A-Level Edexcel Maths Pure Graphs of Functions: The function $f$ is defined by

The function $f$ is defined by $$f(x) = \frac{6}{2x + 5} + \frac{2}{2x - 5} + \frac{60}{4x^2 - 25}, \quad x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5

Question 4

$The-function-$f$-is-defined-by--$$f(x)-=-\frac{6}{2x-+-5}-+-\frac{2}{2x---5}-+-\frac{60}{4x^2---25},-\quad-x->-4.$$--(a)-Show-that-$f(x)-=-\frac{A}{Bx-+-C}$-where-$A$,-$B$-and-$C$-are-constants-to-be-found-Edexcel-A-Level Maths Pure-Question 4-2018-Paper 5.png$

The function $f$ is defined by $$f(x) = \frac{6}{2x + 5} + \frac{2}{2x - 5} + \frac{60}{4x^2 - 25}, \quad x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A... show full transcript

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = \frac{6}{2x + 5} + \frac{2}{2x - 5} + \frac{60}{4x^2 - 25}, \quad x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5

Step 1

Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found.

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Answer

To show the relationship, we first need to combine the fractions in $f(x)$ :

Identify a common denominator for the three fractions:
- The common denominator will be $(2x + 5)(2x - 5)(4x^2 - 25)$ , which can be factored as $(2x + 5)(2x - 5)(2x + 5)(2x - 5)$ .
Rewrite the fractions with this common denominator:

$f(x) = \frac{6(2x - 5)(4x^2 - 25) + 2(2x + 5)(4x^2 - 25) + 60(2x + 5)(2x - 5)}{(2x + 5)(2x - 5)(4x^2 - 25)}$
Simplify the numerator:
- Upon simplifying, we should be able to factor the numerator to match the form $f(x) = \frac{A}{Bx + C}$ , leading to the identification of constants $A$ , $B$ , and $C$ . For instance, this gives: $f(x) = \frac{8}{(2x + 5)(2x - 5)}$ where $A = 8$ , $B = 2$ , and $C = -5$ .

Step 2

Find $f^{-1}(x)$ and state its domain.

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Answer

To find the inverse $f^{-1}(x)$ , we follow these steps:

Set the equation $y = f(x)$ :

$y = \frac{8}{2x - 5}$
Swap $x$ and $y$ to get the inverse:

$x = \frac{8}{2y - 5}$
Solve for $y$ :
- Rearranging gives: $2y - 5 = \frac{8}{x} \implies 2y = \frac{8}{x} + 5 \implies y = \frac{8}{2x} + \frac{5}{2}$
Thus, the inverse function is: $f^{-1}(x) = \frac{4}{x} + \frac{5}{2}$

Now, to state the domain of $f^{-1}(x)$ :

The original function $f(x)$ has a domain for $x > 4$ . Thus, the inverse function has a range corresponding to these values, which implies the domain of $f^{-1}(x)$ is: $x \in \left(0, \frac{8}{3} \right)$ , noting that $f^{-1}(x)$ is not valid for $x = 0$ .

The function $f$ is defined by $$f(x) = \frac{6}{2x + 5} + \frac{2}{2x - 5} + \frac{60}{4x^2 - 25}, \quad x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = \frac{6}{2x + 5} + \frac{2}{2x - 5} + \frac{60}{4x^2 - 25}, \quad x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5

Show that $f(x) = \frac{A}{Bx + C}$ where $A$, $B$ and $C$ are constants to be found.

Find $f^{-1}(x)$ and state its domain.

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