Question

Consider the curve \( C_1 \) given by \( y = e^{-x} \quad \text{for } x \in [0, 10\pi], \) and the curve \( C_2 \) given by \(y = e^{-x}(\sin x + \cos x) \quad \text{for x \in [0, 10\pi]. \) Let \( n \) be the total number of points of intersection of the curves \( C_1 \) and \( C_2 \).
Suppose that \( \alpha_1, \alpha_2, \dots, \alpha_n \in [0, 10\pi] \) are the \( x \)-coordinates of the points of intersection of the curves \( C_1 \) and \( C_2 \) such that \(\alpha_1 < \alpha_2 < \dots < \alpha_n. \)} 15. Then the value of $n$ is {

Hint:

When solving equations like $\sin x + \cos x = 1$ over large intervals, use the periodic nature of the trigonometric functions. There are exactly 2 solutions in each $2\pi$ interval, but check the boundaries of the closed interval carefully.

Answer 1

Correct Answer: 11

Step 1: Understanding the Question:
We need to find the number of solutions for the equation formed by equating the $y$-expressions of two curves in the interval $[0, 10\pi]$.

Step 2: Key Formula or Approach:

• Equate $e^{-x} = e^{-x}(\sin x + \cos x)$.

• Since $e^{-x} > 0$ for all real $x$, we can divide by $e^{-x}$.

Step 3: Detailed Explanation:

• Equation: $e^{-x} = e^{-x}(\sin x + \cos x) \implies 1 = \sin x + \cos x$.

• Express $\sin x + \cos x$ as $\sqrt{2} \sin(x + \pi/4)$.

• $\sqrt{2} \sin(x + \pi/4) = 1 \implies \sin(x + \pi/4) = \frac{1}{\sqrt{2}}$.

• General solutions: $x + \pi/4 = 2k\pi + \pi/4$ or $x + \pi/4 = 2k\pi + 3\pi/4$.

• Case 1: $x = 2k\pi$.
For $x \in [0, 10\pi]$, $k \in \{0, 1, 2, 3, 4, 5\}$.
Values: $0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$. (6 points).

• Case 2: $x = 2k\pi + \pi/2$.
For $x \in [0, 10\pi]$, $k \in \{0, 1, 2, 3, 4\}$.
Values: $\pi/2, 2.5\pi, 4.5\pi, 6.5\pi, 8.5\pi$. (5 points).

• Total points $n = 6 + 5 = 11$.

Step 4: Final Answer:
The total number of intersection points $n$ is 11.