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A-Level Maths – Pure Year 2

Functions, sequences, radians, parametric equations, integration techniques, differential equations, numerical methods.

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12 questions  ·  54 marks  ·  answers included  ·  A-Level Mathematics (Edexcel/AQA/OCR), Year 2
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This worksheet covers:

  • Practice questions covering functions
  • Sequences and series
  • Radians
  • Parametric equations
  • Integration techniques
  • Differential equations
  • And numerical methods
  • Matching the second-year A-Level Pure Maths specification

A calculator is allowed throughout.

Questions

Section A – Skills

1

The function $f(x) = \dfrac{x-1}{x+2}$ is defined for $x eq -2$.(a) Find $f^{-1}(x)$.(b) State the value of $x$ that must be excluded from the domain of $f^{-1}$.

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(a) $f^{-1}(x) = \dfrac{1+2x}{1-x}$.(b) $x = 1$ must be excluded.
[4]
2

A geometric sequence has first term $a = 5$ and common ratio $r = 1.5$.(a) Find the sum of the first $10$ terms, $S_{10}$, giving your answer to 1 decimal place.(b) Find the number of the first term of the sequence that exceeds $200$.

Show answer
(a) $S_{10} = \dfrac{5(1.5^{10}-1)}{1.5-1} = 566.7$.(b) The $11$th term is the first to exceed $200$ (it equals $288.3$; the $10$th term is $192.2$, which does not).
[5]
3

A sector of a circle has radius $8\,\text{cm}$ and angle $1.2$ radians. Find:(a) the arc length of the sector,(b) the area of the sector.

Show answer
(a) Arc length $= r\theta = 8 \times 1.2 = 9.6\,\text{cm}$.(b) Area $= \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(8)^2(1.2) = 38.4\,\text{cm}^2$.
[3]
4

Using the small angle approximations $\sin\theta \approx \theta$ and $\cos\theta \approx 1 – \dfrac{\theta^2}{2}$, show that for small $x$, $\cos(2x) \approx 1 – 2x^2$.

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Using $\cos(2x) \approx 1 – \dfrac{(2x)^2}{2} = 1 – \dfrac{4x^2}{2} = 1 – 2x^2$, as required.
[3]

Section B – Exam-style

5

A curve has parametric equations $x = t^2$, $y = 2t$.(a) Find $\dfrac{dy}{dx}$ in terms of $t$.(b) Find the equation of the tangent to the curve at the point where $t = 3$.

Show answer
(a) $\dfrac{dx}{dt} = 2t$, $\dfrac{dy}{dt} = 2$, so $\dfrac{dy}{dx} = \dfrac{1}{t}$.(b) At $t=3$: point $(9, 6)$, gradient $\dfrac{1}{3}$. Tangent: $y = \dfrac{1}{3}x + 3$.
[5]
6

Use integration by parts to find $\displaystyle\int x e^{2x}\, dx$.

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$\displaystyle\int x e^{2x}\, dx = \left(\dfrac{x}{2} – \dfrac{1}{4}\right)e^{2x} + C$.
[4]
7

Use the identity $\cos(2x) = 1 – 2\sin^2 x$ to find $\displaystyle\int \sin^2 x\, dx$.

Show answer
$\sin^2 x = \dfrac{1-\cos(2x)}{2}$, so $\displaystyle\int \sin^2 x\, dx = \dfrac{x}{2} – \dfrac{\sin(2x)}{4} + C$.
[4]
8

Express $\dfrac{3x+5}{(x+1)(x+2)}$ in partial fractions, and hence find $\displaystyle\int \dfrac{3x+5}{(x+1)(x+2)}\, dx$.

Show answer
$\dfrac{3x+5}{(x+1)(x+2)} = \dfrac{2}{x+1} + \dfrac{1}{x+2}$. Integral $= 2\ln|x+1| + \ln|x+2| + C$.
[5]

Section C – Challenge

Differential equations and numerical methods
9

Solve the differential equation $\dfrac{dy}{dx} = 2xy$, given that $y = 3$ when $x = 0$. Give your answer in the form $y = f(x)$.

Show answer
Separating variables: $\displaystyle\int \dfrac{1}{y}\, dy = \int 2x\, dx \Rightarrow \ln|y| = x^2 + C \Rightarrow y = Ae^{x^2}$. Using $y(0)=3$ gives $A=3$, so $y = 3e^{x^2}$.
[5]
10

Show that the equation $x^3 – x – 2 = 0$ has a root between $x=1$ and $x=2$. Starting with $x_0 = 1.5$, use the Newton-Raphson method to find this root correct to $3$ decimal places after two iterations.

Show answer
$f(1) = -2 < 0$, $f(2) = 4 > 0$, so there is a sign change and hence a root between $1$ and $2$. With $f'(x) = 3x^2-1$: $x_1 = 1.5 – \dfrac{f(1.5)}{f'(1.5)} = 1.522$, $x_2 = 1.521$. The root is $1.521$ (3 d.p.).
[5]
11

Use the trapezium rule with $4$ strips (5 ordinates, $h = 0.5$) to estimate $\displaystyle\int_0^2 \sqrt{1+x^3}\, dx$, giving your answer to 3 decimal places.

Show answer
Using $x = 0, 0.5, 1, 1.5, 2$ with $y$-values $1.000, 1.061, 1.414, 2.092, 3.000$: $\displaystyle\int_0^2 \sqrt{1+x^3}\,dx \approx \dfrac{0.5}{2}\big[(1.000+3.000) + 2(1.061+1.414+2.092)\big] = 3.283$.
[5]
12

The temperature $\theta$ (in $^\circ\text{C}$) of a cooling liquid in a room of temperature $20^\circ\text{C}$ satisfies Newton’s law of cooling, $\theta = 20 + 80e^{-kt}$, where $t$ is time in minutes. The liquid starts at $100^\circ\text{C}$ and cools to $70^\circ\text{C}$ after $5$ minutes.(a) Find the value of $k$, correct to 3 significant figures.(b) Find the temperature after $15$ minutes, correct to 1 decimal place.

Show answer
(a) $70 = 20+80e^{-5k} \Rightarrow e^{-5k}=0.625 \Rightarrow k = \dfrac{-\ln(0.625)}{5} = 0.0940$.(b) $\theta(15) = 20+80e^{-0.0940\times15} = 39.5^\circ\text{C}$.
[6]

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A-Level Maths – Pure Year 1A-Level Maths – MechanicsA-Level Maths – StatisticsA-Level Further Maths – Core Pure

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