Question

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

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

Hint:

Remember: $m_{\text{tangent}} \times m_{\text{normal}} = -1$. Once you find the derivative value as $3$, its negative reciprocal is immediately $-\frac{1}{3}$.

Answer 1

The Correct Option is A

Step 1: Concept
The slope of the tangent ($m_t$) to a curve $y = f(x)$ at a given point is $\frac{dy}{dx}$. The slope of the normal ($m_n$) is perpendicular to the tangent, so $m_n = -\frac{1}{m_t}$.

Step 2: Meaning
We need to find the derivative $\frac{dy}{dx}$, evaluate it at $x = 0$ to get the tangent's slope, and then take the negative reciprocal.

Step 3: Analysis
Given curve: $y = 2x^2 + 3\sin x$. Differentiating with respect to $x$: \[ \frac{dy}{dx} = 4x + 3\cos x \] At $x = 0$: \[ m_t = \left. \frac{dy}{dx} \right|_{x=0} = 4(0) + 3\cos(0) = 3 \] Since the normal is perpendicular to the tangent: \[ m_n = -\frac{1}{m_t} = -\frac{1}{3} \]

Step 4: Conclusion
The slope of the normal to the curve at $x = 0$ is $-\frac{1}{3}$. Final Answer: (A)