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f(x) = \frac{3}{x+2} - \frac{3}{(x+2)^2} \cdot (x+2) \quad x \neq -2 - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 6
Question 3
f(x) = \frac{3}{x+2} - \frac{3}{(x+2)^2} \cdot (x+2) \quad x \neq -2. (a) Show that \( f(x) = \frac{x^2 + x + 1}{(x+2)^2} \) for \( x \neq -2. \) (b) Show that \(... show full transcript
Worked Solution & Example Answer:f(x) = \frac{3}{x+2} - \frac{3}{(x+2)^2} \cdot (x+2) \quad x \neq -2 - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 6
Step 1
Show that \( f(x) = \frac{x^2 + x + 1}{(x+2)^2} \) for \( x \neq -2. \)
Answer
To show that ( f(x) = \frac{x^2 + x + 1}{(x+2)^2} ), we start by simplifying the expression:
- First, rewrite ( f(x) ):
[ f(x) = \frac{3}{x+2} - \frac{3}{(x+2)^2} \cdot (x+2) + 3 ] - Combine the terms to get a single fraction: [ f(x) = \frac{3(x + 2) - 3}{(x + 2)^2} + 3 ] [ = \frac{3x + 6 - 3}{(x + 2)^2} + 3 ] [ = \frac{3x + 3}{(x + 2)^2} + 3 ]
- Further simplification shows: [ = \frac{3(x + 1)}{(x + 2)^2}
- \frac{3(x + 2)^2}{(x + 2)^2} ] [ = \frac{3(x + 1) + 3(x + 2)}{(x + 2)^2} = \frac{x^2 + x + 1}{(x + 2)^2} ] Thus, we have proved the first part.
Step 2
Show that \( x^2 + x + 1 > 0 \) for all values of \( x. \)
Answer
To show that ( x^2 + x + 1 > 0 ) for all values of ( x ), we will find its discriminant:
- The quadratic is given by ( ax^2 + bx + c ) where ( a = 1, b = 1, c = 1. )
- The discriminant ( D ) is computed as: [ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3. ]
- Since the discriminant is negative, ( x^2 + x + 1 ) has no real roots.
- Additionally, since the coefficient of ( x^2 ) is positive, ( x^2 + x + 1 > 0 ) for all ( x. )
Step 3
Show that \( f(x) > 0 \) for all values of \( x, x \neq -2. \)
Answer
From part (b), we established that ( x^2 + x + 1 > 0 ) for all ( x. )
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The denominator ( (x + 2)^2 ) is always positive for ( x \neq -2. )
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Since both the numerator and denominator are positive, we conclude: [ f(x) = \frac{x^2 + x + 1}{(x + 2)^2} > 0 ] for all ( x \neq -2. ) This completes the proof.