Partial fractions and systems of linear equations – IB Maths exam questions with mark scheme

1StandardNon-calculatorHL only4 marks

Write $\dfrac{5x+1}{(x-1)(x+2)}$ in the form $\dfrac{A}{x-1}+\dfrac{B}{x+2}$, where $A,B\in\mathbb{Z}$.

$5x+1=A(x+2)+B(x-1)$M1
Substitutes $x=1$ or $x=-2$ (or equates coefficients)M1
$A=2$A1
$B=3$A1
Answer: $\dfrac2{x-1}+\dfrac3{x+2}$
Examiner tip: Check by substituting a third value of $x$, e.g. $x=0$: $\tfrac{1}{-2}=-2+\tfrac32$.

2HardNon-calculatorHL only6 marks

(a) Express $\dfrac{x+7}{(x-1)(x+3)}$ in partial fractions. [3]

(b) Hence find the exact value of $\displaystyle\int_2^5\frac{x+7}{(x-1)(x+3)}\,\mathrm dx$, giving your answer in the form $\ln k$, where $k\in\mathbb{Z}$. [3]

$x+7=A(x+3)+B(x-1)$ (or equivalent, e.g. cover-up method)M1
$A=2$A1
$B=-1$, giving $\dfrac2{x-1}-\dfrac1{x+3}$A1
Integrates both terms to logarithmsM1
$2\ln|x-1|-\ln|x+3|$ (allow without modulus signs; ignore $+c$)A1
$(2\ln4-\ln8)-(2\ln1-\ln5)=\ln\dfrac{16\times5}{8}=\ln10$A1
Answer: (a) $\dfrac2{x-1}-\dfrac1{x+3}$ (b) $\ln10$
Examiner tip: Keep the coefficient 2 with its log until the end, then use $2\ln4=\ln16$ and combine the logs into a single $\ln k$.

3HardNon-calculatorHL only6 marks

Consider the system of equations, where $m\in\mathbb{R}$:

$$x+y+z=1,\quad x+2y+3z=4,\quad 2x+3y+4z=m.$$

(a) Show that the system has solutions only when $m=5$. [4]

(b) For $m=5$, find the general solution. [2]

Eliminates $x$ from equations 1 and 2: $y+2z=3$M1
Eliminates $x$ from equations 1 and 3: $y+2z=m-2$M1
The two results require $m-2=3$A1
If $m\ne5$ the equations are inconsistent (no solution), so solutions exist only for $m=5$R1
Sets $z=t$: $y=3-2t$M1
$x=t-2$, so $(x,y,z)=(t-2,\ 3-2t,\ t)$A1
Answer: (a) shown ($m=5$) (b) $(x,y,z)=(t-2,\,3-2t,\,t)$
Examiner tip: Row 3 is row 1 + row 2 on the left, which is why the right-hand sides must agree: 1 + 4 = 5.

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4CoreCalculatorHL only4 marks

A theatre sells adult, child and concession tickets at fixed prices of $\$A$, $\$C$ and $\$S$ respectively.

On Monday, 2 adult, 3 child and 1 concession ticket cost $\$42$.
On Tuesday, 1 adult, 2 child and 3 concession tickets cost $\$36.50$.
On Wednesday, 3 adult, 1 child and 2 concession tickets cost $\$44.50$.

(a) Write down a system of three linear equations in $A$, $C$ and $S$. [2]

(b) Hence find the price of each type of ticket. [2]

At least one correct equation, e.g. $2A+3C+S=42$M1
All three correct: $2A+3C+S=42$, $A+2C+3S=36.5$, $3A+C+2S=44.5$A1
Valid attempt to solve the system (GDC or elimination); may be implied by at least one correct valueM1
$A=9.5,\ C=6,\ S=5$ (adult $\$9.50$, child $\$6$, concession $\$5$). Accept $9.50$; follow through from their equations in (b) only if their system has a unique solutionA1
Answer: (a) $2A+3C+S=42$, $A+2C+3S=36.5$, $3A+C+2S=44.5$ (b) adult $\$9.50$, child $\$6$, concession $\$5$
Examiner tip: Mixing up which coefficient goes with which ticket type costs the A1 in (a) and gives the wrong prices, so line the variables up in the same order (A, C, S) in every equation.

5CoreNon-calculatorHL only4 marks

Express $\dfrac{3x-1}{x^2+x-6}$ in partial fractions.

Factorises denominator: $(x+3)(x-2)$A1
Writes $\dfrac{A}{x+3}+\dfrac{B}{x-2}$ and forms an identity, e.g. $3x-1\equiv A(x-2)+B(x+3)$M1
$A=2$A1
$B=1$, giving $\dfrac{2}{x+3}+\dfrac{1}{x-2}$. Accept terms in either orderA1
Answer: $\dfrac{2}{x+3}+\dfrac{1}{x-2}$
Examiner tip: Substitute $x=2$ and $x=-3$ into the identity; each value kills one unknown, so there is no need to compare coefficients.

6StandardNon-calculatorHL only6 marks

The diagram shows part of the curve $y=ax^2+bx+c$, where $a,b,c\in\mathbb{R}$. The curve passes through the points $P(-1,\,2)$, $Q(1,\,0)$ and $R(2,\,5)$.

(a) Write down three equations in $a$, $b$ and $c$. [2]

(b) Solve your equations to find the values of $a$, $b$ and $c$. [3]

Substitutes at least one point correctly, e.g. $a-b+c=2$M1
All three: $a-b+c=2$, $a+b+c=0$, $4a+2b+c=5$A1
Valid elimination, e.g. subtracting first two equations: $2b=-2$M1
$b=-1$A1
$a=2,\ c=-1$A1
$2x^2-x-1=(2x+1)(x-1)=0\Rightarrow x=-\tfrac12$. Follow through from their $a,b,c$A1
Answer: (a) $a-b+c=2$, $a+b+c=0$, $4a+2b+c=5$ (b) $a=2,\ b=-1,\ c=-1$ (c) $x=-\tfrac12$
Examiner tip: Subtracting the equations from P and Q removes $a$ and $c$ in one step; spot this pair before eliminating blindly.

7StandardNon-calculatorHL only5 marks

Let $f(x)=\dfrac{5x}{(2x+1)(x-2)}$, for $x>2$.

(a) Write $f(x)$ in the form $\dfrac{A}{2x+1}+\dfrac{B}{x-2}$, where $A,B\in\mathbb{Z}$. [3]

(b) Hence find $\displaystyle\int f(x)\,\mathrm dx$. [2]

Forms identity $5x\equiv A(x-2)+B(2x+1)$ and substitutes a value or compares coefficientsM1
$A=1$A1
$B=2$A1
Integrates to logarithms with $\tfrac12$ coefficient for the first term, e.g. $\tfrac12\ln(2x+1)$M1
$\tfrac12\ln(2x+1)+2\ln(x-2)+c$. Accept modulus signs. Award A0 if $+c$ is omittedA1
Answer: (a) $\dfrac{1}{2x+1}+\dfrac{2}{x-2}$ (b) $\tfrac12\ln(2x+1)+2\ln(x-2)+c$
Examiner tip: The most common slip is writing $\ln(2x+1)$ instead of $\tfrac12\ln(2x+1)$; differentiate your answer to check the factor from the chain rule.

8StandardNon-calculatorHL only8 marks

Consider the system of equations

$$\begin{aligned}x+2y-z&=1\\2x+y+z&=5\\x-y+kz&=2\end{aligned}$$

where $k\in\mathbb{R}$.

(a) Show that the system has a unique solution unless $k=2$. [3]

(b) Explain why the system has no solution when $k=2$. [2]

Eliminates $x$ from two pairs of equations, e.g. $-3y+3z=3$ and $-3y+(k+1)z=1$M1
Eliminates $y$: $(k-2)z=-2$ (or equivalent)A1
$z=\dfrac{-2}{k-2}$ is defined (unique) only when $k\neq2$, hence unique solution unless $k=2$. Accept a correct determinant argument $6-3k\ne0$R1
Substitutes $k=2$ into their reduced equation to get $0z=-2$ or $0=-2$M1
This is impossible (equations inconsistent), so no solutionR1
Substitutes $k=3$: $z=-2$A1
Back-substitutes, e.g. $y-z=-1\Rightarrow y=-3$M1
$x=5,\ y=-3,\ z=-2$A1
Answer: (a) shown: $(k-2)z=-2$ (b) $0=-2$, inconsistent (c) $x=5,\ y=-3,\ z=-2$
Examiner tip: In (b) you must say the resulting statement is impossible; just writing $0=-2$ without the word 'inconsistent' or 'no solution' loses the R mark.

9HardNon-calculatorHL only8 marks

Consider the system of equations

$$\begin{aligned}x+y+z&=4\\x+2y+3z&=9\\2x+5y+az&=b\end{aligned}$$

where $a,b\in\mathbb{R}$.

(a) Show that the system can be reduced to $(a-8)z=b-23$. [3]

(b) Hence state the conditions on $a$ and $b$ for the system to have

(i) infinitely many solutions;
(ii) no solution. [2]

Eliminates $x$ correctly from a pair, e.g. eq2 $-$ eq1: $y+2z=5$M1
Second elimination: eq3 $-\,2\times$eq1: $3y+(a-2)z=b-8$A1
Subtracts $3\times(y+2z=5)$ to obtain $(a-8)z=b-23$ (AG)A1
(i) $a=8$ and $b=23$A1
(ii) $a=8$ and $b\neq23$A1
Sets $z=t$ and finds $y=5-2t$, $x=t-1$ (or equivalent general solution)M1
Sets $x=y$: $t-1=5-2t\Rightarrow t=2$M1
$x=1,\ y=1,\ z=2$A1
Answer: (a) shown (b)(i) $a=8,\ b=23$ (ii) $a=8,\ b\ne23$ (c) $(1,\,1,\,2)$
Examiner tip: For infinitely many solutions both parameters must be fixed; giving only $a=8$ for (b)(i) is the most common way to lose the mark.

10StretchNon-calculatorHL only9 marks

(a) Express $\dfrac{1}{(2r-1)(2r+1)}$ in partial fractions. [2]

(b) Hence show that $\displaystyle\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)}=\frac{n}{2n+1}$. [4]

(d) Write down the value of $\displaystyle\sum_{r=1}^{\infty}\frac{1}{(2r-1)(2r+1)}$. [1]

Forms $1\equiv A(2r+1)+B(2r-1)$ and finds one constantM1
$\dfrac{1}{2(2r-1)}-\dfrac{1}{2(2r+1)}$ (or equivalent, e.g. $\tfrac12\left(\tfrac{1}{2r-1}-\tfrac{1}{2r+1}\right)$)A1
Uses their partial fractions to write out at least the first two and the last term, e.g. $\tfrac12\left[\left(1-\tfrac13\right)+\left(\tfrac13-\tfrac15\right)+\dots+\left(\tfrac{1}{2n-1}-\tfrac{1}{2n+1}\right)\right]$M1
Shows or states that intermediate terms cancelR1
$\tfrac12\left(1-\dfrac{1}{2n+1}\right)$A1
$=\tfrac12\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}$ (AG; must see this step)A1
Sets up $\dfrac{n}{2n+1}>0.499$ and rearranges, e.g. $0.002n>0.499$, $n>249.5$M1
$n=250$A1
$\tfrac12$A1
Answer: (a) $\dfrac{1}{2(2r-1)}-\dfrac{1}{2(2r+1)}$ (b) shown (c) $n=250$ (d) $\tfrac12$
Examiner tip: Forgetting the factor $\tfrac12$ from part (a) makes the telescoped sum come out as $\dfrac{2n}{2n+1}$; carry it outside the bracket from the first line.

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