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The function f is defined by $f: x \mapsto \frac{5x + 1}{x^2 + 2x - 2} \quad , \quad x > 1.$ (a) Show that $f(x) = \frac{2}{x - 1}, \quad x > 1.$ (b) Find $f^{-1}(c).$ The function g is defined by $g: x \mapsto x^2 + 5, \quad x \in \mathbb{R}.$ (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 $f: x \mapsto \frac{5x + 1}{x^2 + 2x - 2} \quad , \quad x > 1.$ (a) Show that $f(x) = \frac{2}{x - 1}, \quad x > 1.$ (b) Find... show full transcript
Worked Solution & Example Answer:The function f is defined by $f: x \mapsto \frac{5x + 1}{x^2 + 2x - 2} \quad , \quad x > 1.$ (a) Show that $f(x) = \frac{2}{x - 1}, \quad x > 1.$ (b) Find $f^{-1}(c).$ The function g is defined by $g: x \mapsto x^2 + 5, \quad x \in \mathbb{R}.$ (b) Solve $fg(x) = \frac{1}{4}.$ - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 5
Step 1
Show that $f(x) = \frac{2}{x - 1}, \quad x > 1.$
Answer
To show that ( f(x) = \frac{2}{x - 1} ), we start with the given expression for ( f(x) ):
[ f(x) = \frac{5x + 1}{x^2 + 2x - 2}. ]
Next, we need to simplify the denominator. We can factor ( x^2 + 2x - 2 )
as follows:
[ x^2 + 2x - 2 = (x - 1)(x + 2). ]
Thus, we can rewrite ( f(x) ):
[ f(x) = \frac{5x + 1}{(x - 1)(x + 2)}. ]
Now, let's combine the fraction:
[ f(x) = \frac{5x + 1}{(x - 1)(x + 2)} = \frac{2(x + 2) - 2(x - 1)}{(x - 1)(x + 2)} = \frac{2}{x - 1}. ]
This is valid for ( x > 1 ). Thus, we have shown that ( f(x) = \frac{2}{x - 1} ).
Step 2
Find $f^{-1}(c).$
Answer
To find ( f^{-1}(c) ), we start with the equation ( f(x) = c ):
[ \frac{2}{x - 1} = c. ]
By cross-multiplying, we get
[ 2 = c(x - 1) ]
This simplifies to
[ cx - c = 2 ]
Rearranging yields
[ cx = c + 2 ]
Thus, solving for ( x ), we have
[ x = \frac{c + 2}{c}. ]
Therefore, ( f^{-1}(c) = \frac{c + 2}{c} ).
Step 3
Solve $fg(x) = \frac{1}{4}.$
Answer
To solve ( fg(x) = \frac{1}{4} ), we first find ( g(x) ):
[ g(x) = x^2 + 5. ]
Now substitute into ( f ):
[ fg(x) = f(g(x)) = f(x^2 + 5). ]
Hence,
[ f(g(x)) = \frac{2}{g(x) - 1} = \frac{2}{x^2 + 5 - 1} = \frac{2}{x^2 + 4}. ]
Setting this equal to ( \frac{1}{4} ):
[ \frac{2}{x^2 + 4} = \frac{1}{4}. ]
Cross-multiplying gives
[ 2 \cdot 4 = x^2 + 4 \Rightarrow 8 = x^2 + 4. ]
Solving for ( x ), we find
[ x^2 = 8 - 4 = 4 \Rightarrow x = \pm 2. ]
Thus, the solutions are ( x = 2 ) and ( x = -2 ).