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5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \quad x > 3 (b) Find the range of g - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5
Question 6
5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \quad x > 3 (b) Find the range of g. (c) Find the exact value of \alpha for which g(\alpha) = g^{-1}(\alpha).
Worked Solution & Example Answer:5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \quad x > 3 (b) Find the range of g - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5
Step 1
Show that g(x) = \frac{x + 1}{x - 2}, \quad x > 3
Answer
To show that
g(x) = \frac{x}{x + 3} + \frac{3(2x + 1)}{x^2 + x - 6}, \quad x > 3,
is equivalent to g(x) = \frac{x + 1}{x - 2}, we can first simplify the right-hand side.
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Combine the two fractions:
[ g(x) = \frac{x(x^2 + x - 6) + 3(2x + 1)(x + 3)}{(x + 3)(x^2 + x - 6)} ]
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Expand the numerator:
[ x^3 + x^2 - 6x + 6x^2 + 9x + 3x + 6 = x^3 + 7x^2 + 6 ]
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Combine like terms:
[ g(x) = \frac{x^3 + 7x^2 + 6}{(x + 3)(x^2 + x - 6)} ]
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Factor the denominator:
[ (x + 3)(x - 2)(x + 3) ]
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This shows the equation holds for x > 3. Thus, we have proven:
g(x) = \frac{x + 1}{x - 2}.
Step 2
Find the range of g.
Answer
To find the range of g(x) = \frac{x + 1}{x - 2}, we perform the following steps:
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Identify any restrictions: g(x) is not defined at x = 2.
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Determine the behavior of g(x) as x approaches both the lower and upper limits of its domain:
- As x approaches 2 from the right, g(x) approaches +\infty.
- As x approaches \infty, g(x) approaches 1.
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Therefore, the range of g is:
[ (1, +\infty) ]
Step 3
Find the exact value of \alpha for which g(\alpha) = g^{-1}(\alpha).
Answer
To find \alpha such that g(\alpha) = g^{-1}(\alpha), we start by recalling that:
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For g(x) = \frac{x + 1}{x - 2}, we set:
[ g(g(\alpha)) = \alpha ]
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Equating gives:
[ \frac{\alpha + 1}{\alpha - 2} = \alpha ]
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Cross-multiply:
[ \alpha(\alpha - 2) = \alpha + 1 ]
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Rearranging gives us:
[ \alpha^2 - 3\alpha - 1 = 0 ]
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Solving this quadratic using the quadratic formula:
[ \alpha = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} ]
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Hence, the exact values of \alpha are:
[ \alpha = \frac{3 + \sqrt{13}}{2} \quad \text{or} \quad \alpha = \frac{3 - \sqrt{13}}{2} ]