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5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \; x > 3 (b) Find the range of g - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 5
Question 5
5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \; x > 3 (b) Find the range of g. (c) Find the exact value of a for which g(a) = g^{-1}(a).
Worked Solution & Example Answer:5. (a) Show that g(x) = \frac{x + 1}{x - 2}, \; x > 3 (b) Find the range of g - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 5
Step 1
Show that g(x) = \frac{x + 1}{x - 2}, \; x > 3
Answer
To show that ( g(x) ) can be expressed as ( \frac{x + 1}{x - 2} ) for ( x > 3 ), we start with the expression:
[ g(x) = \frac{x}{x + 3} + \frac{3(2x + 1)}{x^2 + x - 6} ]
We factor the denominator of the second term: [ x^2 + x - 6 = (x - 2)(x + 3) ]
Thus, [ \frac{3(2x + 1)}{(x - 2)(x + 3)} ]
Now, we combine the fractions, finding a common denominator: [ g(x) = \frac{x(x - 2) + 3(2x + 1)}{(x - 2)(x + 3)} ]
This can be simplified: [ = \frac{x^2 - 2x + 6x + 3}{(x - 2)(x + 3)} = \frac{x^2 + 4x + 3}{(x - 2)(x + 3)} ]
Factoring the numerator gives: [ = \frac{(x + 1)(x + 3)}{(x - 2)(x + 3)} = \frac{x + 1}{x - 2} \text{ for } x > 3. ]
Therefore, we have established that ( g(x) = \frac{x + 1}{x - 2} ) holds for ( x > 3 ).
Step 2
Step 3
Find the exact value of a for which g(a) = g^{-1}(a).
Answer
To find ( a ) such that ( g(a) = g^{-1}(a) ), we first determine the expression for ( g^{-1}(x) ):
- Start with ( y = \frac{x + 1}{x - 2} ).
- Rearranging gives:
[ y(x - 2) = x + 1 ]
[ yx - 2y = x + 1 ]
[ (y - 1)x = 2y + 1 ]
[ x = \frac{2y + 1}{y - 1} ]
Thus, ( g^{-1}(x) = \frac{2x + 1}{x - 1} ).
Setting ( g(a) = g^{-1}(a) ):
[ \frac{a + 1}{a - 2} = \frac{2a + 1}{a - 1} ]
Cross-multiplying yields:
[ (a + 1)(a - 1) = (2a + 1)(a - 2) ]
Expanding both sides:
[ a^2 - 1 = 2a^2 - 4a + a - 2 ]
[ 0 = a^2 - 5a + 1 ]
Using the quadratic formula, ( a = \frac{5 \pm \sqrt{17}}{2} ).
Thus, the exact values of ( a ) satisfying ( g(a) = g^{-1}(a) ) are:
[ a = \frac{5 + \sqrt{17}}{2} \text{ or } a = \frac{5 - \sqrt{17}}{2}. ]