A-Level Edexcel Maths Pure Modelling with Functions: The function f is defined by f:

The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2}, \quad x > 1.\n (a) Show that f(x) = \frac{2}{x - 1}, \quad x > 1.\n (b) Find f^{-1}(c).\n The function g is defined by g: x ↦ x^2 + 5, \quad x \in \mathbb{R}.\n (b) Solve fg(x) = \frac{1}{4}. - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 5

Question 4

$The-function-f-is-defined-by--f:-x-↦--\frac{5x-+-1}{x^2-+-2x---2},-\quad-x->-1.\n--(a)-Show-that-f(x)-=-\frac{2}{x---1},-\quad-x->-1.\n-(b)-Find-f^{-1}(c).\n--The-function-g-is-defined-by--g:-x-↦-x^2-+-5,-\quad-x-\in-\mathbb{R}.\n--(b)-Solve-fg(x)-=-\frac{1}{4}.-Edexcel-A-Level Maths Pure-Question 4-2005-Paper 5.png$

The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2}, \quad x > 1.\n (a) Show that f(x) = \frac{2}{x - 1}, \quad x > 1.\n (b) Find f^{-1}(c).\n The fu... show full transcript

Worked Solution & Example Answer:The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2}, \quad x > 1.\n (a) Show that f(x) = \frac{2}{x - 1}, \quad x > 1.\n (b) Find f^{-1}(c).\n The function g is defined by g: x ↦ x^2 + 5, \quad x \in \mathbb{R}.\n (b) Solve fg(x) = \frac{1}{4}. - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 5

Step 1

Show that f(x) = \frac{2}{x - 1}, \quad x > 1.

96%

114 rated

Answer

To show that ( f(x) = \frac{2}{x - 1} ), we start with the expression for ( f(x) ):

[ f(x) = \frac{5x + 1}{x^2 + 2x - 2} ]

Next, we factor the denominator:

[ x^2 + 2x - 2 = (x - 1)(x + 2) ]

So, we can rewrite ( f(x) ):

[ f(x) = \frac{5x + 1}{(x - 1)(x + 2)} ]

Now, we simplify this expression. We can express ( 5x + 1 ) in terms of the denominator's factors:

[ 5x + 1 = 2(x - 1) + 2(x + 2) ]

This allows us to rearrange as follows:

[ f(x) = \frac{2(x - 1) + 2(x + 2)}{(x - 1)(x + 2)} = \frac{2}{x - 1} \text{ for } x > 1. ]

Thus, we have shown that ( f(x) = \frac{2}{x - 1} ) for ( x > 1 ).

Step 2

Find f^{-1}(c).

99%

104 rated

Answer

To find ( f^{-1}(c) ), we first set ( y = \frac{2}{x - 1} ).

We want to express it in terms of ( x ). Rearranging gives:

[ y(x - 1) = 2\n yx - y = 2\n yx = 2 + y\n x = \frac{2 + y}{y}.\n]

Thus, ( f^{-1}(c) = \frac{2 + c}{c} ).

Step 3

Solve fg(x) = \frac{1}{4}.

96%

101 rated

Answer

First, we compute ( fg(x) ):

[ f(g(x)) = f(x^2 + 5). ]

Substituting ( g(x) ) into ( f(x) ) gives:

[ f(g(x)) = \frac{5(x^2 + 5) + 1}{(x^2 + 5)^2 + 2(x^2 + 5) - 2}\n = \frac{5x^2 + 26}{(x^2 + 5)^2 + 2x^2 + 10 - 2}. ]

Now, we set this equal to ( \frac{1}{4} ) and solve:

[ \frac{5x^2 + 26}{(x^2 + 5)^2 + 2x^2 + 8} = \frac{1}{4}. ]

Cross-multiplying yields:

[ 4(5x^2 + 26) = (x^2 + 5)^2 + 2x^2 + 8. ]

This leads to a polynomial equation that we can solve for ( x ).

Show that f(x) = \frac{2}{x - 1}, \quad x > 1.

Find f^{-1}(c).

Solve fg(x) = \frac{1}{4}.

Join the A-Level students using SimpleStudy...