# Maths A Level | 10 Success Questions

## Maths A Level – Introduction

In this particular blog post we are going to look at 10 questions from an old exam paper. The old paper prior to the current structure of exams was known as C1 or Core 1. It essentially covered the essentials for algebra that were needed.

These particular topics cover indices, surds, equations and so on. Even though the structure of A Level maths exams has changed since 2017, the knowledge required is pretty much the same. You need to be good at your algebra if you want to be successful with maths at A Level.

We will look at questions and consider the solution that is needed to derive the answer being as detailed as possible.

It should also be noted that the old Core 1 maths paper had a Section A and a Section B part. Section A consisted of shorter questions whereas Section B contained longer, more challenging questions. You still see this format today and in this article we will consider section A only.

## Maths A Level - Questions 1 - 3

Here we have a standard inequality question that you would have seen at GCSE Level. The main method is to treat the inequality as though it was an equation but not to change the inequality sign.

Essentially the techniques required are to subtract 1 from both sides and then to subtract 3x from both sides as well. Finally you can divide by “-5″ as well. The solution is shown here:

\begin{aligned} 1-2 x & <4+3 x \\ -2 x & <4+3 x-1 \\ -2 x-3 x & <4-1 \\ -5 x & <3 \\ x & >\frac{3}{-5} \end{aligned}

Here we have a rearranging formula question where t is to be made the subject. So in order to do this there are a few simple algebraic steps that need to be done.

1st: multiply both sides by 2

2nd: Divide both sides by a

3rd: Square root both sides by a

The complete working is shown here below:

\begin{aligned} & s=\frac{1}{2} a t^2 \\ & 2 s=a t^2 \\ & \frac{2 s}{a}=t^2 \\ & t=\sqrt{\frac{2 s}{a}} \end{aligned}

This is a proof question. The converse just means that opposite so in this case it would be:

“If 2n is an even integer then n is an odd integer”

We need to show if this statement is true or if it is false.

Well if 2n is even then 2n = 2 which means n = 1. This shows that the statement is true when 2n = 2.

However if 2n = 4 then n = 2 which means n is even. This goes against the converse statement and this is therefore false.

## Maths A Level - Questions 4 - 6

This question is making use of the remainder and factor theorem. It is important that with this particular topic you are substituting in the correct numbers as shown and using the correct output values as shown here:

\begin{aligned} & \mathrm{f}(x)=x^3+k x+c \\ & \mathrm{f}(0)=0^3+k \times 0+c \\ & \therefore \underline{6}=c \end{aligned}

\begin{aligned} &\begin{aligned} \mathrm{f}(2) & =2^3+k \times 2+c \\ 0 & =8+2 k+c \end{aligned}\\ &\begin{aligned} 0 & =8+2 k+6 \\ -2 k & =14 \\ k & =-7 \end{aligned} \end{aligned}

You are also making use of the factor theorem. So if x – 2 is a factor of f(x) then f(2) = 0 as shown above to obtain the values of the constants as shown.

Here we have questions relating to indices. For part a) you are simply cube rooting. For part b) the easiest approach is to flip what is inside the brackets and to then change the negative power to a power positive. Solutions for both are shown here:

Part (i)

\begin{aligned} a^3 & =64 x^{12} y^3 \\ & =4^3 x^{12} y^3 \\ & =\left(4 x^4 y\right)^3 \\ \therefore a & =4 x^4 y \end{aligned}

Part (ii)

\begin{aligned} \left(\frac{1}{2}\right)^{-5} & =\left(2^{-1}\right)^{-5} \\ & =2^{(-1) \times(-5)} \\ & =2^5 \\ & =32 \end{aligned}

This is a binomial expansion question and the general term needs to be used. Remember that “coefficient” just means “number in front of”. The result is shown here:

\begin{aligned} { }^5 C_3(3)^{5-3}(-2 x)^3 & =\frac{\lfloor 5}{\lfloor 3 \times\lfloor 5-3}(3)^2(-2 x)^3 \\ & =\frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} \times 9 \times\left(-8 x^3\right) \\ & =-720 x^3 \end{aligned}

So the required coefficient is “-720”.

Of course you could have used Pascal’s Triangle to have expanded the brackets but this would have taken longer OR you could have used the N-C-R approach. The N-C-R approach would be better than Pascal but using the general term is by far the easiest.

## Maths A Level – Questions 7 – 10

A standard equation question that should not cause any real concern. But you want to multiply by the denominator so as to remove the fraction. Then collect like terms and to then solve for x. The answer for which is shown below:

\begin{aligned} & \frac{4 x+5}{2 x}=-3 \\ & 2 x \times \frac{4 x+5}{2 x}=-3 \times 2 x \\ & 4 x+5=-6 x \\ & 10 x+5=0 \\ & 10 x=-5 \\ & x=-\frac{5}{10} \\ & x=-1 / 2 \end{aligned}

Here are two questions regarding surds. For part (i) you are required to simplify the surds and to then subtract. Part (ii) you are required to rationalise the denominator. The solutions for both are shown here:

Part (i)

\begin{aligned} & \sqrt{98}-\sqrt{50} \\ = & \sqrt{49 \times 2}-\sqrt{25 \times 2} \\ = & \sqrt{49} \times \sqrt{2}-\sqrt{25} \times \sqrt{2} \\ = & 7 \sqrt{2}-5 \sqrt{2} \\ = & 2 \sqrt{2} \end{aligned}

Part (ii)

\begin{aligned} & =\frac{6 \sqrt{5}}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}} \\ & =\frac{2 \times 6 \sqrt{5}-6 \sqrt{5} \times \sqrt{5}}{2 \times 2-2 \times \sqrt{5}+2 \times \sqrt{5}-\sqrt{5} \times \sqrt{5}} \\ & =\frac{12 \sqrt{5}-30}{4-5} \\ & =\frac{12 \sqrt{5}}{-1}-\frac{30}{-1} \\ & =30-12 \sqrt{5} \end{aligned}

For part (ii) you need to multiply top and bottom by 2-\sqrt{5}

For part (i) you are simply substituting y for 21 and then substituting for x. You will see in the solution that there are two possible answers and the biggest mistake that is made is not to include

For part (ii) this is a translation question and you need to read the question carefully as you are not required to simplify your answer.

The solutions for both are shown below:

\begin{aligned} & \text { (i) } y=x^2-4 \text { and } y=21 \\ & \therefore x^2-4=21 \\ & x^2=21+4 \\ & x^2=25 \\ & x= \pm 5 \end{aligned}

Part (ii)

\begin{aligned} & y=x^2-4 \\ & y=(x-2)^2-4 \end{aligned}

For part (ii) it is a translation in the x direction so it required to set the following:

\begin{aligned} & x=2 \\ & x-2=0 \end{aligned}

So x is replaced with x – 2.

This is a question that is using algebraic manipulation and in order to “show” the desired result you need to look for the clues in the question. The question states the word “area” and so the formula for the area of a triangle needs to be used as shown using the dimensions given and the value for the area:

\begin{aligned} & 1 / 2 \times \text { base } \times \text { height } \\ & 1 / 2 \times(2 x-3) \times(x+1) \\ & 1 / 2(2 x-3)(x+1)=9 \\ & 1 / 2\left(2 x^2-3 x+2 x-3\right)=9 \\ & 2 x^2-3 x+2 x-3=18 \\ & 2 x^2-x-3=18 \\ & 2 x^2-x-21=0 \end{aligned}

To do the second part of the question it is a case of solving the obtained quadratic to obtain the correct value of x

\begin{aligned} & 2 x^2-x-21=0 \\ & (2 x-7)(x+3)=0 \end{aligned}2 x-7=0 \quad \text { or } \quad x+3=0\therefore \text { Height } \quad x+1=\frac{7}{2}+1=\frac{9}{2} \mathrm{~cm}\text { Base } \quad 2 x-3=\not 2 \times \frac{7}{\not 2}-3=4 \mathrm{~cm}

Because there are two values on x we can only use the positive value as it is not possible to have negative length.

Using the positive value the height and base of the triangle can then be found as seen above.

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