Question 15

Intersection requires sin x + cos x = 1, i.e., √2 sin(x + π/4) = 1, so sin(x + π/4) = 1/√2, giving x + π/4 = π/4 + 2kπ or 3π/4 + 2kπ, i.e., x = 2kπ or x = π/2 + 2kπ. In [0, 10π]: x = 0, 2π, 4π, 6π, 8π, 10π (6 points) and x = π/2, 5π/2, 9π/2, 13π/2, 17π/2 (5 points). Total n = 11.

**Calculating intersections for sine and cosine** I'm figuring out the intersection points for the equation sin(x) + cos(x) = 1 within the range from 0 to 10π. The families of solutions are defined as 2kπ for k from 0 to 5, which gives me six solutions. Then there's also pi/2 + 2kπ for k from 0 to 4, resulting in five more solutions, totaling eleven. I’ll present the final result in JSON format only, as requested.

**Formulating final JSON** I need to solve the question from the image, focusing solely on providing the final JSON without explanation. For Q15, the intersections of curves \( C_1 \) and \( C_2 \) must be determined by solving \( e^{-x} = e^{-x}(\sin x + \cos x) \). After equating and analyzing, I've counted the distinct points in the range [0, 10π]. The total points calculated are 11. Therefore, the final answer in JSON format is: {"answer":11,"unanswered":false}. **Finalizing JSON** I need to confirm the final counts at the endpoints, specifically checking if both 0 and 10π are included. Yes, at \( x = 10π \), the equation \( \sin(0) + \cos(1) \) holds true. I already determined there are a total of 11 points, including these endpoints. To wrap it up, I will present the final answer in JSON format without any additional strings: {"answer":11,"unanswered":false}.

This simplifies to sinx+cosx=1, which gives √2 sin(x+π/4)=1, so sin(x+π/4)=1/√2, yielding x=2kπ or x=π/2+2kπ. Now I need to count these solutions within [0,10π]. For x=2kπ: k=0..5 gives 6 values (0,2π,...,10π). For x=π/2+2kπ: k must satisfy 2kπ≤19π/2, so k=0..4, giving 5 values. Total count n=11. Looking at this problem, I need to find where $e^{-x} = e^{-x}(\sin x + \cos x)$ for $x \in [0, 10\pi]$. **Setting up the equation:** Since $e^{-x} \neq 0$ for all $x$, I can divide both sides: $$1 = \sin x + \cos x$$ **Solving the trigonometric equation:** Using the identity $\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$: $$\sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1 \implies \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$ This gives: $$x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$$ So the solutions are: $$x = 2k\pi \quad \text{or} \quad x = \frac{\pi}{2} + 2k\pi$$ **Finding solutions in $[0, 10\pi]$:** **Case 1:** $x = 2k\pi$ Need $0 \le 2k\pi \le 10\pi \implies 0 \le k \le 5$ This gives $k = 0,1,2,3,4,5$: solutions $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$ — **6 solutions** **Case 2:** $x = \frac{\pi}{2} + 2k\pi$ Need $0 \le \frac{\pi}{2} + 2k\pi \le 10\pi \implies -\frac{1}{4} \le k \le \frac{19}{4} = 4.75$ This gives $k = 0,1,2,3,4$: solutions $x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}, \frac{17\pi}{2}$ — **5 solutions** **Checking for overlap:** The two families $2k\pi$ and $\frac{\pi}{2}+2k\pi$ never coincide since they differ by $\frac{\pi}{2}$ (mod $2\pi$). **Total count:** $$n = 6 + 5 = 11$$ ```json