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Past Papers

## Algebra

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Binomial Expansions

Has not appeared

Factorisation

Linear Equations

Has not appeared

Simultaneous Equations

Cubic Equations

Logarithmic Equations

Exponential Equations

Surd Equations

Modulus Equations

Inequalities

Mathematical Proofs

Distance, Speed & Time

## 2022 Paper 1 Question 1

(a) Find the two values of $$m\in\mathbb{Z}$$ for which the following equation in $$x$$ has exactly one solution:

\begin{align}3x^2-mx+3=0\end{align}

$$m=-6$$ or $$m=6$$

Solution

\begin{align}b^2-4ac=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-m)^2-4(3)(3)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m^2=36\end{align}

$$m=-6$$ or $$m=6$$

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Explain why the following equation in $$x$$ has no real solutions:

\begin{align}(2x+3)^2+7=0\end{align}

The discriminant is less than zero.

Solution

\begin{align}(2x+3)^2+7=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x^2+12x+9+7=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x^2+12x+16=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b^2-4ac&=12^2-4(4)(16)\\&=-112\end{align}

As the discriminant is less than zero, the roots are complex.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c)

(i) Show that $$x=-1$$ is not a solution of $$3x^2+2x+5=0$$.

(ii) Find the remainder when $$3x^2+2x+5$$ is divided by $$x+1$$.

That is, find the value of $$c$$ when $$3x^2+2x+5$$ is written in the form

\begin{align}3x^2+2x+5=(x+1)(ax+b)+c\end{align}

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

(ii) $$6$$

Solution

(i)

\begin{align}3x^2+2x+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3(-1)^2+2(-1)+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3-2+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6\end{align}

As this is not equal to zero, it is not a solution.

(ii)

\begin{align}3x^2+2x+5=(x+1)(ax+b)+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+2x+5=ax^2+bx+ax+b+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+2x+5=ax^2+(a+b)x+(b+c)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=a\end{align}

\begin{align}2=a+b\end{align}

\begin{align}5=b+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=a\end{align}

\begin{align}2=3+b\end{align}

\begin{align}5=b+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=a\end{align}

\begin{align}-1=b\end{align}

\begin{align}5=b+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=a\end{align}

\begin{align}-1=b\end{align}

\begin{align}5=-1+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=a\end{align}

\begin{align}-1=b\end{align}

\begin{align}6=c\end{align}

Therefore, the remainder is $$6$$.

Video Walkthrough

## 2022 Paper 1 Question 2(b)

(b) The diagram shows the graph of a function $$f(x)=ax^2+bx+c$$, where $$a,b,c\in\mathbb{Z}$$.
Three regions on the diagram are marked K, L, and N.
Each of these regions is bounded by the $$x$$-axis, the graph of$$f(x)$$, and two vertical lines.

(i) The area of region K is $$538$$ square units. Use integration of $$f(x)$$ to show that:

\begin{align}4a+3b+3c=807\end{align}

(ii) The areas of the three regions K, L, and N give the following three equations (including the equation from part (b)(i)):

\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}

Solve these equations to find the values of $$a$$, $$b$$ and $$c$$.

(ii) $$a=-12$$, $$b=60$$ and $$c=225$$

Solution

(i)

\begin{align}\int_0^2 f(x)\,dx=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int_0^2 (ax^2+bx+c)\,dx=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int_0^2ax^2\,dx+\int_0^2bx\,dx+\int_0^2c\,dx=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a\int_0^2x^2\,dx+b\int_0^2x\,dx+c\int_0^2\,dx=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left.\frac{ax^3}{3}\right|_0^2+\left.\frac{bx^2}{2}\right|_0^2+\left.cx\right|_0^2=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left[\frac{a(2^3)}{3}-\frac{a(0^3)}{3})\right]+\left[\frac{b(2^2)}{2}-\frac{b(0^2)}{2}\right]+[c(2)-c(0)]=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{8a}{3}+\frac{4b}{2}+2c=538\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8a+6b+6c=1614\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4a+3b+3c=807\end{align}

as required.

(ii)

\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}

\begin{align}\downarrow\end{align}

\begin{align}24a+6b&=72\\48a+6b&=-216\end{align}

\begin{align}\downarrow\end{align}

\begin{align}24a=-288\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a=-12\end{align}

and

\begin{align}24a+6b=72\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b&=\frac{72-24a}{6}\\&=\frac{72-24(-12)}{6}\\&=60\end{align}

and

\begin{align}4a+3b+3c&=807\end{align}

\begin{align}\downarrow\end{align}

\begin{align}c&=\frac{807-4a-3b}{3}\\&=\frac{807-4(-12)-3(60)}{3}\\&=225\end{align}

Video Walkthrough

## 2022 Paper 1 Question 4(b)

(b) The first three terms in an arithmetic sequence are as follows, where $$k\in\mathbb{R}$$:

\begin{align}5e^{-k},&&13,&&5e^k\end{align}

(i) By letting $$y=e^k$$ in this arithmetic sequence, show that:

\begin{align}5y^2-26y+5=0\end{align}

(ii) Use the equation in $$y$$ in part (b)(i) to find the two possible values of $$k$$.
Give each value in the form $$\ln p$$ or $$-\ln p$$, where $$p\in\mathbb{N}$$.

(ii) $$k=-\ln5$$ or $$k=\ln5$$

Solution

(i)

\begin{align}13-5e^{-k}=5e^k-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}13-\frac{5}{y}=5y-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}13y-5=5y^2-13y\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5y^2-26y+5=0\end{align}

as required.

(ii)

\begin{align}5y^2-26y+5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5y-1)(y-5)=0\end{align}

\begin{align}\downarrow\end{align}

$$y=\dfrac{1}{5}$$ or $$y=5$$

\begin{align}\downarrow\end{align}

$$e^k=\dfrac{1}{5}$$ or $$e^k=5$$

\begin{align}\downarrow\end{align}

$$k=\ln\left(\dfrac{1}{5}\right)$$ or $$k=\ln5$$

\begin{align}\downarrow\end{align}

$$k=-\ln5$$ or $$k=\ln5$$

Video Walkthrough

## 2022 Paper 1 Question 5(b)

(b) $$f(x)=2x^3-21x^2+40x+63$$, where $$x\in\mathbb{R}$$.

(i) $$x+1$$ is a factor of $$f(x)$$. Find the three values of $$x$$ for which $$f(x)=0$$.

(ii) Find the range of values of $$x$$ for which $$f'(x)$$ is negative, correct to $$2$$ decimal places.

(i) $$x=-1$$, $$x=4.5$$ or $$x=7$$

(ii) $$1.14<x<5.86$$

Solution

(i)

$\require{enclose} \begin{array}{rll} 2x^2-23x+63\phantom{000000}\, \\[-3pt] x+1 \enclose{longdiv}{\,2x^3-21x^2+40x+63} \\[-3pt] \underline{2x^3+2x^2\phantom{00000000000}\,\,} \\[-3pt] -23x^2+40x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-23x^2-23x\phantom{00}\,}\\[-3pt]\phantom{00}63x+63\\[-3pt]\phantom{00}\underline{63x+63}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}2x^3-21x^2+40x+63=(x+1)(2x^2-23x+63)\end{align}

$\,$

Factors

\begin{align}(x+1)(2x^2-23x+63)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+1)(2x-9)(x-7)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-1$$, $$x=4.5$$ or $$x=7$$

(ii)

Derivative

\begin{align}f'(x)=2x^3-21x^2+40x+63\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f^{\prime\prime}(x)=6x^2-42x+40\end{align}

$\,$

Inequality

\begin{align}6x^2-42x+40<0\end{align}

Roots:

$$a=6$$, $$b=-42$$ and $$c=40$$

\begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=\frac{-(-42)\pm\sqrt{(-42)^2-4(6)(40)}}{2(6)}\\&=1.137…\mbox{ or }5.862…\\&\approx1.14\mbox{ or } 5.86\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.14<x<5.86\end{align}

Video Walkthrough

## 2022 Paper 1 Question 10(e)

If we learn a skill, and then don’t practise it, how well we can do it usually decreases over time.
For example, if you learn to play the guitar, and then don’t play the guitar for a number of months, you will probably not be as good at playing the guitar the first time you try it again.

This effect can be modelled by the following equation:

\begin{align}A=B(t+1)^c\end{align}

where $$A$$ is a measure of how well the skill can be done at a certain time ($$t=0$$);
$$B$$ is a measure of how well the skill can be done $$t$$ months later, without practising;
$$c$$ is a constant; and $$A,B,c,t\in\mathbb{R}$$.

(e)

(i) Write $$c$$ in terms of $$\log_{10} A$$, $$\log_{10} B$$ and $$\log_{10} (t+1)$$.

(ii) A student got $$80\%$$ on a guitar exam.
After two years of not playing the guitar, the student got $$47\%$$ on the same exam.

Use this to find the value of $$c$$ in the model above, correct to $$3$$ decimal places.

(i) $$c=\dfrac{\log_{10}A-\log_{10}B}{\log_{10}(t+1)}$$

(ii) $$0.165$$

Solution

(i)

\begin{align}A=B(t+1)^c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\log_{10}A&=\log_{10}[B(t+1)^c]\\&=\log_{10}B+\log_{10}(t+1)^c\\&=\log_{10}B+c\log_{10}(t+1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}c=\frac{\log_{10}A-\log_{10}B}{\log_{10}(t+1)}\end{align}

(ii)

\begin{align}c&=\frac{\log_{10}80-\log_{10}47}{\log_{10}(24+1)}\\&\approx0.165\end{align}

Video Walkthrough

## 2021 Paper 1 Question 2

(a) Given that $$x=-3$$ is a solution to $$|x+p|=5$$, find the two values of $$p$$, where $$p\in\mathbb{Z}$$.

$$p=-2$$ or $$p=8$$

Solution

\begin{align}|-3+p|=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-3+p)^2=5^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}9-6p+p^2=25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p^2-6p-16=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(p+2)(p-8)=0\end{align}

\begin{align}\downarrow\end{align}

$$p=-2$$ or $$p=8$$

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) $$(x+4)$$ is a factor of $$f(x)=x^3+qx^2-22x+56$$, where $$x\in\mathbb{R}$$ and $$q\in\mathbb{Z}$$.

Show that $$q=-5$$, and find the three roots of $$f(x)$$.

$$x=-4$$, $$x=2$$ and $$x=7$$.

Solution

\begin{align}f(-4)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-4)^3+q(-4)^2-22(-4)+56=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-64+16q+88+56=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}16q+80=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q&=-\frac{80}{16}\\&=-5\end{align}

and

$\require{enclose} \begin{array}{rll} x^2-9x+14\phantom{000000}\, \\[-3pt] x+4 \enclose{longdiv}{\,x^3-5x^2-22x+56} \\[-3pt] \underline{x^3+4x^2\phantom{0000000000}\,\,} \\[-3pt] -9x^2-22x\phantom{0000}\,\\[-3pt]\phantom{00000}\underline{-9x^2-36x\phantom{0000}\,}\\[-3pt]\phantom{00}14x+56\\[-3pt]\phantom{00}\underline{14x+56}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}x^3-5x^2-22x+56&=(x+4)(x^2-9x+14)\\&=(x+4)(x-2)(x-7)\end{align}

Therefore, the roots are $$x=-4$$, $$x=2$$ and $$x=7$$.

Video Walkthrough

## 2021 Paper 1 Question 3(b)

(b)

(i) Given that $$f(x)=3x^2+8x-35$$, where $$x\in\mathbb{R}$$, find the two roots of $$f(x)=0$$.

(ii) Hence or otherwise, solve the equation $$3^{2m+1}=35-8(3^m)$$, where $$m\in\mathbb{R}$$.
Give your answer in the form $$m=\log_3p-q$$, where $$p,q\in\mathbb{N}$$.

$$x=\dfrac{7}{3}$$ or $$x=-5$$

$$\log_3 7-1$$

Solution

(i)

\begin{align}3x^2+8x-35=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(3x-7)(x+5)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=\dfrac{7}{3}$$ or $$x=-5$$

(ii)

$$3^m=\dfrac{7}{3}$$ or $$3^m=-5$$

Since it is not possible for $$3^m=-5$$, the first root is the only possibility.

\begin{align}3^m=\frac{7}{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m&=\log_3\left(\frac{7}{3}\right)\\&=\log_3 7-\log_3 3\\&=\log_3 7-1\end{align}

Video Walkthrough

## 2021 Paper 2 Question 5(b)

(b) At 9.00 a.m. a delivery van leaves a factory.
It travels towards its destination at an average speed of $$60\mbox{ km/h}$$.
One hour and $$45$$ minutes later a second van leaves the factory on the same route.
It travels at an average speed of $$95\mbox{ km/h}$$.
Both vans arrive at their destination at the same time.
Find at what time they arrive.

$$13:45$$

Solution

\begin{align}d_1=d_2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_1t_1=v_2t_2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}60t_1=95(t_1-1.75)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}60t_1=95t_1-166.25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t_1&=\frac{166.25}{95-60}\\&=4.75\mbox{ h} \end{align}

Therefore, the time that they arrive at is $$4.75$$ hours past $$9:00$$, i.e. it is $$13:45$$.

Video Walkthrough

## 2021 Paper 2 Question 7(b)

The diagram (Triangle $$ABC$$) shows the $$3$$ sections of a level triathlon course.
In order to complete the triathlon, each contestant must swim $$4\mbox{ km}$$ from $$C$$ to $$B$$, cycle from $$B$$ to $$A$$, and then run $$28\mbox{ km}$$ from $$A$$ to $$C$$.
Mary can cycle at an average speed of $$25\mbox{ km/hour}$$.
It takes her $$1$$ hour and $$12$$ minutes to cycle from $$B$$ to $$A$$.

(b) On average, Mary can run $$5.6$$ times as fast as she can swim.
It takes her $$4.8$$ hours to complete the course.
Find her average swimming speed in $$\mbox{ km/h}$$.

$$2.5\mbox{ km/h}$$

Solution

\begin{align}T_s+T_r&=4.8-1.2\\&=3.6\mbox{ h}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{d_s}{v_s}+\frac{d_r}{v_r}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4}{v_s}+\frac{28}{5.6v_s}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4(5.6)+28}{5.6v_s}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4(5.6)+28=(3.6)(5.6v_s)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50.4=20.16v_s\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_s&=\frac{50.4}{20.16}\\&=2.5\mbox{ km/h}\end{align}

Video Walkthrough

## 2020 Paper 1 Question 1

(a) $$f(x)=x^2+5x+p$$ where $$x\in\mathbb{R}$$, $$-3\leq p\leq8$$, and $$p\in\mathbb{Z}$$.

(i) Find the value of $$p$$ for which $$x+3$$ is a factor of $$f(x)$$.

(ii) Find the value of $$p$$ for which $$f(x)$$ has roots which differ by $$3$$.

(iii) Find the two values of $$p$$ for which the graph of $$f(x)$$ will not cross the $$x$$-axis.

(i) $$p=6$$

(ii) $$p=4$$

(iii) $$p=7$$ or $$p=8$$

Solution

(i)

\begin{align}f(-3)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-3)^2+5(-3)+p=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p&=-9+15\\&=6\end{align}

(ii)

\begin{align}x^2+5x+p&=(x-\alpha)(x-\alpha-3)\\&=x^2-\alpha x-3x-\alpha x+\alpha^2+3\alpha\\&=x^2+(-3-2\alpha)x+(\alpha^2+3\alpha)\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}5=-3-2\alpha\end{align}

\begin{align}p=\alpha^2+3\alpha\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\alpha=-4\end{align}

\begin{align}p&=\alpha^2+3\alpha\\&=(-4)^2+3(-4)\\&=4\end{align}

(iii)

\begin{align}b^2-4ac<0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5^2-4(1)(p)<0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}25-4p<0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-4p<-25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4p>25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p>6.75\end{align}

Therefore, $$p=7$$ or $$p=8$$ (since $$-3\leq p\leq8$$).

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Find the range of values of $$x$$ for which $$|2x+5|-1\leq0$$, where $$x\in\mathbb{R}$$.

$$-3\leq x\leq -2$$

Solution

\begin{align}|2x+5|\leq 1\end{align}

\begin{align}(2x+5)^2\leq 1^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x^2+20x+25\leq1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x^2+20x+24\leq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2+5x+6\leq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+3)(x+2)\leq 0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-3\leq x\leq -2\end{align}

Video Walkthrough

## 2020 Paper 1 Question 3(b)

(b) The real variables $$y$$ and $$x$$ are related by $$y=5x^2$$.

(i) The equation $$y=5x^2$$ can be rewritten in the form $$\log_5y=a+b\log_5x$$.
Find the value of $$a$$ and the value of $$b$$.

(ii) Hence, or otherwise, find the real values of $$y$$ for which

\begin{align}\log_5y=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}

(i) $$a=1$$ and $$b=2$$

(ii) $$y=\dfrac{1}{5}$$ and $$y=3{,}125$$

Solution

(i)

\begin{align}\log_5y&=\log_55x^2\\&=\log_55+\log_5x^2\\&=1+2\log_5x\end{align}

\begin{align}\downarrow\end{align}

$$a=1$$ and $$b=2$$

(ii)

\begin{align}\log_5y=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\log_5x^2=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\log_5x^2=\log_55^2+\log_5\left(\frac{126}{25}x-1\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\log_5x^2=\log_55^2+\log_5(5^2)\left(\frac{126}{25}x-1\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5x^2=126x-25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5x^2-126x+25=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5x-1)(x-25)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=\dfrac{1}{5}$$ or $$x=25$$

When $$x=\dfrac{1}{5}$$:

\begin{align}y&=5x^2\\&=5\left(\frac{1}{5}\right)^2\\&=\frac{1}{5}\end{align}

and when $$x=25$$:

\begin{align}y&=5x^2\\&=5(25^2)\\&=3{,}125\end{align}

Video Walkthrough

## 2020 Paper 1 Question 7(c)

(a) A number of the form $$1+2+3+…+n$$ is sometimes called a triangular number because
it can be represented as an equilateral triangle.
The diagram below shows the first three terms in the sequence of triangular numbers.

(i) Complete the table below to list the next five triangular numbers.

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

(ii) The $$n$$th triangular number can be found directly using the formula

\begin{align}T_n=\frac{n(n+1)}{2}\end{align}

Is $$1275$$ a triangular number? Give a reason for your answer.

(i)

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

$$10$$

$$15$$

$$21$$

$$28$$

$$36$$

(ii) Yes

Solution

(i)

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

$$10$$

$$15$$

$$21$$

$$28$$

$$36$$

(ii)

\begin{align}\frac{n(n+1)}{2}=1275\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n^2+n=2550\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n^2+n-2550=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(n-50)(n+51)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n=50\end{align}

Therefore, $$1275$$ is a triangular number.

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(b)

(i) The $$(n+1)$$th triangular number can be written as $$T_{n+1}=T_n+(n+1)$$, where $$n\in\mathbb{N}$$.

Write the expression $$\dfrac{n(n+1)}{2}+(n+1)$$ as a single fraction in its simplest form.

(ii) Prove that the sum of any two consecutive triangular numbers will always be a square number (a number in the form $$k^2$$, where $$k\in\mathbb{N}$$).

(iii) Two consecutive triangular numbers sum to $$12{,}544$$.
Find the smaller of these two numbers.

(i) $$\dfrac{(n+1)(n+2)}{2}$$

(iii) $$6{,}216$$

Solution

(i)

\begin{align}T_{n+1}&=T_n+(n+1)\\&=\frac{n(n+1)}{2}+(n+1)\\&=\frac{n(n+1)+2(n+1)}{2}\\&=\frac{(n+1)(n+2)}{2}\end{align}

(ii)

\begin{align}T_{n+1}+T_n&=\frac{(n+1)(n+2)}{2}+\frac{n(n+1)}{2}\\&=\frac{(n+1)(2n+2)}{2}\\&=2\frac{(n+1)(n+1)}{2}\\&=(n+1)^2\end{align}

(iii)

\begin{align}(n+1)^2=12{,}544\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n+1=112\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n=111\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_{111}&=\frac{(111)(111+1)}{2}\\&=6{,}216\end{align}

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(c) Some numbers are both triangular and square, for example $$36$$.
Leonhard Euler (1778) discovered the following formula for these numbers

\begin{align}N_k=\left(\frac{(3+2\sqrt{2})^k-(3-2\sqrt{2})^k}{4\sqrt{2}}\right)^2\end{align}

where $$N_k$$ is the $$k$$th number that is both triangular and square.
Use Euler’s formula to find $$N_3$$, the third number that is both triangular and square.

$$1{,}225$$

Solution

\begin{align}N_3&=\left(\frac{(3+2\sqrt{2})^3-(3-2\sqrt{2})^3}{4\sqrt{2}}\right)^2\\&=1{,}225\end{align}

Video Walkthrough

## 2020 Paper 1 Question 9(e)

The number of bacteria in the early stages of a growing colony of bacteria can be approximated
using the function:

\begin{align}N(t)=450e^{0.065t}\end{align}

where $$t$$ is the time, measured in hours, since the colony started to grow, and $$N(t)$$ is the number of bacteria in the colony at time $$t$$.

(e) The number of bacteria in the early stages of a different colony of bacteria can be approximated using the function:

\begin{align}P(t)=220e^{0.17t}\end{align}

where $$P(t)$$ is the number of bacteria and $$t$$ is measured in hours.
Assume that both colonies start growing at the same time.
Find the time, to the nearest hour, at which the number of bacteria in both colonies will be equal.

$$7\mbox{ hr}$$

Solution

\begin{align}N(t)=P(t)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}450e^{0.065t}=220e^{0.17t}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{e^{0.065t}}{e^{0.17t}}=\frac{220}{450}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e^{-0.105t}=\frac{220}{450}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-0.105t=\ln\left(\frac{220}{450}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=-\frac{1}{0.105}\ln\left(\frac{220}{450}\right)\\&\approx7\mbox{ hr}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 1

(a) In the expansion of $$(2x+1)(x^2+px+4)$$, where $$p\in\mathbb{N}$$, the coefficient of $$x$$ is twice the coefficient of $$x^2$$. Find the value of $$p$$.

$$p=2$$

Solution

\begin{align}(2x+1)(x^2+px+4)&=2x^3+2px^2+8x+x^2+px+4\\&=2x^3+(2p+1)x^2+(8+p)x+4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8+p=2(2p+1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8+p=4p+2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3p=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p=2\end{align}

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(b) Solve the equation $$\dfrac{3}{2x+1}+\dfrac{2}{5}=\dfrac{2}{3x-1}$$ where $$x\neq-\dfrac{1}{2},\dfrac{1}{3}$$, and $$x\in\mathbb{R}$$.

$$x=-3$$ or $$x=\dfrac{3}{4}$$

Solution

\begin{align}\frac{3}{2x+1}+\frac{2}{5}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{3(5)+2(2x+1)}{(2x+1)(5)}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4x+17}{10x+5}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(4x+17)(3x-1)=2(10x+5)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x^2-4x+51x-17=20x+10\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x^2+27x-27=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+9x-9=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+3)(4x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-3$$ or $$x=\dfrac{3}{4}$$

Video Walkthrough

## 2019 Paper 1 Question 3(a) & (b)

(a) Factorise fully: $$3xy-9x+4y-12$$

$$(3x+4)(y-3)$$

Solution

\begin{align}3xy-9x+4y-12&=3x(y-3)+4(y-3)\\&=(3x+4)(y-3)\end{align}

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(b) $$g(x)=3x\ln x-9x+4\ln x-12$$.
Using your answer to part (a) or otherwise, solve $$g(x)=0$$.

$$x=-\dfrac{4}{3}$$ or $$x=e^3$$

Solution

\begin{align}(3x+4)(\ln x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$3x+4=0$$ or $$\ln x-3=0$$

\begin{align}\downarrow\end{align}

$$x=-\dfrac{4}{3}$$ or $$x=e^3$$

Video Walkthrough

## 2019 Paper 1 Question 6(b)

(b) Prove, using contradiction, that $$\sqrt{2}$$ is not a rational number.

Solution

Assume $$\sqrt{2}$$ is a rational number, i.e. that

\begin{align}\sqrt{2}=\frac{a}{b}\end{align}

where $$a,b\in\mathbb{Z}$$ and the fraction is written in its simplest form. Hence

\begin{align}a^2=2b^2\end{align}

and therefore $$a$$ is even, i.e. $$a=2m$$ for some $$m\in\mathbb{Z}$$.

\begin{align}(2m)^2=2b^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b^2=2m^2\end{align}

and therefore $$b$$ is even, i.e. $$b=2n$$ for some $$n\in\mathbb{Z}$$.

\begin{align}\sqrt{2}&=\frac{a}{b}\\&=\frac{2m}{2n}\end{align}

However, we stated that the fraction was already in its simplest form – a contraction.

Hence, $$\sqrt{2}$$ is irrational.

Video Walkthrough

## 2018 Paper 1 Question 1

(a) Solve the simultaneous equations.

\begin{align}2x+3y-z&=-4\\3x+2y+2z&=14\\x-3z&=-13\end{align}

$$x=2$$ and $$y=-1$$ and $$z=5$$

Solution

\begin{align}2x+3y-z&=-4\\3x+2y+2z&=14\\x-3z&=-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x+6y-2z&=-8\\9x+6y+6z&=42\\x-3z&=-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5x+8z=50\\x-3z&=-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5x+8z=50\\5x-15z&=-65\end{align}

\begin{align}\downarrow\end{align}

\begin{align}23z=115\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z&=\frac{115}{23}\\&=5\end{align}

and

\begin{align}x-3z=-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=3z-13\\&=3(5)-13\\&=2\end{align}

and

\begin{align}2x+3y-z=-4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{-4-2x+z}{3}\\&=\frac{-4-2(2)+5}{3}\\&=-1\end{align}

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(b) Solve the inequality $$\dfrac{2x-3}{x+2}\geq3$$, where $$x\in\mathbb{R}$$ and $$x\neq-2$$.

$$-9\leq x<2$$

Solution

\begin{align}\frac{2x-3}{x+2}\geq3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(2x-3)(x+2)\geq3(x+2)^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x^2+4x-3x-6\geq3x^2+12x+12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-x^2-11x-18\geq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2+11x+18\leq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+9)(x+2)\leq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-9\leq x<-2 \end{align}

Video Walkthrough

## 2018 Paper 1 Question 2(b)

(b) If ݂$$f(x)=x^3-17x^2+80x-64$$, $$x\in\mathbb{R}$$, show that ݂$$f(1)=0$$, and find another value of $$x$$ for which ݂$$f(x)=0$$.

$$x=8$$

Solution

\begin{align}f(1)&=1^3-17(1^2)+80(1)-64\\&=1-17+80-64\\&=0\end{align}

as required.

$\require{enclose} \begin{array}{rll} x^2-16x+64\phantom{000000}\, \\[-3pt] x-1 \enclose{longdiv}{\,x^3-17x^2+80x-64} \\[-3pt] \underline{x^3-x^2\phantom{00000000000}\,\,} \\[-3pt] -16x^2+80x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-16x^2+16x\phantom{00}\,}\\[-3pt]\phantom{00}64x-64\\[-3pt]\phantom{00}\underline{64x-64}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}f(x)&=x^3-17x^2+80x-64\\&=(x-1)(x^2-16x+64)\\&=(x-1)(x-8)^2\end{align}

Therefore, another value of $$x$$ in which $$f(x)=0$$ is $$x=8$$.

Video Walkthrough

## 2018 Paper 2 Question 3(b)

(b) Find $$a$$, $$b$$, $$c$$ and $$d$$, if $$\dfrac{(n+3)!(n+2)!}{(n+1)!(n+1)!}=an^3+bn^2+cn+d$$, where $$a$$, $$b$$, $$c$$, and $$d\in\mathbb{N}$$.

$$a=1$$, $$b=7$$, $$c=16$$ and $$d=12$$

Solution

\begin{align}\frac{(n+3)!(n+2)!}{(n+1)!(n+1)!}&=\frac{(n+3)(n+2)(n+1)!(n+2)(n+1)!}{(n+1)!(n+1)!}\\&=(n+3)(n+2)(n+2)\\&=(n+3)(n^2+4n+4)\\&=n^3+4n^2+4n+3n^2+12n+12\\&=n^3+7n^2+16n+12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a=1 && b=7 && c=16 && d=12\end{align}

Video Walkthrough

## 2017 Paper 1 Question 1

(a) Write the function $$f(x)=2x^2-7x-10$$, where $$x\in\mathbb{R}$$, in the form $$a(x+h)^2+k$$, where $$a$$, $$h$$ and $$k\in\mathbb{Q}$$.

$$f(x)=2\left(x-\dfrac{7}{4}\right)^2-\dfrac{129}{8}$$

Solution

\begin{align}f(x)&=2x^2-7x-10\\&=2\left(x^2-\frac{7}{2}x-5\right)\\&=2\left[\left(x-\frac{7}{4}\right)^2-\frac{129}{16}\right]\\&=2\left(x-\frac{7}{4}\right)^2-\frac{129}{8}\end{align}

Video Walkthrough
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(b) Hence, write the minimum point of ݂$$f$$.

$$\left(\dfrac{7}{4},-\dfrac{129}{8}\right)$$

Solution

$$\left(\dfrac{7}{4},-\dfrac{129}{8}\right)$$

Video Walkthrough

## 2017 Paper 1 Question 5(a)

The function $$f$$ is such that $$f(x)=2x^3+5x^2-4x-3$$, where $$x\in\mathbb{R}$$.

(a) Show that $$x=-3$$ is a root of ݂$$f(x)$$ and find the other two roots.

$$x=-3$$, $$x=-\dfrac{1}{2}$$ and $$x=1$$

Solution

\begin{align}f(-3)&=2(-3)^3+5(-3)^2-4(-3)-3\\&=-54+45+12-3\\&=0\end{align}

Therefore, $$x=-3$$ is a root.

$\require{enclose} \begin{array}{rll} 2x^2-x-1\phantom{000000}\, \\[-3pt] x+3 \enclose{longdiv}{\,2x^3+5x^2-4x-3} \\[-3pt] \underline{2x^3+6x^2\phantom{00000000}\,\,} \\[-3pt] -x^2-4x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-x^2-3x\phantom{00}\,}\\[-3pt]\phantom{00}-x-3\\[-3pt]\phantom{00}\underline{-x-3}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}f(x)&=(x+3)(2x^2-x-1)\\&=(x+3)(2x+1)(x-1)\end{align}

\begin{align}\downarrow\end{align}

$$x=-3$$, $$x=-\dfrac{1}{2}$$ and $$x=1$$

Video Walkthrough

## 2017 Paper 2 Question 2(d)

An experiment measures the fuel consumption at various speeds for a particular model of car.
The data collected are shown in Table $$1$$ below.

Speed (km/hour) $$40$$ $$48$$ $$56$$ $$64$$ $$88$$ $$96$$ $$112$$

Fuel consumption (km/litre)

$$21$$

$$16$$

$$18$$

$$16$$

$$13$$

$$11$$

$$9$$

(d) Mary drove from Cork to Dublin at an average speed of $$96\mbox{ km/h}$$.
Jane drove the same journey at an average speed of $$112\mbox{ km/h}$$.
Each travelled $$260\mbox{ km}$$ and paid $$132.9$$ cents per litre for the fuel.
Both used the model of car used to generate the data in Table $$1$$.

(i) Find how much longer it took Mary to complete the journey.

(ii) Based on the data in Table $$1$$ and their average speeds, find how much more Jane spent on fuel during the course of this journey.

(i) $$23\mbox{ min}$$

(ii) $$6.98\mbox{ euro}$$

Solution

(i)

\begin{align}t_M-t_J&=\frac{d_M}{s_M}-\frac{d_J}{s_J}\\&=\frac{260}{96}-\frac{260}{112}\\&=0.3689..\mbox{ hr}\\&\approx23\mbox{ min}\end{align}

(ii)

\begin{align}\left(\frac{260}{9}\times1.329\right)-\left(\frac{260}{11}\times1.329\right)\approx6.98\mbox{ euro}\end{align}

Video Walkthrough

## 2016 Paper 1 Question 2

(a) Find the range of values of $$x$$ for which $$|x-4|\geq2$$, where $$x\in\mathbb{R}$$.

$$x\leq2$$ and $$x\geq6$$

Solution

\begin{align}|x-4|\geq 2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|x-4|^2\geq 4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x-4)^2\geq 4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x-4)^2\geq 4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2-8x+16\geq4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2-8x+12\geq0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x-2)(x-6)\geq0\end{align}

\begin{align}\downarrow\end{align}

$$x\leq2$$ and $$x\geq6$$

Video Walkthrough
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(b) Solve the simultaneous equations:

\begin{align}x^2+xy+2y^2&=4\\2x+3y&=-1\end{align}

$$x=\dfrac{17}{11}$$ and $$y=-\dfrac{15}{11}$$ or $$x=-2$$ and $$y=1$$

Solution

\begin{align}x^2+xy+2y^2&=4\\2x+3y&=-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2+xy+2y^2=4\end{align}

\begin{align}x=\frac{-1-3y}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{-1-3y}{2}\right)^2+\left(\frac{-1-3y}{2}\right)y+2y^2=4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1+6y+9y^2}{4}+\frac{-y-3y^2}{2}+2y^2=4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1+6y+9y^2-2y-6y^2+8y^2=16\end{align}

\begin{align}\downarrow\end{align}

\begin{align}11y^2+4y-15=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(11y+15)(y-1)=0\end{align}

\begin{align}\downarrow\end{align}

$$y=-\dfrac{15}{11}$$ or $$y=1$$

and

\begin{align}x=\frac{-1-3y}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{-1-3\left(-\dfrac{15}{11}\right)}{2}\end{align}

or

\begin{align}x=\frac{-1-3(1)}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{-1+\frac{45}{11}}{2}\end{align}

or

\begin{align}x=\frac{-4}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{\frac{34}{11}}{2}\end{align}

or

\begin{align}x=-2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{17}{11}\end{align}

or

\begin{align}x=-2\end{align}

Therefore, the solutions are $$x=\dfrac{17}{11}$$ and $$y=-\dfrac{15}{11}$$ or $$x=-2$$ and $$y=1$$.

Video Walkthrough

## 2016 Paper 1 Question 3(b)

$$f(x)=\dfrac{2}{e^x}$$ and $$g(x)=e^x-1$$

(b) ݂Solve $$f(x)=g(x)$$ using algebra.

$$x=\ln2$$

Solution

\begin{align}\frac{2}{e^x}=e^x-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e^x(e^x-1)=2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(e^x)^2-e^x-2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y^2-y-2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(y+1)(y-2)=0\end{align}

$$y=-1$$ or $$y=2$$

\begin{align}\downarrow\end{align}

$$e^x=-1$$ or $$e^x=2$$

\begin{align}\downarrow\end{align}

$$x=\ln(-1)$$ or $$x=\ln2$$

\begin{align}\downarrow\end{align}

$$x=\ln2$$

Video Walkthrough

## 2016 Paper 1 Question 4(b)

(b) Given $$\log_a2=p$$ and $$\log_a3=q$$, where ܽ$$a>0$$, write each of the following in terms of $$p$$ and $$q$$:

(i) $$\log_a\dfrac{8}{3}$$

(ii) $$\log_a\dfrac{9a^2}{16}$$

(i) $$3p-q$$

(ii) $$2(q+1)-4p$$

Solution

(i)

\begin{align}\log_a\frac{8}{3}&=\log_a8-\log_a3\\&=\log_a2^3-\log_a3\\&=3\log_a2-\log_a3\\&=3p-q\end{align}

(ii)

\begin{align}\log_a\frac{9a^2}{16}&=\log_a9a^2-\log_a16\\&=\log_a(3a)^2-\log_a2^4\\&=2\log_a3a-4\log_a2\\&=2(\log_a3+\log_aa)-4\log_a2\\&=2(q+1)-4p\end{align}

Video Walkthrough

## 2015 Paper 1 Question 2

Solve the equation $$x^3-3x^2-9x+11=0$$.
Write any irrational solution in the form $$a+b\sqrt{c}$$, where $$a,b,c\in\mathbb{Z}$$.

$$x=1$$, $$x=1-2\sqrt{3}$$ and $$1+2\sqrt{3}$$.

Solution

\begin{align}f(1)&=1^3-3(1^2)-9(1)+11\\&=1-3-9+11\\&=0\end{align}

\begin{align}\downarrow\end{align}

$$(x-1)$$ is a factor

\begin{align}\downarrow\end{align}

$\require{enclose} \begin{array}{rll} x^2-2x-11\phantom{000000}\, \\[-3pt] x-1 \enclose{longdiv}{\,x^3-3x^2-9x+11} \\[-3pt] \underline{x^3-x^2\phantom{0000000000}\,\,} \\[-3pt] -2x^2-9x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-2x^2+2x\phantom{00}\,}\\[-3pt]\phantom{00}-11x+11\\[-3pt]\phantom{00}\underline{-11x+11}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}(x^2-2x-11)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-11)}}{2(1)}\\&=\frac{2\pm\sqrt{48}}{2}\\&=\frac{2\pm4\sqrt{3}}{2}\\&=1\pm2\sqrt{3}\end{align}

Therefore, the solutions are $$x=1$$, $$x=1-2\sqrt{3}$$ and $$1+2\sqrt{3}$$.

Video Walkthrough

## 2015 Paper 1 Question 5(a)

(a) Solve the equation $$x=\sqrt{x+6}$$, $$x\in\mathbb{R}$$.

$$x=3$$

Solution

\begin{align}x=\sqrt{x+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2=x+6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2-x-6=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+2)(x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-2$$ or $$x=3$$

$\,$

$$\mathbf{x=-2}$$

\begin{align}-2=\sqrt{-2+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2=\sqrt{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2=2\end{align}

\begin{align}\downarrow\end{align}

Ignored

$\,$

$$\mathbf{x=3}$$

\begin{align}3=\sqrt{3+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=\sqrt{9}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=3\end{align}

Therefore, the only solution is $$x=3$$.

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