question 11 (essay worth 12 points)\n(inverse trigonometric functions | trigonometric equations hc)\npart…

Explanation:

Part A (i)

Step1: Identify monotonic intervals

The inverse function of $k$ can be constructed on intervals where $k$ is strictly monotonic (either increasing or decreasing). From the graph, $k(x)$ appears to be $k(x) = 5 \cos(x)$. This function is strictly decreasing on $[0, \pi]$ and strictly increasing on $[\pi, 2\pi]$.

Step2: Determine the length of the intervals

The length of the interval $[0, \pi]$ is $\pi - 0 = \pi$. The length of the interval $[\pi, 2\pi]$ is $2\pi - \pi = \pi$.

Step3: Find the maximum length

The largest interval on which the inverse can be constructed corresponds to half the period of the sinusoidal function. The period is $2\pi$, so half the period is $\pi$. The length of the largest interval is $\pi$.

(ii)

Step1: Determine the equation for k(x)

From the graph, the amplitude is $A = 5$, the vertical shift is $D = 0$, the period is $T = 2\pi$, so $B = \frac{2\pi}{T} = 1$. Since the maximum is at $x=0$, the function is $k(x) = 5 \cos(x)$.

Step2: Restrict the domain for arccosine

To define the inverse $h(x)$ using arccosine, we restrict the domain of $k(x)$ to $[0, \pi]$, where $k(x)$ is one-to-one and its range is $[-5, 5]$. The range of $\arccos(u)$ is $[0, \pi]$.

Step3: Derive the inverse function h(x)

Let $y = k(x) = 5 \cos(x)$. Swap $x$ and $y$: $x = 5 \cos(y)$. Solve for $y$: $\cos(y) = \frac{x}{5}$. Apply arccosine: $y = \arccos\left(\frac{x}{5}\right)$. So, $h(x) = \arccos\left(\frac{x}{5}\right)$.

(iii)

Step1: Relate cosine to sine

Use the identity $\cos(y) = \sin\left(\frac{\pi}{2} - y\right)$. We have $k(x) = 5 \cos(x)$. Let $y = k(x)$. We need an inverse $g(x)$ in terms of arcsine.

Step2: Restrict the domain for arcsine

Consider the inverse $g(x)$ such that its range corresponds to the restricted domain of $k(x)$ used in (ii), which is $[0, \pi]$. Let $y = g(x)$. We want $y \in [0, \pi]$. From $x = 5 \cos(y)$, we use $\cos(y) = \sin(\frac{\pi}{2}-y)$. So $x = 5 \sin(\frac{\pi}{2}-y)$.

Step3: Derive the inverse function g(x)

Solve for $y$: $\sin\left(\frac{\pi}{2} - y\right) = \frac{x}{5}$. Apply arcsine: $\frac{\pi}{2} - y = \arcsin\left(\frac{x}{5}\right)$. Note that for $y \in [0, \pi]$, $\frac{\pi}{2}-y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, which is the range of arcsine. Solve for $y$: $y = \frac{\pi}{2} - \arcsin\left(\frac{x}{5}\right)$. So, $g(x) = \frac{\pi}{2} - \arcsin\left(\frac{x}{5}\right)$.

Part B (i)

Step1: Set up the inverse function derivation

Let $y = f(x) = 4.1 \tan(0.5x)$. To find the inverse $f^{-1}(x)$, swap $x$ and $y$. $x = 4.1 \tan(0.5y)$

Step2: Solve for y

Divide by 4.1: $\tan(0.5y) = \frac{x}{4.1}$. Apply arctangent: $0.5y = \arctan\left(\frac{x}{4.1}\right)$. Multiply by 2: $y = 2 \arctan\left(\frac{x}{4.1}\right)$.

Step3: Write the inverse function

The equation for the inverse function is $f^{-1}(x) = 2 \arctan\left(\frac{x}{4.1}\right)$.

(ii)

Step1: Set up the equation

We need to solve $f(x) = 5$ for $x$ in the interval $(3\pi, 5\pi)$. $4.1 \tan(0.5x) = 5$

Step2: Solve for tan(0.5x)

$\tan(0.5x) = \frac{5}{4.1}$

Step3: Find the general solution for x

$0.5x = \arctan\left(\frac{5}{4.1}\right) + n\pi$, where $n$ is an integer. $x = 2 \arctan\left(\frac{5}{4.1}\right) + 2n\pi$

Step4: Find the solution in the interval (3π, 5π)

Let $\alpha = \arctan\left(\frac{5}{4.1}\right) \approx 0.8834$. So $x \approx 2(0.8834) + 2n\pi = 1.7668 + 2n\pi$. We need $3\pi < x < 5\pi$, which is approximately $9.4248 < x < 15.7080$. For $n=0$, $x \approx 1.7668$. For $n=1$, $x \approx 1.7668 + 2\pi \approx 8.0500$. For $n=2$, $x \approx 1.7668 + 4\pi \approx 14.3332$. This is in the interval. For $n=3$, $x \approx 1.7668 + 6\pi \approx 20.6164$. The value of $x$ is $2 \arctan\left(\frac{5}{4.1}\right) + 4\pi$.

(iii)

Step1: Set up the equation

We need to find all solutions to $f(x) = -5$. $4.1 \tan(0.5x) = -5$

Step2: Solve for tan(0.5x)

$\tan(0.5x) = -\frac{5}{4.1}$

Step3: Find the general solution for x

$0.5x = \arctan\left(-\frac{5}{4.1}\right) + n\pi$, where $n$ is an integer. $x = 2 \arctan\left(-\frac{5}{4.1}\right) + 2n\pi$ Alternatively, since $\arctan(-u) = -\arctan(u)$: $x = -2 \arctan\left(\frac{5}{4.1}\right) + 2n\pi$

Answer:

Part A (i) The length of the largest interval is $\pi$. (ii) $h(x) = \arccos\left(\frac{x}{5}\right)$ (iii) $g(x) = \frac{\pi}{2} - \arcsin\left(\frac{x}{5}\right)$

Part B (i) $f^{-1}(x) = 2 \arctan\left(\frac{x}{4.1}\right)$ (ii) $x = 2 \arctan\left(\frac{5}{4.1}\right) + 4\pi$ (iii) $x = 2 \arctan\left(-\frac{5}{4.1}\right) + 2n\pi$, where $n$ is any integer. (or $x = -2 \arctan\left(\frac{5}{4.1}\right) + 2n\pi$, where $n$ is any integer)