SECTION B Question 1 Suppose the functions \( f, g, h, r \) and \( \ell \) are defined as follows: \[ \begin{array}{l} f(x)=\frac{1}{3} \log _{3} \frac{1}{4}+\log _{3} x \\ g(x)=\sqrt{(x-3)^{2}} \\ h(x)=5 x-2 x^{2} \\ r(x)=2^{1 x+1}-2^{x-2} \\ \ell(x)=\frac{1}{\sqrt{x}} \end{array} \] 1.1 Write down \( D_{f} \), the domain of \( f \) and then solve the equation \( f(x)=-\log _{\frac{1}{1}} \sqrt[3]{x} \). 1.2 Write down \( D_{p} \) and then solve the equation \( g(x)=\frac{x}{2} \). 1.3 Write down \( D_{h} \) and then solve the inequality \( 2 \geq h(x) \). 1.4 Write down \( D_{r} \) and then solve the equation \( r(x)=0 \). 1.5 Write down \( D_{\text {re }} \) without first calculating \( (r \cdot \ell)(x) \). 1.6 Write down \( D_{\frac{1}{n}} \) without first calculating \( \frac{1}{9}(x) \). [30]

Let's solve each part of the question step by step. ### 1.1 Domain of \( f \) and solving \( f(x) = -\log_{\frac{1}{1}} \sqrt[3]{x} \) **Domain of \( f \):** The function \( f(x) = \frac{1}{3} \log_{3} \frac{1}{4} + \log_{3} x \) is defined when \( x > 0 \) because the logarithm is only defined for positive values. Thus, the domain \( D_f \) is: \[ D_f = (0, \infty) \] **Solving the equation:** We need to solve: \[ f(x) = -\log_{\frac{1}{1}} \sqrt[3]{x} \] Since \( \log_{\frac{1}{1}} \) is undefined, we will assume it was meant to be \( -\log_{3} \sqrt[3]{x} \). So we rewrite the equation: \[ \frac{1}{3} \log_{3} \frac{1}{4} + \log_{3} x = -\log_{3} \sqrt[3]{x} \] Using the property of logarithms, we can express \( -\log_{3} \sqrt[3]{x} \) as: \[ -\log_{3} x^{1/3} = -\frac{1}{3} \log_{3} x \] Thus, the equation becomes: \[ \frac{1}{3} \log_{3} \frac{1}{4} + \log_{3} x = -\frac{1}{3} \log_{3} x \] Combining the logarithmic terms: \[ \frac{1}{3} \log_{3} \frac{1}{4} + \frac{4}{3} \log_{3} x = 0 \] Multiplying through by 3: \[ \log_{3} \frac{1}{4} + 4 \log_{3} x = 0 \] This simplifies to: \[ 4 \log_{3} x = -\log_{3} \frac{1}{4} \] Using the property of logarithms: \[ \log_{3} x^4 = \log_{3} 4 \] Thus: \[ x^4 = 4 \implies x = \sqrt[4]{4} = 2^{1/2} = \sqrt{2} \] ### 1.2 Domain of \( g \) and solving \( g(x) = \frac{x}{2} \) **Domain of \( g \):** The function \( g(x) = \sqrt{(x-3)^{2}} \) is defined for all real numbers since the square root is defined for all real values. Thus, the domain \( D_g \) is: \[ D_g = (-\infty, \infty) \] **Solving the equation:** We need to solve: \[ g(x) = \frac{x}{2} \] This simplifies to: \[ |x - 3| = \frac{x}{2} \] This gives us two cases to consider: **Case 1:** \( x - 3 = \frac{x}{2} \) \[ 2(x - 3) = x \implies 2x - 6 = x \implies x = 6 \] **Case 2:** \( x - 3 = -\frac{x}{2} \) \[ 2(x - 3) = -x \implies 2x - 6 = -x \implies 3x = 6 \implies x = 2 \] Thus, the solutions are \( x = 6 \) and \( x = 2 \). ### 1.3 Domain of \( h \) and solving \( 2 \geq h(x) \) **Domain of \( h \):** The function \( h(x) = 5x - 2x^2 \) is a polynomial and is defined for all real numbers. Thus, the domain \( D_h \) is: \[ D_h = (-\infty, \infty) \] **Solving the inequality:** We need to solve: \[ 2 \geq 5x - 2x^2 \] Rearranging gives: \[ 2x^2 - 5x + 2 \geq 0 \] We can find the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \] This gives us the roots: \[ x = 2 \quad \text{and} \quad x = \frac{1}{2} \] Now we test intervals around the roots \( \frac{1}{2} \) and \( 2 \): - For \( x < \frac{1}{2} \), choose \( x = 0 \): \( 2(0)^2 - 5(0) + 2 = 2 \geq 0 \) (True) - For \( \frac{1}{2} < x < 2 \), choose \( x = 1 \): \( 2(1)^2 - 5(1) + 2 = -1 \geq 0 \) (False) - For \( x > 2 \), choose \( x = 3 \): \( 2(3)^2 - 5(3) + 2 = 2 \geq 0 \) (True) Thus, the solution to the inequality is: \[ x \in (-\infty, \frac{1}{2}] \cup [2, \infty) \] ### 1.4 Domain of \( r \) and solving \( r(x) = 0 \) **Domain of \( r \):** The function \( r(x) = 2^{1x+1} - 2^{x-2} \) is defined for all real numbers since it is an exponential function. Thus, the domain \( D_r \) is: \[ D_r = (-\infty, \infty) \] **Solving the equation:** We need to solve: \[ r(x) = 0 \implies 2^{1x+1} - 2^{x-2} = 0 \] This can be rewritten as: \[ 2^{x+1} = 2^{x-2} \] Since the bases are the same