Question

Let \( f_n(x) \) be the \(n^{\text{th}}\) derivative of \( f(x) \). The least value of \( n \) such that \[ f_n(x) = f_{n+1}(x) \] where \[ f(x) = x^2 + e^x \] is: 

• A. 4
• B. 5
• C. 2
• D. 3
• E. 6

Hint:

For any function \( f(x) = P(x) + e^{x} \) where \( P(x) \) is a polynomial of degree \( k \), the condition \( f_{n} = f_{n+1} \) will first be satisfied when \( n = k + 1 \).

Answer 1

The Correct Option is D

Concept: The derivative of a polynomial \( x^{k} \) eventually becomes zero after \( k+1 \) differentiations. However, the derivative of the exponential function \( e^{x} \) is always \( e^{x} \). We need to find the point where the polynomial part disappears, leaving only the repeating exponential part.

Step 1: Calculate successive derivatives of \( f(x) \).
Given \( f(x) = x^{2} + e^{x} \):
• First derivative (\( f_{1} \)): \( \frac{d}{dx}(x^{2} + e^{x}) = 2x + e^{x} \)
• Second derivative (\( f_{2} \)): \( \frac{d}{dx}(2x + e^{x}) = 2 + e^{x} \)
• Third derivative (\( f_{3} \)): \( \frac{d}{dx}(2 + e^{x}) = 0 + e^{x} = e^{x} \)
• Fourth derivative (\( f_{4} \)): \( \frac{d}{dx}(e^{x}) = e^{x} \)

Step 2: Identify the least \( n \) for \( f_{n} = f_{n+1} \).
From our calculations: \( f_{3} = e^{x} \) and \( f_{4} = e^{x} \). Thus, \( f_{3} = f_{4} \). The least value of \( n \) for which this equality holds is 3.