The point \( P(1,0) \) lies on the curve \( y=\sin \left(\frac{14 \pi}{x}\right) \). (a) If \( Q \) is the point \( \left(x, \sin \left(\frac{14 \pi}{x}\right)\right) \), find the slope of the secant line \( P Q \) (correct to four decimal places) for the following values o (i) 2 (ii) 1.5 (iii) 1.4 (iv) 1.3 (v) 1.2 (xi) 0.9 (vi) 0.5 (vi) 0.8 (viii) 0.6 (vi) 0.7 0

To find the slope of the secant line \( PQ \) for the points \( P(1, 0) \) and \( Q\left(x, \sin\left(\frac{14\pi}{x}\right)\right) \), we use the formula for the slope of a secant line: \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sin\left(\frac{14\pi}{x}\right) - 0}{x - 1} = \frac{\sin\left(\frac{14\pi}{x}\right)}{x - 1} \] We will calculate this slope for the specified values of \( x \). ### (i) \( x = 2 \) \[ \text{slope} = \frac{\sin\left(\frac{14\pi}{2}\right)}{2 - 1} = \frac{\sin(7\pi)}{1} = 0 \] ### (ii) \( x = 1.5 \) \[ \text{slope} = \frac{\sin\left(\frac{14\pi}{1.5}\right)}{1.5 - 1} = \frac{\sin\left(\frac{28\pi}{3}\right)}{0.5} \] Calculating \( \frac{28\pi}{3} \) modulo \( 2\pi \): \[ \frac{28\pi}{3} = 9\pi + \frac{1\pi}{3} \quad \text{(since } 9\pi \text{ is } 4.5 \text{ full cycles of } 2\pi\text{)} \] Thus, \( \sin\left(\frac{28\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). So, \[ \text{slope} = \frac{\frac{\sqrt{3}}{2}}{0.5} = \sqrt{3} \approx 1.7321 \] ### (iii) \( x = 1.4 \) \[ \text{slope} = \frac{\sin\left(\frac{14\pi}{1.4}\right)}{1.4 - 1} = \frac{\sin(10\pi)}{0.4} = \frac{0}{0.4} = 0 \] ### (iv) \( x = 1.3 \) \[ \text{slope} = \frac{\sin\left(\frac{14\pi}{1.3}\right)}{1.3 - 1} = \frac{\sin\left(\frac{14\pi}{1.3}\right)}{0.3} \] Calculating \( \frac{14\pi}{1.3} \): \[ \frac{14\pi}{1.3} \approx 34.9 \quad \text{(approximately)} \] Finding \( 34.9 \mod 2\pi \): \[ 34.9 \div 6.2832 \approx 5.56 \quad \Rightarrow \quad 34.9 - 5 \cdot 2\pi \approx 34.9 - 31.4 \approx 3.5 \] Thus, \( \sin(34.9) \approx -0.3508 \). So, \[ \text{slope} \approx \frac{-0.3508}{0.3} \approx -1.1693 \] ### (v) \( x = 1.2 \) \[ \text{slope} = \frac{\sin\left(\frac{14\pi}{1.2}\right)}{1.2 - 1} = \frac{\sin(11.6667\pi)}{0.2} \] Calculating \( 11.6667 \mod 2\pi \): \[ 11.6667 \div 6.2832 \approx 1.86 \quad \Rightarrow \quad 11.6667 - 1 \cdot 2\pi \approx 11.6667 - 6.2832 \approx 5.3835 \] Thus, \( \sin(5.3835) \approx -0.7071 \). So, \[ \text{slope} \approx \frac{-0.7071}{0.2} \approx -3.5355 \] ### (vi) \( x =