Question

The slope of the tangent to the curve \( y = 2x^P + 3\sin x \) at \( x = 0 \) is:

• A. \( 3 \)
• B. \( -1 \)
• C. \( 0 \)
• D. \( -3 \)

Hint:

To find the slope of the tangent at a point: - Differentiate the given function. - Substitute the value of \( x \) into the derivative. - Ensure any powers of \( x \) do not make the derivative undefined at that point.

Answer 1

The Correct Option is C

Step 1: Differentiate the function \( y = 2x^P + 3\sin x \) with respect to \( x \) to get the slope of the tangent. \[ \frac{dy}{dx} = \frac{d}{dx}(2x^P) + \frac{d}{dx}(3\sin x) = 2P x^{P-1} + 3\cos x \] Step 2: Evaluate the derivative at \( x = 0 \): - \( \lim_{x \to 0} 2P x^{P-1} \): This is defined only if \( P>1 \). If \( P = 1 \), the term becomes 2. If \( P<1 \), the derivative diverges. Assuming \( P = 2 \) (most plausible integer for the given form), then: \[ \frac{dy}{dx} = 2 \cdot 2 \cdot x^{2-1} + 3\cos x = 4x + 3\cos x \] At \( x = 0 \): \[ \left. \frac{dy}{dx} \right|_{x=0} = 4(0) + 3\cos(0) = 0 + 3 = 3 \] But the correct answer is marked as (C) \( 0 \), which suggests either: - \( P = 0 \): then \( x^0 = 1 \), derivative of constant is 0 - \( P = 1 \): then \( 2x \rightarrow \) derivative = 2 - Best assumption for the slope to be 0 is \( P = 1 \), but that leads to slope 2 + 3 = 5. So to get slope 0, perhaps \( P = 0 \) and original function is: \[ y = 2x^0 + 3\sin x = 2 + 3\sin x \Rightarrow \frac{dy}{dx} = 3\cos x \] Then, \( \frac{dy}{dx}|_{x=0} = 3\cos(0) = 3 \) None of these lead to 0 unless \( P = 1 \) and a cancellation occurs. We must assume that \( P = 2 \), as initially tried, then: \[ \frac{dy}{dx} = 4x + 3\cos x \Rightarrow \frac{dy}{dx}|_{x=0} = 0 + 3 = 3 \] So Answer: (A) 3 (Seems like there was an error in the source you gave — you can correct the answer or clarify the value of \( P \).)