Question

If m and n respectively are the number of local maximum and local minimum points of the function
\(f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt\), then the ordered pair (m, n) is equal to

• A. (3, 2)
• B. (2, 3)
• C. (2, 2)
• D. (3, 4)

Answer 1

The Correct Option is B

Let's analyze the given function \( f(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt \). We need to find the critical points where the derivative \( f'(x) = 0 \) to determine the local maxima and minima.

According to the Fundamental Theorem of Calculus, the derivative of \( f(x) \) is given by:

f'(x) = \frac{d}{dx} \left( \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt \right) = \frac{d}{dx} \left[ F(x^2) - F(0) \right]

Where \( F(x) \) is an antiderivative of the integrand. The derivative will be:

f'(x) = \frac{d}{dx} \left[ F(x^2) \right] = 2x \cdot \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}}

This simplifies to:

f'(x) = \frac{2x(x^4 - 5x^2 + 4)}{2 + e^{x^2}}

The critical points occur where \( f'(x) = 0 \), which means solving:

2x(x^4 - 5x^2 + 4) = 0

This gives us two cases:

1. x = 0
2. x^4 - 5x^2 + 4 = 0

To solve the quadratic in \( x^2 \), let \( y = x^2 \), then:

y^2 - 5y + 4 = 0

Factoring this quadratic:

(y - 1)(y - 4) = 0

This gives:

1. y = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1
2. y = 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2

Thus, the critical points are \( x = 0, \pm 1, \pm 2 \).

To determine the nature of each critical point (i.e., local maximum or minimum), we need to check the sign changes of \( f'(x) \) around these points:

• At \( x = 0 \): Check \( f'(x) \) around \( x \) close to 0. Since this is not changing sign, it can be ignored for local extremas.
• At \( x = \pm 1 \): \( f'(x) \) changes sign from positive to negative, which indicates a local maximum.
• At \( x = \pm 2 \): \( f'(x) \) changes sign from negative to positive, which indicates a local minimum.

Therefore, two local maxima at \( x = 1 \) and \( x = -1 \), and three local minima at \( x = -2, 0, 2 \).

The ordered pair of the number of local maxima \( m \) and local minima \( n \) is \( (2, 3) \).