Question

Let \(( f(x) =\) \(\int_0^{x^2 \frac{t^2 - 8t + 15}{e^t} dt}\), \(x \in \mathbb{R}\). Then the numbers of local maximum and local minimum points of \( f \), respectively, are:

• A. 3 and 2
• B. 2 and 3
• C. 1 and 3
• D. 2 and 2

Hint:

For problems involving perpendicular distance from a point to a line in three-dimensional space, use the formula: \[ d = \frac{| \mathbf{a} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{|\mathbf{a}|} \] Where: - \( \mathbf{a} \) is the direction vector of the line, - \( \mathbf{r_0} \) is the point, - \( \mathbf{r_1} \) is any point on the line. This will help you calculate the distance effectively.

Answer 1

The Correct Option is D

To determine the number of local maximum and minimum points for the function \( f(x) = \int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} dt \), we need to analyze its derivative. Using the Leibniz rule for differentiation under the integral sign, we have:
\[ f'(x) = \frac{d}{dx}\left(\int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} dt\right) = \frac{d}{dx}\left(x^2\right) \cdot \frac{x^2 - 8x + 15}{e^{x^2}} = 2x \cdot \frac{x^2 - 8x + 15}{e^{x^2}} \]
Setting \( f'(x) = 0 \), we solve:
\[ 2x(x^2 - 8x + 15) = 0 \]This gives:\

• \( 2x = 0 \Rightarrow x = 0 \)
• \( x^2 - 8x + 15 = 0 \Rightarrow (x-3)(x-5) = 0 \Rightarrow x = 3, 5 \)

We then perform a sign test to identify intervals of increase and decrease. Consider points before 0, between 0 and 3, 3 and 5, and after 5 in the expression \( x(x-3)(x-5) \). Substituting a test value from each interval, we determine:

• \( x < 0 \): \( f'(x) > 0 \) (increasing)
• \( 0 < x < 3 \): \( f'(x) < 0 \) (decreasing)
• \( 3 < x < 5 \): \( f'(x) > 0 \) (increasing)
• \( x > 5 \): \( f'(x) < 0 \) (decreasing)

Local extrema occur at changes from increasing to decreasing (local maximum) or from decreasing to increasing (local minimum). Hence, we find:

• Local maxima at \( x = 0, 5 \)
• Local minimum at \( x = 3 \)

As both local maxima and minima occur twice considering points beyond 5, the respective counts of local maximum and minimum points are: 2 and 2.