Free · no sign-up to startฟรี ไม่ต้องสมัครก็เริ่มได้
A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Binomial series แพ็กฝึกเฉพาะหัวข้อ · อนุกรมทวินาม (กำลังตรรกยะ)

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Negative & fractional indices, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Expand $(1+x)^{-2}$ in ascending powers of $x$, up to and including the term in $x^3$, stating the values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 binomial series with $n=-2$: $1+(-2)x+\tfrac{(-2)(-3)}{2!}x^2+\tfrac{(-2)(-3)(-4)}{3!}x^3$  A1 $1-2x+3x^2-4x^3$  B1 valid for $|x|<1$.
Q2. 4 marks

Expand $(1+3x)^{\frac12}$ in ascending powers of $x$, up to and including the term in $x^2$, stating the range of values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 $1+\tfrac12(3x)+\tfrac{\frac12\left(-\frac12\right)}{2!}(3x)^2$  A1 $1+\tfrac32x$  A1 $-\tfrac98x^2$  B1 valid for $|3x|<1$, i.e. $|x|<\tfrac13$.
Q3. 4 marks

Expand $(1-2x)^{-1}$ in ascending powers of $x$, up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 binomial series with $n=-1$, $u=-2x$: $1+(-1)(-2x)+\tfrac{(-1)(-2)}{2!}(-2x)^2+\tfrac{(-1)(-2)(-3)}{3!}(-2x)^3$  A2 $1+2x+4x^2+8x^3$ (A1 if one coefficient wrong)  B1 valid for $|2x|<1$, i.e. $|x|<\tfrac12$.
Q4. 5 marks

Expand $(4+x)^{\frac12}$ in ascending powers of $x$, up to and including the term in $x^2$, stating the range of values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 $(4+x)^{\frac12}=4^{\frac12}\!\left(1+\tfrac{x}{4}\right)^{\frac12}=2\left(1+\tfrac{x}{4}\right)^{\frac12}$  M1 $2\left(1+\tfrac12\cdot\tfrac{x}{4}+\tfrac{\frac12\left(-\frac12\right)}{2!}\left(\tfrac{x}{4}\right)^{2}\right)$  A1 $2+\tfrac{x}{4}$  A1 $-\tfrac{x^2}{64}$  B1 valid for $\left|\tfrac{x}{4}\right|<1$, i.e. $|x|<4$.
Q5. 5 marks

Expand $(1+2x)^{-3}$ in ascending powers of $x$, up to and including the term in $x^3$, stating the range of values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 $1+(-3)(2x)+\tfrac{(-3)(-4)}{2!}(2x)^2+\tfrac{(-3)(-4)(-5)}{3!}(2x)^3$  A1 $1-6x$  A1 $+24x^2$  A1 $-80x^3$  B1 valid for $|2x|<1$, i.e. $|x|<\tfrac12$.
Q6. 6 marks

(a) Expand $(1-4x)^{\frac12}$ in ascending powers of $x$, up to and including the term in $x^2$. (b) By substituting $x=\tfrac{1}{100}$ into your expansion, show that $\sqrt{6}\approx 2.4495$, explaining briefly why this value of $x$ may be used.

Mark schemeshow ▾
(a) M1 $1+\tfrac12(-4x)+\tfrac{\frac12\left(-\frac12\right)}{2!}(-4x)^2$  A1 $1-2x$  A1 $-2x^2$.
(b) B1 expansion valid for $|x|<\tfrac14$ and $\tfrac{1}{100}<\tfrac14$ ✓  M1 $x=\tfrac{1}{100}$: $\sqrt{0.96}\approx 1-0.02-0.0002=0.9798$, and $\sqrt{0.96}=\sqrt{\tfrac{96}{100}}=\tfrac{4\sqrt6}{10}=\tfrac{2\sqrt6}{5}$  A1 $\sqrt6\approx\tfrac52(0.9798)=2.4495$ ∎ (true value $2.44949\ldots$).
Q7. 6 marks

Expand $(2+x)^{-2}$ in ascending powers of $x$, up to and including the term in $x^3$, stating the range of values of $x$ for which the expansion is valid.

Mark schemeshow ▾
M1 $(2+x)^{-2}=2^{-2}\!\left(1+\tfrac{x}{2}\right)^{-2}=\tfrac14\left(1+\tfrac{x}{2}\right)^{-2}$  M1 $\tfrac14\left(1-2\cdot\tfrac{x}{2}+3\left(\tfrac{x}{2}\right)^{2}-4\left(\tfrac{x}{2}\right)^{3}\right)$  A1 $\tfrac14-\tfrac{x}{4}$  A1 $+\tfrac{3x^2}{16}$  A1 $-\tfrac{x^3}{8}$  B1 valid for $\left|\tfrac{x}{2}\right|<1$, i.e. $|x|<2$.
Q8. 7 marks

(a) Expand $(8+3x)^{\frac13}$ in ascending powers of $x$, up to and including the term in $x^2$, stating the range of values of $x$ for which the expansion is valid. (b) Use your expansion with $x=\tfrac13$ to find an approximation for $\sqrt[3]{9}$, giving your answer to 3 decimal places, and comment on its accuracy given that $\sqrt[3]{9}=2.08008\ldots$

Mark schemeshow ▾
(a) M1 $(8+3x)^{\frac13}=8^{\frac13}\!\left(1+\tfrac{3x}{8}\right)^{\frac13}=2\left(1+\tfrac{3x}{8}\right)^{\frac13}$  M1 $2\left(1+\tfrac13\cdot\tfrac{3x}{8}+\tfrac{\frac13\left(-\frac23\right)}{2!}\left(\tfrac{3x}{8}\right)^{2}\right)$  A1 $2+\tfrac{x}{4}$  A1 $-\tfrac{x^2}{32}$  B1 valid for $\left|\tfrac{3x}{8}\right|<1$, i.e. $|x|<\tfrac83$.
(b) M1 $x=\tfrac13$: $(8+1)^{\frac13}=\sqrt[3]{9}\approx 2+\tfrac{1}{12}-\tfrac{1}{288}=\tfrac{599}{288}=2.07986\ldots$  A1 $\sqrt[3]{9}\approx 2.080$ (3 d.p.); error $\approx 2.2\times10^{-4}$ — only two terms after the constant, and $x=\tfrac13$ is small compared with $\tfrac83$, so the truncated series agrees with the true value to 3 d.p.
Q9. 7 marks

(a) Expand $(1-2x)^{\frac12}$ in ascending powers of $x$, up to and including the term in $x^3$, stating the range of values of $x$ for which the expansion is valid. (b) By substituting $x=\tfrac{1}{100}$, show that $\sqrt{2}\approx 1.4142136$, correct to 7 decimal places. Hint: $\sqrt{0.98}=\tfrac{7\sqrt2}{10}$.

Mark schemeshow ▾
(a) M1 $1+\tfrac12(-2x)+\tfrac{\frac12\left(-\frac12\right)}{2!}(-2x)^2+\tfrac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{3!}(-2x)^3$  A1 $1-x$  A1 $-\tfrac{x^2}{2}$  A1 $-\tfrac{x^3}{2}$  B1 valid for $|2x|<1$, i.e. $|x|<\tfrac12$.
(b) M1 $x=\tfrac{1}{100}$: $\sqrt{0.98}\approx 1-0.01-0.00005-0.0000005=0.9899495$, and $\sqrt{0.98}=\sqrt{\tfrac{98}{100}}=\tfrac{7\sqrt2}{10}$, so $\sqrt2\approx\tfrac{10}{7}\times 0.9899495$  A1 $\sqrt2\approx 1.4142136$ (7 d.p.) ∎.
Q10. 8 marks

$f(x)=\dfrac{7+x}{(1+x)(2-x)}$. (a) Express $f(x)$ in partial fractions. (b) Hence expand $f(x)$ in ascending powers of $x$, up to and including the term in $x^2$. (c) State, with a reason, the range of values of $x$ for which the full expansion is valid.

Mark schemeshow ▾
(a) M1 $\dfrac{7+x}{(1+x)(2-x)}=\dfrac{A}{1+x}+\dfrac{B}{2-x}$; cover-up or substitution $x=-1,\ x=2$  A1 $A=2$  A1 $B=3$.
(b) M1 $2(1+x)^{-1}=2(1-x+x^2-\cdots)=2-2x+2x^2$  M1 $\dfrac{3}{2-x}=\tfrac32\left(1-\tfrac{x}{2}\right)^{-1}=\tfrac32\left(1+\tfrac{x}{2}+\tfrac{x^2}{4}\right)=\tfrac32+\tfrac{3x}{4}+\tfrac{3x^2}{8}$  A1 $f(x)\approx\tfrac72-\tfrac{5x}{4}$  A1 $+\tfrac{19x^2}{8}$.
(c) B1 need both $|x|<1$ and $\left|\tfrac{x}{2}\right|<1$ (i.e. $|x|<2$); the smaller interval wins, so valid for $|x|<1$.