Edexcel IAL October 2021 P4 (WMA14/01) Q3 Partial Fractions, Differentiation
\( g(x) = \frac{3x^3 + 8x^2 – 3x – 6}{x(x+3)} \equiv Ax + B + \frac{C}{x} + \frac{D}{x+3} \)
(a) Find the values of the constants A, B, C and D.
A curve has equation
\( y = g(x) \quad x > 0 \)
Using the answer to part (a),
(b) find \( g'(x) \).(2)
(c) Hence, explain why \( g'(x) > 3 \) for all values of \( x \) in the domain of \( g \).(1)
Solution to Part a: Partial Fractions for \( g(x) = \frac{3x^3 + 8x^2 – 3x – 6}{x(x+3)} \)
Key Concepts Used:
- Partial fraction decomposition
- Polynomial division
Step-by-Step Working:
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Set Up the General Form
We express \( g(x) \) as:
\( \frac{3x^3 + 8x^2 – 3x – 6}{x(x+3)} = Ax + B + \frac{C}{x} + \frac{D}{x+3} \)
Why are we solving it using Partial Fractions?
The numerator’s degree (3) is higher than the denominator’s degree (2), so we include a linear term \( Ax + B \) in addition to the partial fractions. -
Combine Terms on the Right Side
Multiply through by the denominator \( x(x+3) \):
\( 3x^3 + 8x^2 – 3x – 6 = (Ax + B)x(x+3) + C(x+3) + Dx \)
Expand the right-hand side:
\( (Ax + B)(x^2 + 3x) + Cx + 3C + Dx \)
\( = Ax^3 + 3Ax^2 + Bx^2 + 3Bx + Cx + 3C + Dx \)
\( = Ax^3 + (3A + B)x^2 + (3B + C + D)x + 3C \)
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Equate Coefficients
Match coefficients with the left-hand side \( 3x^3 + 8x^2 – 3x – 6 \):
\( x^{3} \) term:
\( A=3 \)
\( x^{2} \) term:
\( 3A+B=8 \)
\( 3(3)+B=8\Rightarrow B=-1 \)
x term:
\( 3B+C+D=-3 \)
\( 3(-1)+C+D=-3\Rightarrow C+D=0 \)
Constant term:
\( 3C=-6\Rightarrow C=-2 \)
Substitute \( C=-2 \) into \( C+D=0 \):
\( -2+D=0\Rightarrow D=2 \)
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Final Form:
\( g(x)=3x-1-\frac{2}{x}+\frac{2}{x+3} \)
Identified Constants:
\( A=3,\,B=-1,\,C=-2,\,D=2 \)
Final Answer:
\( A=3,\,B=-1,\,C=-2,\,D=2 \)
Solution to Part b: Finding \( g'(x) \)
Key Concepts Used:
- Differentiation
Step-by-Step Working:
Differentiate \( g(x) \) using the expression from part (a).
\( g(x) = 3x – 1 – 2x^{-1} + 2(x + 3)^{-1} \)
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Differentiate Term-by-Term
\( \frac{d}{dx}(3x) = 3 \)
\( \frac{d}{dx}(-1) = 0 \)
\( \frac{d}{dx}(-2x^{-1}) = 2x^{-2} \)
\( \frac{d}{dx}(2(x + 3)^{-1}) = -2(x + 3)^{-2} \)
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Combined Differentiation of each term gives us
\( g'(x) = 3 + \frac{2}{x^2} – \frac{2}{(x + 3)^2} \)
Final Answer:
\( g'(x) = 3 + \frac{2}{x^2} – \frac{2}{(x + 3)^2} \)
Solution to Part c: Explaining why \( g'(x) > 0 \)
For now, let’s just consider \( \frac{2}{x^2} – \frac{2}{(x+3)^2} \), which can be simplified as
\( \frac{2(x+3)^2 – 2x^2}{x^2(x+3)^2} \)
Now, since the domain of \( g(x) \) is given as \( x > 0 \), so it’s obvious that \( x + 3 > x \) as we add 3 more in \( x \).
Moreover, since \( x + 3 > x \rightarrow (x + 3)^2 > x^2 \)
With this it may be concluded that the fraction \( \frac{2(x+3)^2 – 2x^2}{x^2(x+3)^2} \) or \( \frac{2}{x^2} – \frac{2}{(x+3)^2} \) would be positive as from a larger value \( (2(x+3)^2) \) a smaller value \( (2x^2) \) is subtracted.
Now, since \( \frac{2}{x^2} – \frac{2}{(x+3)^2} \) is positive, the expression of \( g'(x) = 3 + \frac{2}{x^2} – \frac{2}{(x+3)^2} \) would be greater than 3 because a positive fraction is getting added to 3.
\( g'(x) = 3 + (\text{positive quantity}) > 3 \)
Hence, \( g'(x) > 3 \) for all \( x \) in the domain of \( g \).
Final Answer:
\( g'(x) > 3 \) because \( \frac{2}{x^2} – \frac{2}{(x+3)^2} > 0 \) for all \( x \) in the domain of \( g(x) \).