1. **State the problem:** We have three functions: - $f(x)=\frac{4x+5}{1-2x}$ - $g(x)=3x+5$ - $h(x)=px^2+q-1$ We need to find values and expressions as per the sub-questions. 2. **(a)(i) Find $f\left(-\frac{1}{2}\right)$:** $$f\left(-\frac{1}{2}\right)=\frac{4\left(-\frac{1}{2}\right)+5}{1-2\left(-\frac{1}{2}\right)}=\frac{-2+5}{1+1}=\frac{3}{2}$$ 3. **(a)(ii) Find $g(-2)$:** $$g(-2)=3(-2)+5=-6+5=-1$$ 4. **(a)(iii) Find $q$ given $h(0)=-4$:** Substitute $x=0$ into $h(x)$: $$h(0)=p(0)^2+q-1=q-1=-4$$ Solve for $q$: $$q-1=-4 \implies q=-3$$ 5. **(a)(iv) Find $x$ making $f$ undefined:** $f$ is undefined when the denominator is zero: $$1-2x=0 \implies x=\frac{1}{2}$$ 6. **(b)(i) Find $f^{-1}(x)$ (inverse of $f$):** Start with $y=\frac{4x+5}{1-2x}$ and solve for $x$ in terms of $y$. $$y=\frac{4x+5}{1-2x} \implies y(1-2x)=4x+5$$ $$y - 2xy = 4x + 5$$ Group $x$ terms: $$y - 5 = 4x + 2xy = x(4+2y)$$ Solve for $x$: $$x=\frac{y-5}{4+2y}$$ Thus, $$f^{-1}(x)=\frac{x-5}{4+2x}$$ 7. **(b)(ii) Find $(fg)(x) = f(g(x))$:** Substitute $g(x)$ into $f$: $$f(g(x))=f(3x+5) = \frac{4(3x+5)+5}{1-2(3x+5)} = \frac{12x+20+5}{1 - 6x - 10} = \frac{12x+25}{-6x - 9}$$ Simplify denominator: $$-6x -9 = -3(2x +3)$$ So, $$(fg)(x) = \frac{12x+25}{-6x - 9}$$ 8. **(c) Find $x$ such that $f^{-1}(x) = -1$:** Substitute and solve: $$-1 = \frac{x-5}{4 + 2x}$$ Multiply both sides: $$-1(4 + 2x) = x - 5$$ $$-4 - 2x = x - 5$$ $$-4 + 5 = x + 2x$$ $$1 = 3x \implies x = \frac{1}{3}$$ **Final answers:** - $f\left(-\frac{1}{2}\right) = \frac{3}{2}$ - $g(-2) = -1$ - $q = -3$ - $f$ undefined at $x=\frac{1}{2}$ - $f^{-1}(x) = \frac{x-5}{4+2x}$ - $(fg)(x) = \frac{12x+25}{-6x - 9}$ - $f^{-1}(x) = -1$ when $x=\frac{1}{3}$