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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Partial fractions แพ็กฝึกเฉพาะหัวข้อ · เศษส่วนย่อย

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Partial fractions, 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

Express $\dfrac{5x+7}{(x+1)(x+3)}$ in partial fractions.

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M1 $\dfrac{5x+7}{(x+1)(x+3)}=\dfrac{A}{x+1}+\dfrac{B}{x+3}$, so $5x+7=A(x+3)+B(x+1)$  A1 $x=-1$: $2=2A$, $A=1$  A1 $x=-3$: $-8=-2B$, $B=4$; answer $\dfrac{1}{x+1}+\dfrac{4}{x+3}$.
Q2. 4 marks

Express $\dfrac{7x-2}{(x-2)(x+1)}$ in the form $\dfrac{A}{x-2}+\dfrac{B}{x+1}$, where $A$ and $B$ are constants to be found.

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M1 $7x-2=A(x+1)+B(x-2)$  M1 substitute $x=2$ and $x=-1$ (or compare coefficients)  A1 $x=2$: $12=3A$, $A=4$  A1 $x=-1$: $-9=-3B$, $B=3$; answer $\dfrac{4}{x-2}+\dfrac{3}{x+1}$.
Q3. 4 marks

Express $\dfrac{3x+5}{(x+2)^2}$ in the form $\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}$, where $A$ and $B$ are constants to be found.

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M1 $3x+5=A(x+2)+B$  M1 substitute $x=-2$ and compare $x$ coefficients  A1 comparing $x$: $A=3$  A1 $x=-2$: $B=-1$; answer $\dfrac{3}{x+2}-\dfrac{1}{(x+2)^2}$.
Q4. 5 marks

Express $\dfrac{x^2+8x+9}{(x+1)(x+2)^2}$ in the form $\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$, where $A$, $B$ and $C$ are constants to be found.

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M1 $x^2+8x+9=A(x+2)^2+B(x+1)(x+2)+C(x+1)$  M1 substitute $x=-1$, $x=-2$, then compare $x^2$ coefficients  A1 $x=-1$: $2=A$, so $A=2$  A1 $x=-2$: $-3=-C$, so $C=3$  A1 $x^2$: $1=A+B$, so $B=-1$; answer $\dfrac{2}{x+1}-\dfrac{1}{x+2}+\dfrac{3}{(x+2)^2}$.
Q5. 5 marks

Express $\dfrac{x^2+5x+8}{(x+1)(x+2)}$ in the form $P+\dfrac{Q}{x+1}+\dfrac{R}{x+2}$, where $P$, $Q$ and $R$ are constants to be found.

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M1 improper fraction: divide by $(x+1)(x+2)=x^2+3x+2$  A1 $x^2+5x+8=(x^2+3x+2)+(2x+6)$, so $P=1$ with remainder $2x+6$  M1 $2x+6=Q(x+2)+R(x+1)$  A1 $x=-1$: $4=Q$  A1 $x=-2$: $2=-R$, $R=-2$; answer $1+\dfrac{4}{x+1}-\dfrac{2}{x+2}$.
Q6. 6 marks

(a) Express $\dfrac{3x+11}{(x-3)(x+1)}$ in partial fractions. (b) Hence find $\displaystyle\int \frac{3x+11}{(x-3)(x+1)}\,dx$.

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(a) M1 $3x+11=A(x+1)+B(x-3)$  A1 $x=3$: $20=4A$, $A=5$  A1 $x=-1$: $8=-4B$, $B=-2$; so $\dfrac{5}{x-3}-\dfrac{2}{x+1}$.
(b) M1 integrate each term to a $\ln$ form  A1 $5\ln|x-3|$  A1 $-2\ln|x+1|+c$ (constant required).
Q7. 6 marks

Express $\dfrac{4x^2-5x-11}{(x-1)(x+2)(x-3)}$ in partial fractions.

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B1 correct form $\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{x-3}$  M1 $4x^2-5x-11=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)$  M1 substitute $x=1$, $x=-2$, $x=3$  A1 $x=1$: $-12=-6A$, $A=2$  A1 $x=-2$: $15=15B$, $B=1$  A1 $x=3$: $10=10C$, $C=1$; answer $\dfrac{2}{x-1}+\dfrac{1}{x+2}+\dfrac{1}{x-3}$.
Q8. 7 marks

(a) Show that $\dfrac{2x^3+5x^2-x-2}{(x-1)(x+3)}$ can be written in the form $ax+b+\dfrac{C}{x-1}+\dfrac{D}{x+3}$, stating the values of $a$, $b$, $C$ and $D$. (b) Hence find $\displaystyle\int \frac{2x^3+5x^2-x-2}{(x-1)(x+3)}\,dx$.

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(a) M1 divide by $(x-1)(x+3)=x^2+2x-3$  A1 quotient $2x+1$, i.e. $a=2$, $b=1$, remainder $3x+1$  M1 $\dfrac{3x+1}{(x-1)(x+3)}$: $3x+1=C(x+3)+D(x-1)$  A1 $x=1$: $4=4C$, $C=1$  A1 $x=-3$: $-8=-4D$, $D=2$.
(b) M1 integrate term by term  A1 $x^2+x+\ln|x-1|+2\ln|x+3|+c$.
Q9. 7 marks

$f(x)=\dfrac{4x^2+7x-3}{(x+1)^2(x-2)}$. (a) Express $f(x)$ in the form $\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x-2}$. (b) Hence show that $\displaystyle\int_3^6 f(x)\,dx=\ln 112+\frac{3}{14}$.

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(a) M1 $4x^2+7x-3=A(x+1)(x-2)+B(x-2)+C(x+1)^2$; substitute $x=-1$, $x=2$, compare $x^2$  A1 $x=-1$: $-6=-3B$, $B=2$  A1 $x=2$: $27=9C$, $C=3$  A1 $x^2$: $4=A+C$, $A=1$; so $f(x)=\dfrac{1}{x+1}+\dfrac{2}{(x+1)^2}+\dfrac{3}{x-2}$.
(b) M1 integrate: $\ln|x+1|-\dfrac{2}{x+1}+3\ln|x-2|$  A1 substitute limits: $\left(\ln 7-\tfrac{2}{7}+3\ln 4\right)-\left(\ln 4-\tfrac{1}{2}+3\ln 1\right)$  A1 $=\ln 7+2\ln 4+\tfrac{3}{14}=\ln 112+\tfrac{3}{14}$ ∎.
Q10. 8 marks

$g(x)=\dfrac{4+5x}{(1-x)(1+2x)}$. (a) Express $g(x)$ in the form $\dfrac{A}{1-x}+\dfrac{B}{1+2x}$. (b) Hence find the binomial expansion of $g(x)$ in ascending powers of $x$, up to and including the term in $x^2$. (c) State the range of values of $x$ for which the full expansion is valid.

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(a) M1 $4+5x=A(1+2x)+B(1-x)$  A1 $x=1$: $9=3A$, $A=3$  A1 $x=-\tfrac12$: $\tfrac32=\tfrac32 B$, $B=1$; so $g(x)=\dfrac{3}{1-x}+\dfrac{1}{1+2x}$.
(b) M1 $3(1-x)^{-1}=3(1+x+x^2+\cdots)$  M1 $(1+2x)^{-1}=1-2x+4x^2-\cdots$  A1 either expansion correct  A1 $g(x)=4+x+7x^2+\cdots$.
(c) B1 valid for $|x|<\tfrac12$ (the stricter of $|x|<1$ and $|2x|<1$).