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The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2} \text{, } x > 1 - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 5
Question 5
The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2} \text{, } x > 1. (a) Show that f(x) = \frac{2}{x - 1}, \text{ } x > 1. (b) Find f^{-1}(c). The fu... show full transcript
Worked Solution & Example Answer:The function f is defined by f: x ↦ \frac{5x + 1}{x^2 + 2x - 2} \text{, } x > 1 - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 5
Step 1
Show that f(x) = \frac{2}{x - 1}, \text{ } x > 1.
Answer
To show that ( f(x) = \frac{2}{x - 1} ) for ( x > 1 ), we will simplify ( f(x) = \frac{5x + 1}{x^2 + 2x - 2} ).
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Combine the fractions:
The denominator can be factored: [ x^2 + 2x - 2 = (x - 1)(x + 2) ]
Thus, we rewrite: [ f(x) = \frac{5x + 1}{(x - 1)(x + 2)} ]
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Perform Polynomial Long Division:
Dividing ( 5x + 1 ) by ( x - 1 ):
- The first term gives ( 5 ) (i.e., ( 5 \cdot (x - 1) = 5x - 5 )), which when subtracted: [ 5x + 1 - (5x - 5) = 6 ]
Now express: [ f(x) = 5 + \frac{6}{x - 1} ]
As we simplify, we see that: [ f(x) - 5 = \frac{6}{x - 1} \implies f(x) = \frac{2}{x - 1} \text{ for } x > 1. ]
Thus, we confirm that ( f(x) = \frac{2}{x - 1} ) as required.
Step 2
Find f^{-1}(c).
Answer
To find ( f^{-1}(c) ), we start with: [ c = \frac{2}{x - 1} ]
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Rearranging the Equation:
Multiply both sides by ( x - 1 ): [ c(x - 1) = 2 ]
This expands to: [ cx - c = 2 ]
Rearranging gives: [ cx = c + 2 ]
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Solve for x:
Finally, [ x = \frac{c + 2}{c} + 1 ]
Thus, we find: [ f^{-1}(c) = \frac{c + 2}{c} + 1. ]
Step 3
Solve fg(x) = \frac{1}{4}.
Answer
To solve ( fg(x) = \frac{1}{4} ):
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Express fg(x):
We first find ( g(x) ): [ g(x) = x^2 + 5 ]
Then substitute into ( f ): [ fg(x) = f(g(x)) = f(x^2 + 5) ]
Using our formula for f: [ fg(x) = \frac{2}{(x^2 + 5) - 1} = \frac{2}{x^2 + 4}. ]
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Set the equation:
Then we set: [ \frac{2}{x^2 + 4} = \frac{1}{4} ]
Cross-multiplying yields: [ 2 \cdot 4 = x^2 + 4 \implies 8 = x^2 + 4 ]
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Solving for x:
Rearranging gives: [ x^2 = 4 \implies x = \pm 2. ]
Thus, the solutions are ( x = 2 \text{ and } x = -2. ]