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A-Level Maths – Pure Year 1

Algebra, coordinate geometry, binomial expansion, trigonometry, differentiation, integration, vectors.

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12 questions  ·  55 marks  ·  answers included  ·  A-Level Mathematics (Edexcel/AQA/OCR, Year 1 / AS Pure)
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This worksheet covers:

  • Practice questions covering AS Pure content
  • Algebra
  • Coordinate geometry
  • The binomial expansion
  • Trigonometry
  • Differentiation
  • Integration
  • Vectors

Section A is non-calculator; a calculator may be used for Sections B and C.

Questions

Section A – Algebra and Coordinate Geometry

Non-calculator
1

Solve the quadratic equation $2x^2-7x-4=0$, giving your answers as exact fractions or integers.

Show answer
$(2x+1)(x-4)=0$, so $x=-\frac{1}{2}$ or $x=4$.
[3]
2

The points $A(-2,5)$ and $B(4,-7)$ lie on a line $l$.(a) Find the equation of $l$ in the form $y=mx+c$.(b) Find the equation of the perpendicular bisector of $AB$, giving your answer in the form $y=mx+c$.

Show answer
(a) Gradient $m=\dfrac{-7-5}{4-(-2)}=-2$, so $y=-2x+1$.(b) Midpoint $(1,-1)$; perpendicular gradient $=\dfrac{1}{2}$, so $y=\frac{1}{2}x-\frac{3}{2}$.
[6]
3

Find the first four terms, in ascending powers of $x$, of the binomial expansion of $(1+2x)^5$.

Show answer
$(1+2x)^5=1+10x+40x^2+80x^3+\ldots$
[4]
4

Solve $2\sin x-1=0$ for $0\leq x\leq 360$, giving your answers in degrees.

Show answer
$\sin x=\frac{1}{2}$, so $x=30$ or $x=150$.
[3]

Section B – Differentiation and Integration

Calculator allowed
5

A curve has equation $y=x^3-4x^2+2$.(a) Find $\dfrac{dy}{dx}$.(b) Find the equation of the tangent to the curve at the point where $x=1$.

Show answer
(a) $\dfrac{dy}{dx}=3x^2-8x$.(b) At $x=1$: gradient $=-5$, $y=-1$, so tangent is $y=-5x+4$.
[5]
6

Find the value of $\displaystyle\int_{1}^{3}(3x^2-2x+1)\,dx$.

Show answer
$\left[x^3-x^2+x\right]_1^3=(27-9+3)-(1-1+1)=21-1=20$.
[4]
7

A curve has equation $y=x^3-3x^2-9x+5$.(a) Find the coordinates of the stationary points.(b) Determine the nature of each stationary point.

Show answer
(a) $\dfrac{dy}{dx}=3x^2-6x-9=0 \Rightarrow x=-1,\,x=3$; points $(-1,10)$ and $(3,-22)$.(b) $\dfrac{d^2y}{dx^2}=6x-6$: at $x=-1$, $\dfrac{d^2y}{dx^2}=-12<0$ (maximum); at $x=3$, $\dfrac{d^2y}{dx^2}=12>0$ (minimum).
[6]
8

Solve $2\sin^2x-\sin x-1=0$ for $0\leq x\leq 360$, giving all solutions in degrees.

Show answer
Factorising: $(2\sin x+1)(\sin x-1)=0$, so $\sin x=1$ or $\sin x=-\frac{1}{2}$. This gives $x=90,\,210,\,330$.
[5]

Section C – Vectors and Extended Problems

Calculator allowed
9

The vector $\mathbf{a}=3\mathbf{i}-4\mathbf{j}$.(a) Find $|\mathbf{a}|$.(b) Find the unit vector in the direction of $\mathbf{a}$.

Show answer
(a) $|\mathbf{a}|=\sqrt{3^2+(-4)^2}=5$.(b) $\hat{\mathbf{a}}=\dfrac{3}{5}\mathbf{i}-\dfrac{4}{5}\mathbf{j}$.
[3]
10

Two lines have vector equations $l_1:\ \mathbf{r}=(1,2,3)+t(1,-1,2)$ and $l_2:\ \mathbf{r}=(1,-3,6)+s(2,3,1)$. Show that $l_1$ and $l_2$ intersect, and find the coordinates of the point of intersection.

Show answer
Setting components equal: $1+t=1+2s$, $2-t=-3+3s$, $3+2t=6+s$. The first two equations give $t=2,\,s=1$, which also satisfies the third ($3+4=6+1=7$), so the lines intersect. Substituting $t=2$ into $l_1$ gives the point $(3,0,7)$.
[5]
11

An open box is made from a square sheet of card of side $20\,\text{cm}$ by cutting a square of side $x$ cm from each corner and folding up the sides.(a) Show that the volume of the box is $V=x(20-2x)^2$.(b) Find the value of $x$ that maximises $V$, and find this maximum volume, correct to 1 decimal place.

Show answer
(a) The base has side $(20-2x)$ and height $x$, so $V=x(20-2x)^2$.(b) $\dfrac{dV}{dx}=(20-2x)^2-4x(20-2x)=(20-2x)(20-6x)=0$, giving $x=10$ (rejected, gives $V=0$) or $x=\dfrac{10}{3}$. Since $\dfrac{d^2V}{dx^2}<0$ there, this is a maximum: $x=\dfrac{10}{3}\approx 3.3\,\text{cm}$, with $V_{\max}=\dfrac{16000}{27}\approx 592.6\,\text{cm}^3$.
[7]
12

Show that $x=2$ is a root of $f(x)=x^3-3x^2-4x+12$, and hence factorise $f(x)$ completely.

Show answer
$f(2)=8-12-8+12=0$, so $(x-2)$ is a factor. Dividing: $f(x)=(x-2)(x^2-x-6)=(x-2)(x-3)(x+2)$.
[4]

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