Question

If a number is drawn at random from the set $\{1, 3, 5, 7, \dots, 59\}$, then the probability that it lies in the interval in which the function $f(x) = x^3 - 16x^2 + 20x - 5$ is strictly decreasing is:

• A. $\frac{1}{5}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Hint:

To determine the intervals where a function is strictly decreasing, find the derivative and solve the inequality $f'(x)<0$. Then, count the number of favorable outcomes in the given set and calculate the probability.

Answer 1

The Correct Option is D

We are given the function $f(x) = x^3 - 16x^2 + 20x - 5$, and we are asked to find the probability that a number drawn at random from the set $\{1, 3, 5, 7, \dots, 59\}$ lies in the interval where the function is strictly decreasing. 
Step 1: The function $f(x)$ is strictly decreasing where its derivative is negative. We first find the derivative of $f(x)$: $$f'(x) = 3x^2 - 32x + 20$$ Step 2: To find the intervals where the function is strictly decreasing, we solve the inequality $f'(x)<0$. First, solve the equation $f'(x) = 0$ to find the critical points: $$3x^2 - 32x + 20 = 0$$ Using the quadratic formula: $$x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(3)(20)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 240}}{6} = \frac{32 \pm \sqrt{784}}{6} = \frac{32 \pm 28}{6}$$ Thus, the critical points are: $$x = \frac{32 + 28}{6} = 10 \quad {and} \quad x = \frac{32 - 28}{6} = \frac{4}{6} = \frac{2}{3}$$ Step 3: Now, to determine the sign of $f'(x)$, we test the intervals formed by the critical points: $(-\infty, \frac{2}{3})$, $(\frac{2}{3}, 10)$, and $(10, \infty)$.
- For $x \in (\frac{2}{3}, 10)$, the function is strictly decreasing, as $f'(x)<0$.
Step 4: We are interested in the set $\{1, 3, 5, 7, \dots, 59\}$. The numbers in this set correspond to the odd integers between 1 and 59. These numbers are: $$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59$$ Thus, the total number of elements in the set is 30. 
Step 5: The function is strictly decreasing in the interval $(1, 10)$, which corresponds to the odd numbers $3, 5, 7, 9$. Thus, there are 4 numbers in the set where the function is strictly decreasing.
Step 6: The probability that a randomly selected number from the set lies in the interval where the function is strictly decreasing is the ratio of favorable outcomes to total outcomes: $$P(strictly decreasing) = \frac{4}{30} = \frac{2}{15}$$ Thus, the correct answer is $\frac{1}{6}$.