The function f is defined by $$f(x) = 4 + 3 imes 2^{-x}, ext{ where } x ext{ is a real number}$$ 10 (a) Using set notation, state the range of f 10 (b) (i) Using set notation, state the domain of f^{-1} 10 (b) (ii) Find an expression for f^{-1}(x) 10 (c) The function g is defined by $$g(x) = 5 - rac{1}{ ext{√}x}, ext{ where } x ext{ is a real number: } x > 0$$ 10 (c) (i) Find an expression for gf(x) 10 (c) (ii) Solve the equation gf(x) = 2, giving your answer in an exact form. - AQA - A-Level Maths: Mechanics - Question 10 - 2017 - Paper 1

The range of the function $f$ serves as the domain for the inverse function $f^{-1}$. Therefore, the domain of $f^{-1}$ is:

$\{ x : x > 4, x \in \mathbb{R} \}$.

To find $f^{-1}(x)$, we start with the equation:

$y = 4 + 3 \times 2^{-x}$

We can solve this for $x$:

1. Rearranging gives: $y - 4 = 3 \times 2^{-x}$
2. Dividing by 3: $\frac{y - 4}{3} = 2^{-x}$
3. Taking logarithms: $-x = \log_2\left(\frac{y - 4}{3}\right)$
4. Thus, $x = -\log_2\left(\frac{y - 4}{3}\right)$

Hence, the expression for $f^{-1}(x)$ is:

$f^{-1}(x) = -\log_2\left(\frac{x - 4}{3}\right)$.

To find $gf(x)$, we start with:

$g(f(x)) = g(4 + 3 \times 2^{-x})$

Substituting this into the expression for $g(x)$:

$g(f(x)) = 5 - \frac{1}{\sqrt{f(x)}}$ The expression becomes: $gf(x) = 5 - \frac{1}{\sqrt{4 + 3 \times 2^{-x}}}$.

To solve $gf(x) = 2$:

1. Setting up the equation: $5 - \frac{1}{\sqrt{4 + 3 \times 2^{-x}}} = 2$
2. Rearranging gives: $\frac{1}{\sqrt{4 + 3 \times 2^{-x}}} = 3$
3. Inverting both sides: $\sqrt{4 + 3 \times 2^{-x}} = \frac{1}{3}$
4. Squaring both sides leads to: $4 + 3 \times 2^{-x} = \frac{1}{9}$
5. Rearranging gives: $3 \times 2^{-x} = \frac{1}{9} - 4$ Thus, $\frac{1}{9} - 4 = -\frac{35}{9}$ $6 \times 2^{-x} = -\frac{35}{27}$
6. Solving for $x$ involves logarithms and leads to: $x = -\log_2\left(\frac{35}{54}\right)$.