1. **State the problem:** We have two curves: - Curve C1: $y = \sin\left(x + \frac{\pi}{8}\right)$ - Curve C2: $y = -5\sin\left(x + \frac{\pi}{8}\right)$ We need to find: (a) The $x$-coordinate of the minimum turning point $P$ on C1 in $0 \leq x < 2\pi$. (b) The number of roots of $\sin\left(x + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right)$ in $0 \leq x < 101\pi$. (c) The coordinates of the minimum turning point $Q$ on C2 in $0 \leq x < 2\pi$. 2. **Formula and rules:** - The minimum turning points of $y = \sin u$ occur where $u = \frac{3\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$. - The sine function has period $2\pi$. - The equation $\sin A = \sin B$ has solutions $A = B + 2k\pi$ or $A = \pi - B + 2k\pi$. 3. **(a) Find $x$-coordinate of minimum turning point $P$ on C1:** Let $u = x + \frac{\pi}{8}$. Minimum turning points satisfy: $$u = \frac{3\pi}{2} + 2k\pi$$ For $0 \leq x < 2\pi$, find $k$ such that $x$ is in this interval. Solve for $x$: $$x = u - \frac{\pi}{8} = \frac{3\pi}{2} + 2k\pi - \frac{\pi}{8} = \frac{12\pi}{8} - \frac{\pi}{8} + 2k\pi = \frac{11\pi}{8} + 2k\pi$$ For $k=0$, $$x = \frac{11\pi}{8} \approx 4.32$$ Since $0 \leq x < 2\pi \approx 6.28$, this is valid. **Answer (a):** $x = \frac{11\pi}{8}$ 4. **(b) Number of roots of $\sin\left(x + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right)$ in $0 \leq x < 101\pi$:** Set $u = x + \frac{\pi}{8}$. Equation becomes: $$\sin u = \sin \frac{\pi}{8}$$ Solutions for $u$ are: $$u = \frac{\pi}{8} + 2k\pi \quad \text{or} \quad u = \pi - \frac{\pi}{8} + 2k\pi = \frac{7\pi}{8} + 2k\pi$$ We want $x$ in $[0, 101\pi)$, so: $$0 \leq x < 101\pi \implies 0 \leq u - \frac{\pi}{8} < 101\pi \implies \frac{\pi}{8} \leq u < 101\pi + \frac{\pi}{8}$$ Count integer $k$ such that $u$ is in this range for each solution: For $u = \frac{\pi}{8} + 2k\pi$: $$\frac{\pi}{8} \leq \frac{\pi}{8} + 2k\pi < 101\pi + \frac{\pi}{8}$$ Subtract $\frac{\pi}{8}$: $$0 \leq 2k\pi < 101\pi$$ Divide by $2\pi$: $$0 \leq k < \frac{101\pi}{2\pi} = 50.5$$ So $k = 0,1,2,\ldots,50$ (51 values). For $u = \frac{7\pi}{8} + 2k\pi$: $$\frac{\pi}{8} \leq \frac{7\pi}{8} + 2k\pi < 101\pi + \frac{\pi}{8}$$ Subtract $\frac{7\pi}{8}$: $$\frac{\pi}{8} - \frac{7\pi}{8} \leq 2k\pi < 101\pi + \frac{\pi}{8} - \frac{7\pi}{8}$$ $$-\frac{6\pi}{8} \leq 2k\pi < 101\pi - \frac{6\pi}{8}$$ $$-\frac{3\pi}{4} \leq 2k\pi < 101\pi - \frac{3\pi}{4}$$ Divide by $2\pi$: $$-\frac{3}{8} \leq k < \frac{101\pi - \frac{3\pi}{4}}{2\pi} = \frac{101 - \frac{3}{8}}{2} = \frac{101 - 0.375}{2} = \frac{100.625}{2} = 50.3125$$ Since $k$ is integer, $k = 0,1,2,\ldots,50$ (51 values). Total roots = $51 + 51 = 102$. **Answer (b):** 102 roots because each solution branch contributes 51 roots in the interval. 5. **(c) Coordinates of minimum turning point $Q$ on C2:** Curve C2: $y = -5\sin\left(x + \frac{\pi}{8}\right)$ Minimum turning points of $y = -5\sin u$ occur where $-5\sin u$ is minimum. Since $\sin u$ minimum is $-1$, $-5\sin u$ maximum is $5$, minimum is $-5 \times (-1) = 5$. Minimum turning points of $y = -5\sin u$ correspond to maximum points of $\sin u$. Maximum points of $\sin u$ occur at: $$u = \frac{\pi}{2} + 2k\pi$$ For $0 \leq x < 2\pi$: $$x = u - \frac{\pi}{8} = \frac{\pi}{2} + 2k\pi - \frac{\pi}{8} = \frac{4\pi}{8} - \frac{\pi}{8} + 2k\pi = \frac{3\pi}{8} + 2k\pi$$ For $k=0$, $$x = \frac{3\pi}{8}$$ Calculate $y$ at $x = \frac{3\pi}{8}$: $$y = -5 \sin\left(\frac{3\pi}{8} + \frac{\pi}{8}\right) = -5 \sin\left(\frac{4\pi}{8}\right) = -5 \sin\left(\frac{\pi}{2}\right) = -5 \times 1 = -5$$ **Answer (c):** Coordinates of $Q$ are $\left(\frac{3\pi}{8}, -5\right)$.