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f(x) = \[ \frac{-3}{x+2} + \frac{3}{(x+2)^2} \cdot (x \neq -2) \] (a) Show that \( f(x) = \frac{x^2 + x + 1}{(x+2)^2} \) for \( x \neq -2 \) - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 6
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![f(x)-=--\[-\frac{-3}{x+2}-+-\frac{3}{(x+2)^2}-\cdot-(x-\neq--2)-\]---(a)-Show-that-\(-f(x)-=-\frac{x^2-+-x-+-1}{(x+2)^2}-\)-for-\(-x-\neq--2-\)-Edexcel-A-Level Maths Pure-Question 3-2007-Paper 6.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/MathsPure_2007_6_1_QP_2_2007.jpg)
f(x) = \[ \frac{-3}{x+2} + \frac{3}{(x+2)^2} \cdot (x \neq -2) \] (a) Show that \( f(x) = \frac{x^2 + x + 1}{(x+2)^2} \) for \( x \neq -2 \). (b) Show that \( x^... show full transcript
Worked Solution & Example Answer:f(x) = \[ \frac{-3}{x+2} + \frac{3}{(x+2)^2} \cdot (x \neq -2) \] (a) Show that \( f(x) = \frac{x^2 + x + 1}{(x+2)^2} \) for \( x \neq -2 \) - 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 from the definition of ( f(x) ):
[ f(x) = \frac{-3}{x+2} + \frac{3}{(x+2)^2} ]
First, find a common denominator for the terms:
[ f(x) = \frac{-3(x+2) + 3}{(x+2)^2} ]
Now simplify the numerator:
[ -3(x+2) + 3 = -3x - 6 + 3 = -3x - 3 ]
Thus,
[ f(x) = \frac{-3(x + 1)}{(x + 2)^2} ]
Now, factor out the negative sign:
[ = \frac{3(-x^2 - x - 1)}{(x + 2)^2} = \frac{x^2 + x + 1}{(x+2)^2} ]
Therefore, ( f(x) = \frac{x^2 + x + 1}{(x+2)^2} ) holds for all ( x \neq -2 ).
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 need to analyze the quadratic expression. The discriminant ( D ) of the quadratic equation ( ax^2 + bx + c ) is given by:
[ D = b^2 - 4ac ]
Here, ( a = 1, b = 1, c = 1 ), so:
[ D = 1^2 - 4(1)(1) = 1 - 4 = -3 ]
Since the discriminant is negative, the quadratic does not cross the x-axis, meaning it does not have real roots. Thus, since the coefficient of ( x^2 ) is positive, ( x^2 + x + 1 > 0 ) for all values of ( x ).
Step 3
Show that f(x) > 0 for all values of x, x ≠ -2.
Answer
From part (b), we concluded that ( x^2 + x + 1 > 0 ) for all ( x ). Since the denominator ( (x+2)^2 ) is always positive for all ( x \neq -2 ), we have:
[ f(x) = \frac{x^2 + x + 1}{(x + 2)^2} > 0 ]
Thus, ( f(x) > 0 ) for all values of ( x ) except ( -2 ).