1: Differentiate the curve.
Given: \[ y = 2x^{2} + 3\sin x \] Differentiate with respect to $x$: \[ \frac{dy}{dx} = 4x + 3\cos x \]

Step 2: Find slope of tangent at $x=0$.
\[ \frac{dy}{dx}\Big|_{x=0} = 4(0) + 3\cos(0) = 3 \] So, slope of tangent at $x=0$ is $3$.

Step 3: Find slope of normal.
The slope of the normal is the negative reciprocal of the slope of the tangent. \[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{3} \]

Step 4: Match with options.
The correct answer is option (C) $-\dfrac{1}{3}$.