Figure 1.6: Extended Binomial Convergence of $(1-x)^{-1/2}$ for $|x| < 1$
1. Partial Fractions Decomposition (Syllabus AHL 1.11)
A rational expression $\frac{P(x)}{Q(x)}$ where $\deg(P) < \deg(Q)$ (a proper rational function) can be split into a sum of simpler fractions according to the algebraic structure of the denominator factors:
| Factor in Denominator $Q(x)$ | Partial Fraction Form | Method of Finding Numerators |
|---|---|---|
| Distinct Linear Factor: $(ax + b)$ | $\frac{A}{ax + b}$ | Heaviside Cover-Up: Evaluate $P(x)/Q'(x)$ or cover the factor and substitute $x = -b/a$. |
| Repeated Linear Factor: $(ax + b)^2$ | $\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$ | Substitute $x = -b/a$ to obtain $B$, then equate coefficients or substitute $x=0$ to obtain $A$. |
| Repeated Linear Factor (Cubic): $(ax + b)^3$ | $\frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \frac{C}{(ax + b)^3}$ | Substitute $x = -b/a$ for $C$, then equate coefficients of descending powers. |
| Irreducible Quadratic Factor: $(ax^2 + bx + c)$ with $\Delta < 0$ | $\frac{Ax + B}{ax^2 + bx + c}$ | Numerator must be linear ($Ax + B$). Equate coefficients of $x^2, x$, and constant terms. |
Crucial Rule for Improper Rational Functions: If $\deg(P) \ge \deg(Q)$, polynomial long division is mandatory first:
$$\frac{P(x)}{Q(x)} = \text{Quotient}(x) + \frac{R(x)}{Q(x)}, \quad \text{where } \deg(R) < \deg(Q)$$Only the remainder fraction $\frac{R(x)}{Q(x)}$ is subsequently decomposed into partial fractions!
2. Extended Binomial Theorem for Fractional and Negative Powers (Syllabus AHL 1.10)
When the exponent $r \in \mathbb{Q}$ is negative or fractional, the expansion of $(1 + x)^r$ does not terminate; it forms an infinite Maclaurin series:
$$(1 + x)^r = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \dots + \binom{r}{n} x^n + \dots$$where the generalized binomial coefficient is defined for all $r \in \mathbb{R}$ and $n \in \mathbb{N}$ as:
$$\binom{r}{n} = \frac{r(r-1)(r-2)\dots(r-n+1)}{n!}$$Radius & Interval of Convergence: By the ratio test, this infinite series converges if and only if:
$$|x| < 1 \iff -1 < x < 1$$3. Expanding Generalized Binomials $(a + bx)^r$
When the first term is $a \neq 1$, you must factor out $a^r$ before applying the formula:
$$(a + bx)^r = \left[a \left(1 + \frac{b}{a}x\right)\right]^r = a^r \left(1 + \frac{b}{a}x\right)^r$$The series expansion is valid if and only if:
$$\left|\frac{b}{a}x\right| < 1 \iff |x| < \left|\frac{a}{b}\right|$$Worked Examples
Worked Example 1: Repeated Linear Factor Decomposition
Question: Express $f(x) = \frac{4x^2 + 5x + 7}{(x + 2)(x - 1)^2}$ in partial fractions. [5 marks]
Solution:
Set up the partial fraction template with a linear and repeated linear term: [M1]
$$\frac{4x^2 + 5x + 7}{(x + 2)(x - 1)^2} = \frac{A}{x + 2} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$$Multiply both sides by $(x + 2)(x - 1)^2$: [M1]
$$4x^2 + 5x + 7 = A(x - 1)^2 + B(x + 2)(x - 1) + C(x + 2)$$Substitute $x = 1$ to isolate $C$: [A1]
$$4(1)^2 + 5(1) + 7 = A(0) + B(0) + C(1 + 2) \implies 16 = 3C \implies C = \frac{16}{3} \quad [\text{A1}]$$Substitute $x = -2$ to isolate $A$: [A1]
$$4(-2)^2 + 5(-2) + 7 = A(-2 - 1)^2 \implies 16 - 10 + 7 = 9A \implies 13 = 9A \implies A = \frac{13}{9}$$Equate coefficients of $x^2$ to find $B$: [M1]
$$4 = A + B \implies B = 4 - A = 4 - \frac{13}{9} = \frac{23}{9} \quad [\text{A1}]$$Check with $x = 0$: Constant term on LHS is $7$. On RHS: $A(1) - 2B + 2C = \frac{13}{9} - \frac{46}{9} + \frac{32}{3} = -\frac{33}{9} + \frac{96}{9} = \frac{63}{9} = 7$ (verified!).
Final decomposition:
Worked Example 2: Irreducible Quadratic Factor
Question: Find the partial fraction decomposition of $g(x) = \frac{x^2 - x + 14}{(x - 2)(x^2 + 4)}$. [5 marks]
Solution:
Because $(x^2 + 4)$ has no real roots ($\Delta = -16 < 0$), set up the decomposition with a linear numerator over the quadratic factor: [M1]
$$\frac{x^2 - x + 14}{(x - 2)(x^2 + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 4}$$Multiply through by the common denominator: [M1]
$$x^2 - x + 14 = A(x^2 + 4) + (Bx + C)(x - 2)$$Substitute $x = 2$: [A1]
$$2^2 - 2 + 14 = A(2^2 + 4) + 0 \implies 16 = 8A \implies A = 2$$Equate coefficients of $x^2$: [M1]
$$1 = A + B \implies 1 = 2 + B \implies B = -1 \quad [\text{A1}]$$Equate constant terms (or substitute $x = 0$): [M1]
$$14 = 4A - 2C \implies 14 = 4(2) - 2C \implies 14 = 8 - 2C \implies 2C = -6 \implies C = -3 \quad [\text{A1}]$$Final decomposition:
$$\frac{x^2 - x + 14}{(x - 2)(x^2 + 4)} = \frac{2}{x - 2} - \frac{x + 3}{x^2 + 4}$$Worked Example 3: Improper Fraction & Long Division
Question: Express $h(x) = \frac{x^3 + 2x^2 - 3x + 1}{x^2 + x - 2}$ in partial fractions. [6 marks]
Solution:
Step 1: Check degrees & divide: $\deg(\text{num}) = 3 > \deg(\text{den}) = 2$. Division is mandatory: [M1]
$$\begin{array}{r} x + 1\phantom{)} \\ x^2 + x - 2{\overline{\smash{\big)}\,x^3 + 2x^2 - 3x + 1\phantom{)}}} \\ \underline{-(x^3 + x^2 - 2x)\phantom{+1)}} \\ x^2 - x + 1\phantom{)} \\ \underline{-(x^2 + x - 2)} \\ -2x + 3\phantom{)} \end{array}$$Quotient is $x + 1$ and remainder is $-2x + 3$. [A1]
$$h(x) = x + 1 + \frac{-2x + 3}{x^2 + x - 2} = x + 1 + \frac{-2x + 3}{(x + 2)(x - 1)}$$Step 2: Decompose proper remainder fraction: [M1]
$$\frac{-2x + 3}{(x + 2)(x - 1)} = \frac{C}{x + 2} + \frac{D}{x - 1} \implies -2x + 3 = C(x - 1) + D(x + 2)$$Substitute $x = 1$: $-2(1) + 3 = D(3) \implies 1 = 3D \implies D = \frac{1}{3}$. [A1]
Substitute $x = -2$: $-2(-2) + 3 = C(-3) \implies 7 = -3C \implies C = -\frac{7}{3}$. [A1]
Final form: [A1]
$$h(x) = x + 1 - \frac{7}{3(x + 2)} + \frac{1}{3(x - 1)}$$Worked Example 4: Extended Binomial of $(4 - x)^{-1/2}$ & Approximation
Question:
(a) Find the binomial expansion of $f(x) = (4 - x)^{-1/2}$ in ascending powers of $x$ up to and including the term in $x^2$. [4 marks]
(b) State the interval of convergence for this series. [1 mark]
(c) By substituting $x = 2$ into your expansion, estimate $\sqrt{2}$ as a rational fraction. [3 marks]
Solution:
(a) Expansion: Factor out $4^{-1/2} = \frac{1}{2}$: [M1]
$$(4 - x)^{-1/2} = 4^{-1/2} \left(1 - \frac{x}{4}\right)^{-1/2} = \frac{1}{2} \left(1 - \frac{x}{4}\right)^{-1/2}$$Apply $(1 + u)^{-1/2} = 1 + \left(-\frac{1}{2}\right)u + \frac{(-1/2)(-3/2)}{2!}u^2 + \dots$ with $u = -\frac{x}{4}$: [M1]
$$= 1 - \frac{1}{2}\left(-\frac{x}{4}\right) + \frac{3}{8}\left(-\frac{x}{4}\right)^2 = 1 + \frac{x}{8} + \frac{3x^2}{128} \quad [\text{A1}]$$Multiply through by the external factor $\frac{1}{2}$: [A1]
$$(4 - x)^{-1/2} = \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} + \dots$$(b) Convergence Interval: $\left|-\frac{x}{4}\right| < 1 \iff |x| < 4$ (or $-4 < x < 4$). [A1]
(c) Approximation of $\sqrt{2}$:
For $x = 2 \in (-4, 4)$, notice that $(4 - 2)^{-1/2} = 2^{-1/2} = \frac{1}{\sqrt{2}}$, so $\sqrt{2} = \frac{1}{f(2)}$. [M1]
Since $f(2) = \frac{1}{\sqrt{2}}$, we have $\sqrt{2} \approx \frac{64}{43} \approx 1.488$. [A1]
(Better accuracy is achieved with smaller $|x|$, e.g., $x = \frac{2}{9}$ or using $\sqrt{\frac{1+x}{1-x}}$).
Worked Example 5: Combining Partial Fractions & Binomial Series
Question: Given $f(x) = \frac{2x - 13}{(x - 1)(x - 4)}$:
(a) Express $f(x)$ in partial fractions. [3 marks]
(b) Hence obtain the series expansion of $f(x)$ in ascending powers of $x$ up to and including the term in $x^2$. [4 marks]
(c) State the exact range of values of $x$ for which the expansion is valid. [1 mark]
Solution:
(a) Partial fractions: [M1]
$$\frac{2x - 13}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4} \implies 2x - 13 = A(x - 4) + B(x - 1)$$Let $x = 1$: $2(1) - 13 = A(1 - 4) \implies -11 = -3A \implies A = \frac{11}{3}$. [A1]
Let $x = 4$: $2(4) - 13 = B(4 - 1) \implies -5 = 3B \implies B = -\frac{5}{3}$. [A1]
(b) Series Expansion: Expand both terms using $(1 - u)^{-1} = 1 + u + u^2 + \dots$: [M1]
$$(1 - x)^{-1} = 1 + x + x^2 + \dots \quad (\text{valid for } |x| < 1)$$ $$\left(1 - \frac{x}{4}\right)^{-1} = 1 + \frac{x}{4} + \frac{x^2}{16} + \dots \quad (\text{valid for } |x| < 4)$$Multiply by external coefficients and combine like powers: [M1]
$$f(x) = -\frac{11}{3}(1 + x + x^2) + \frac{5}{12}\left(1 + \frac{x}{4} + \frac{x^2}{16}\right)$$- Constant term: $-\frac{11}{3} + \frac{5}{12} = -\frac{44}{12} + \frac{5}{12} = -\frac{39}{12} = -\frac{13}{4}$. [A1]
- Linear term ($x$): $-\frac{11}{3}x + \frac{5}{48}x = \left(-\frac{176}{48} + \frac{5}{48}\right)x = -\frac{171}{48}x = -\frac{57}{16}x$. [A1]
- Quadratic term ($x^2$): $-\frac{11}{3}x^2 + \frac{5}{192}x^2 = \left(-\frac{704}{192} + \frac{5}{192}\right)x^2 = -\frac{699}{192}x^2 = -\frac{233}{64}x^2$.
Expansion: $f(x) = -\frac{13}{4} - \frac{57}{16}x - \frac{233}{64}x^2 + \dots$
(c) Convergence Range: The combined series converges for the intersection of both validity intervals: $|x| < 1 \cap |x| < 4 \implies |x| < 1$. [A1]
Practice Exercises (10 Tiered Questions)
Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.
Express $\frac{7x - 1}{(x - 3)(x + 1)}$ in partial fractions.
Show Full Solution
$\frac{7x - 1}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}$. [M1]
$7x - 1 = A(x + 1) + B(x - 3)$.
Let $x = 3$: $7(3) - 1 = A(4) \implies 20 = 4A \implies A = 5$. [A1]
Let $x = -1$: $7(-1) - 1 = B(-4) \implies -8 = -4B \implies B = 2$. [A1]
Result: $\frac{5}{x - 3} + \frac{2}{x + 1}$. [A1]
Expand $(1 + 3x)^{-1}$ up to the term in $x^3$ and state the interval of convergence.
Show Full Solution
Using $(1 + u)^{-1} = 1 - u + u^2 - u^3 + \dots$ with $u = 3x$: [M1]
$(1 + 3x)^{-1} = 1 - (3x) + (3x)^2 - (3x)^3 + \dots = 1 - 3x + 9x^2 - 27x^3 + \dots$ [A1]
Convergence requires $|3x| < 1 \iff |x| < \frac{1}{3}$. [A1]
Expand $(1 + x)^{1/3}$ in ascending powers of $x$ up to the term in $x^2$.
Show Full Solution
$(1 + x)^r = 1 + rx + \frac{r(r-1)}{2}x^2 + \dots$ with $r = \frac{1}{3}$: [M1]
First term: $1$.
$x$-term: $\frac{1}{3}x$. [A1]
$x^2$-term: $\frac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2}x^2 = -\frac{1}{9}x^2$. [A1]
Result: $1 + \frac{1}{3}x - \frac{1}{9}x^2 + \dots$ (valid for $|x| < 1$).
Find the partial fraction decomposition of $\frac{3x^2 + 5x + 4}{(x + 2)(x + 1)^2}$.
Show Full Solution
$\frac{3x^2 + 5x + 4}{(x + 2)(x + 1)^2} = \frac{A}{x + 2} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$. [M1]
$3x^2 + 5x + 4 = A(x + 1)^2 + B(x + 2)(x + 1) + C(x + 2)$.
Let $x = -1$: $3 - 5 + 4 = C(1) \implies C = 2$. [A1]
Let $x = -2$: $12 - 10 + 4 = A(-1)^2 \implies A = 6$. [A1]
Compare $x^2$ coefficients: $3 = A + B \implies 3 = 6 + B \implies B = -3$. [A1]
Result: $\frac{6}{x + 2} - \frac{3}{x + 1} + \frac{2}{(x + 1)^2}$.
Decompose $\frac{x^2 - x + 14}{(x - 2)(x^2 + 4)}$ into partial fractions.
Show Full Solution
$\frac{x^2 - x + 14}{(x - 2)(x^2 + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 4}$. [M1]
$x^2 - x + 14 = A(x^2 + 4) + (Bx + C)(x - 2)$.
Let $x = 2$: $4 - 2 + 14 = A(8) \implies 16 = 8A \implies A = 2$. [A1]
Compare $x^2$ coefficients: $1 = A + B \implies 1 = 2 + B \implies B = -1$. [A1]
Compare constant terms: $14 = 4A - 2C \implies 14 = 8 - 2C \implies 2C = -6 \implies C = -3$. [A1]
Result: $\frac{2}{x - 2} - \frac{x + 3}{x^2 + 4}$.
Express $\frac{x^3 + 2x^2 - 3x + 1}{x^2 + x - 2}$ in the form $Ax + B + \frac{C}{x + 2} + \frac{D}{x - 1}$.
Show Full Solution
Because $\deg(\text{num}) = 3 > \deg(\text{den}) = 2$, divide polynomials first: [M1]
$(x^3 + 2x^2 - 3x + 1) \div (x^2 + x - 2)$: quotient is $x + 1$, remainder is $-2x + 3$. [A1]
So $\frac{x^3 + 2x^2 - 3x + 1}{x^2 + x - 2} = x + 1 + \frac{-2x + 3}{(x + 2)(x - 1)}$.
Now decompose $\frac{-2x + 3}{(x + 2)(x - 1)} = \frac{C}{x + 2} + \frac{D}{x - 1}$: [M1]
$-2x + 3 = C(x - 1) + D(x + 2)$.
Let $x = 1$: $1 = 3D \implies D = \frac{1}{3}$. [A1]
Let $x = -2$: $7 = -3C \implies C = -\frac{7}{3}$. [A1]
Result: $x + 1 - \frac{7}{3(x + 2)} + \frac{1}{3(x - 1)}$.
Find the coefficient of $x^3$ in the expansion of $\frac{1 + 2x}{(1 - 2x)^2}$ for $|x| < \frac{1}{2}$.
Show Full Solution
$(1 - 2x)^{-2} = 1 + (-2)(-2x) + \frac{(-2)(-3)}{2}(-2x)^2 + \frac{(-2)(-3)(-4)}{6}(-2x)^3 + \dots$ [M1]
$= 1 + 4x + 3(4x^2) + (-4)(-8x^3) = 1 + 4x + 12x^2 + 32x^3 + \dots$ [A1]
Multiply by $(1 + 2x)$: [M1]
$(1 + 2x)(1 + 4x + 12x^2 + 32x^3 + \dots)$
Terms contributing to $x^3$: $1(32x^3) + 2x(12x^2) = 32x^3 + 24x^3 = 56x^3$. [A1]
The coefficient of $x^3$ is $56$. [A1]
Given $f(x) = \frac{2x - 13}{(x - 1)(x - 4)}$:
(a) Express $f(x)$ in partial fractions.
(b) Hence find the series expansion of $f(x)$ up to the term in $x^2$, and state the interval of convergence for the series.
Show Full Solution
(a) Partial fractions: [M1]
$\frac{2x - 13}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4} \implies 2x - 13 = A(x - 4) + B(x - 1)$.
Let $x = 1$: $-11 = -3A \implies A = \frac{11}{3}$. [A1]
Let $x = 4$: $-5 = 3B \implies B = -\frac{5}{3}$. [A1]
$f(x) = -\frac{11}{3}(1 - x)^{-1} + \frac{5}{12}\left(1 - \frac{x}{4}\right)^{-1}$.
(b) Binomial expansions: [M1]
• $(1 - x)^{-1} = 1 + x + x^2 + \dots$ (valid for $|x| < 1$).
• $\left(1 - \frac{x}{4}\right)^{-1} = 1 + \frac{x}{4} + \frac{x^2}{16} + \dots$ (valid for $|x| < 4$).
Combine: [M1]
Constant: $-\frac{11}{3} + \frac{5}{12} = -\frac{13}{4}$.
$x$-term: $-\frac{11}{3}x + \frac{5}{48}x = -\frac{57}{16}x$.
$x^2$-term: $-\frac{11}{3}x^2 + \frac{5}{192}x^2 = -\frac{233}{64}x^2$.
Expansion: $-\frac{13}{4} - \frac{57}{16}x - \frac{233}{64}x^2 + \dots$ [A1]
Valid for the intersection of domains: $|x| < 1$. [A1]
Find a formula for the general term (in terms of $n$) in the expansion of $(1 - x)^{-3}$ in ascending powers of $x$.
Show Full Solution
The general coefficient of $x^n$ in $(1 + u)^r$ with $u = -x$ and $r = -3$ is: [M1]
$\frac{r(r-1)(r-2)\dots(r-n+1)}{n!} (-x)^n = \frac{(-3)(-4)(-5)\dots(-(n+2))}{n!} (-1)^n x^n$. [M1]
Factor out $(-1)^n$ from the numerator:
$(-1)^n \frac{3 \times 4 \times 5 \dots \times (n+2)}{n!} (-1)^n x^n = (-1)^{2n} \frac{(n+2)! / 2!}{n!} x^n$. [M1]
Since $(-1)^{2n} = 1$ and $\frac{(n+2)!}{2! n!} = \frac{(n+2)(n+1)}{2}$:
General term: $\frac{(n+1)(n+2)}{2} x^n$. [A1]
Use the first three terms of the expansion of $(1 + x)^{1/3}$ with $x = -\frac{1}{27}$ to estimate $\sqrt[3]{26}$ to 5 decimal places.
Show Full Solution
Notice that $26 = 27 - 1 = 27\left(1 - \frac{1}{27}\right)$. [M1]
So $26^{1/3} = 27^{1/3}\left(1 - \frac{1}{27}\right)^{1/3} = 3\left(1 - \frac{1}{27}\right)^{1/3}$. [A1]
Using $(1 - u)^{1/3} \approx 1 - \frac{1}{3}u - \frac{1}{9}u^2$ with $u = \frac{1}{27}$: [M1]
• First term: $1$
• Second term: $-\frac{1}{3(27)} = -\frac{1}{81} \approx -0.0123457$
• Third term: $-\frac{1}{9(27^2)} = -\frac{1}{6561} \approx -0.0001524$
Sum inside bracket: $1 - 0.0123457 - 0.0001524 = 0.9875019$. [A1]
Multiply by 3: $3 \times 0.9875019 = 2.9625057 \approx 2.96251$. [A1]
(Calculator true value: $26^{1/3} \approx 2.962496\dots$, accurate to 4 decimal places!).
Formula Booklet References
Syllabus AHL 1.10: Extension of the Binomial Theorem
$$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots, \quad n \in \mathbb{Q}, \; |x| < 1$$Key Formulas & Identities
1. Expanding $(a + bx)^n$
$$(a + bx)^n = a^n \left(1 + \frac{b}{a}x\right)^n, \quad \text{valid for } |x| < \left|\frac{a}{b}\right|$$2. Partial Fraction Decomposition Templates
- Linear: $\frac{A}{ax+b}$
- Repeated linear: $\frac{A}{ax+b} + \frac{B}{(ax+b)^2}$
- Irreducible quadratic: $\frac{Ax+B}{ax^2+bx+c}$
- Always check degree: If $\deg(\text{num}) \ge \deg(\text{den})$, long division first!
3. Important Negative Power Binomial Series
- $(1 - x)^{-1} = 1 + x + x^2 + x^3 + \dots = \sum_{n=0}^\infty x^n$
- $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots = \sum_{n=0}^\infty (-1)^n x^n$
- $(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=0}^\infty (n+1)x^n$
- $(1 - x)^{-3} = 1 + 3x + 6x^2 + 10x^3 + \dots = \sum_{n=0}^\infty \frac{(n+1)(n+2)}{2}x^n$
Common Exam Pitfalls
Forgetting to Factor Out $a^n$
You cannot apply the booklet formula directly to $(4 + x)^{1/2}$! You must first factor out $4^{1/2} = 2$. Writing $1 + \frac{1}{2}(4+x)$ is an automatic zero-mark error.
Domain of Convergence for Combined Series
When adding two binomial series (e.g. from partial fractions), the domain of the combined series is the stricter (smaller) of the two intervals. If one requires $|x| < 1$ and the other $|x| < 3$, the overall range is $|x| < 1$.
Linear Numerator for Irreducible Quadratics
Whenever decomposing over an irreducible quadratic $(ax^2 + bx + c)$, the numerator must be linear ($Ax + B$). Assuming a constant numerator $A$ loses all subsequent marks.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
For what values of $x$ does the series expansion of $(3 - 2x)^{-1/2}$ converge?
What is the correct partial fraction template for $\frac{x^2 + 1}{(x - 1)^2(x^2 + 4)}$?
The first two non-trivial terms of $(1 + ax)^n$ are $1 - 6x + 24x^2 + \dots$. What are the values of $a$ and $n$?
Why must polynomial division be carried out on $\frac{x^2 + 3}{x^2 - 1}$ before decomposing into partial fractions?
Section B: Examination-Style Free Response Questions
Consider the rational function $f(x) = \frac{5}{(1 - x)(2 + 3x)}$.
(a) Express $f(x)$ in partial fractions. [3 marks]
(b) Hence obtain the expansion of $f(x)$ in ascending powers of $x$ up to and including the term in $x^2$. [3 marks]
Show Full Markscheme
(a) Partial fractions: [3 marks]
$\frac{5}{(1 - x)(2 + 3x)} = \frac{A}{1 - x} + \frac{B}{2 + 3x}$ [M1]
$5 = A(2 + 3x) + B(1 - x)$
Let $x = 1$: $5 = A(5) \implies A = 1$ [A1].
Let $x = -\frac{2}{3}$: $5 = B\left(1 - \left(-\frac{2}{3}\right)\right) = B\left(\frac{5}{3}\right) \implies B = 3$ [A1].
$f(x) = \frac{1}{1 - x} + \frac{3}{2 + 3x}$. [AG]
(b) Series expansion: [3 marks]
• $\frac{1}{1 - x} = (1 - x)^{-1} = 1 + x + x^2 + \dots$ [A1]
• $\frac{3}{2 + 3x} = \frac{3}{2}\left(1 + \frac{3}{2}x\right)^{-1} = \frac{3}{2}\left(1 - \frac{3}{2}x + \frac{9}{4}x^2 - \dots\right) = \frac{3}{2} - \frac{9}{4}x + \frac{27}{8}x^2 - \dots$ [A1]
Adding the two expansions: [M1]
Constant: $1 + \frac{3}{2} = \frac{5}{2}$.
$x$-term: $1 - \frac{9}{4} = -\frac{5}{4}$.
$x^2$-term: $1 + \frac{27}{8} = \frac{35}{8}$.
$f(x) = \frac{5}{2} - \frac{5}{4}x + \frac{35}{8}x^2 + \dots$ [A1].
Let $g(x) = \sqrt{\frac{1 + x}{1 - x}}$.
(a) Write $g(x)$ in the form $(1 + x)(1 - x^2)^{-1/2}$. [2 marks]
(b) Find the series expansion of $(1 - x^2)^{-1/2}$ up to the term in $x^4$. [2 marks]
(c) Hence find the Maclaurin series expansion of $g(x)$ up to the term in $x^3$. [2 marks]
(d) State the range of validity of this expansion. [1 mark]
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(a) Algebraic simplification: [2 marks]
$g(x) = \sqrt{\frac{1+x}{1-x}} = \frac{\sqrt{1+x}}{\sqrt{1-x}} = \frac{\sqrt{1+x}\sqrt{1+x}}{\sqrt{1-x}\sqrt{1+x}} = \frac{1+x}{\sqrt{1-x^2}} = (1+x)(1-x^2)^{-1/2}$ [M1][A1].
(b) Expansion of $(1 - x^2)^{-1/2}$: [2 marks]
Let $u = -x^2$. $(1 + u)^{-1/2} = 1 - \frac{1}{2}u + \frac{3}{8}u^2 - \dots$ [M1]
$= 1 - \frac{1}{2}(-x^2) + \frac{3}{8}(-x^2)^2 = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \dots$ [A1].
(c) Expansion of $g(x)$: [2 marks]
$g(x) = (1 + x)\left(1 + \frac{1}{2}x^2 + \dots\right) = 1 + \frac{1}{2}x^2 + x + \frac{1}{2}x^3 = 1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^3 + \dots$ [M1][A1].
(d) Validity: [1 mark]
$| -x^2 | < 1 \iff x^2 < 1 \iff |x| < 1$ [A1].