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.
(5)
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 Question 3(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:
1. Set Up the General Form
\[\frac{3x^3 + 8x^2 – 3x – 6}{x(x+3)} = Ax + B + \frac{C}{x} + \frac{D}{x+3}\]
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.
2. 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:
\[(Ax + B)(x^2 + 3x) + Cx + 3C + Dx\]
\[= Ax^3 + 3Ax^2 + Bx^2 + 3Bx + Cx + 3C + Dx\]
\[= Ax^3 + (3A + B)x^2 + (3B + C + D)x + 3C\]
3. 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 \( C + D = 0 \):
\[-2 + D = 0 \Rightarrow D = 2\]
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 Question 3(b): Finding \( g'(x) \)
Differentiate \( g(x) \) using the expression from part (a).
\[g(x) = 3x – 1 – 2x^{-1} + 2(x + 3)^{-1}\]
1. Differentiate Term-by-Term
\[\frac{d}{dx}(3x) = 3\]
\[\frac{d}{dx}(-1) = 0\]
\[\frac{d}{dx}(-2x^{-1}) = 2x^{-2}\]
\[\frac{d}{dx}(2(x + 3)^{-1}) = -2(x + 3)^{-2}\]
2. 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 Question 3(c): Explaining why \( g'(x) > 3 \)
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 to \( x \).
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 \((2(x+3)^2)\) we subtract the smaller value \((2x^2)\).
Now, the expression of \( g'(x) = 3 + \frac{2}{x^2} – \frac{2}{(x+3)^2} \) would be greater than 3 because a positive fraction \(\frac{2}{x^2} – \frac{2}{(x+3)^2}\) 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:
\[g'(x) > 3 \text{ because } \frac{2}{x^2} – \frac{2}{(x + 3)^2} > 0 \text{ for all } x \text{ in the domain of } g(x).\]
