The problem presents three parts, A, B, and C, involving functions $g(x)$, $h(x)$, $j(x)$, $k(x)$, and $m(x)$. (A) We are given $g(x) = 3\ln x - \frac{1}{2}\ln x$ and $h(x) = \frac{\sin^2 x - 1}{\cos x}$. We need to rewrite $g(x)$ as a single natural logarithm and rewrite $h(x)$ in terms of $\cos x$ only. (B) We are given $j(x) = 2\sin x \cos x - \cos x$ and $k(x) = 8e^{3x} - e$. We need to solve $j(x) = 0$ for $x$ in $[0, \frac{\pi}{2}]$ and $k(x) = 3e$ for $x$. (C) We are given $m(x) = \cos(2x) + 4$. We need to find the values of $x$ such that $m(x) = \frac{9}{2}$.

1. Problem Description

The problem presents three parts, A, B, and C, involving functions $g(x)$, $h(x)$, $j(x)$, $k(x)$, and $m(x)$.

(A) We are given $g(x) = 3\ln x - \frac{1}{2}\ln x$ and $h(x) = \frac{\sin^2 x - 1}{\cos x}$. We need to rewrite $g(x)$ as a single natural logarithm and rewrite $h(x)$ in terms of $\cos x$ only.

(B) We are given $j(x) = 2\sin x \cos x - \cos x$ and $k(x) = 8e^{3x} - e$. We need to solve $j(x) = 0$ for $x$ in $[0, \frac{\pi}{2}]$ and $k(x) = 3e$ for $x$.

(C) We are given $m(x) = \cos(2x) + 4$. We need to find the values of $x$ such that $m(x) = \frac{9}{2}$.

2. Solution Steps

(A)(i)

We need to rewrite $g(x) = 3\ln x - \frac{1}{2}\ln x$ as a single natural logarithm.

$g(x) = 3\ln x - \frac{1}{2}\ln x = \ln(x^3) - \ln(\sqrt{x}) = \ln\left(\frac{x^3}{\sqrt{x}}\right) = \ln\left(\frac{x^3}{x^{1/2}}\right) = \ln(x^{3 - 1/2}) = \ln(x^{5/2})$

(A)(ii)

We need to rewrite $h(x) = \frac{\sin^2 x - 1}{\cos x}$ in terms of $\cos x$ only.

Since $\sin^2 x + \cos^2 x = 1$, we have $\sin^2 x - 1 = -\cos^2 x$.

Therefore, $h(x) = \frac{-\cos^2 x}{\cos x} = -\cos x$.

(B)(i)

We need to solve $j(x) = 2\sin x \cos x - \cos x = 0$ for $x$ in $[0, \frac{\pi}{2}]$.

$j(x) = \cos x (2\sin x - 1) = 0$.

So either $\cos x = 0$ or $2\sin x - 1 = 0$, which means $\sin x = \frac{1}{2}$.

If $\cos x = 0$, then $x = \frac{\pi}{2}$.

If $\sin x = \frac{1}{2}$, then $x = \frac{\pi}{6}$.

Both solutions are in the interval $[0, \frac{\pi}{2}]$.

(B)(ii)

We need to solve $k(x) = 8e^{3x} - e = 3e$ for $x$.

$8e^{3x} - e = 3e$

$8e^{3x} = 4e$

$e^{3x} = \frac{4e}{8} = \frac{e}{2}$

Taking the natural logarithm of both sides:

$3x = \ln\left(\frac{e}{2}\right) = \ln e - \ln 2 = 1 - \ln 2$

$x = \frac{1 - \ln 2}{3}$

(C)

We need to find $x$ such that $m(x) = \cos(2x) + 4 = \frac{9}{2}$.

$\cos(2x) + 4 = \frac{9}{2}$

$\cos(2x) = \frac{9}{2} - 4 = \frac{9}{2} - \frac{8}{2} = \frac{1}{2}$

$2x = \cos^{-1}\left(\frac{1}{2}\right)$

$2x = \frac{\pi}{3} + 2\pi n$ or $2x = -\frac{\pi}{3} + 2\pi n$ for any integer $n$.

$x = \frac{\pi}{6} + \pi n$ or $x = -\frac{\pi}{6} + \pi n$ for any integer $n$.

3. Final Answer

(A)(i) $\ln(x^{5/2})$

(A)(ii) $-\cos x$

(B)(i) $x = \frac{\pi}{6}, \frac{\pi}{2}$

(B)(ii) $x = \frac{1 - \ln 2}{3}$

(C) $x = \frac{\pi}{6} + n\pi$, $x = -\frac{\pi}{6} + n\pi$ where $n$ is any integer.