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Either paper 2 marks
A sequence of numbers is formed by the iterative process
\(u_{n+1} = 8 - \frac{ 3 }{ u_n }\) with \(u_1 = 6\)
Work out the values of \(u_2\) and \(u_3\).
(a)The value of \(u_2\)
(b)The value of \(u_3\)
Show answer
(a) \(7.5\)
(b) \(7.6\)Replace \(u_n\) by \(u_1 = 6\):
\(u_2 = 8 - \frac{ 3 }{ 6 }\)
\(u_2 = 7.5\)
Now feed \(u_2 = 7.5\) back in as the new \(u_n\):
\(u_3 = 8 - \frac{ 3 }{ 7.5 }\)
\(u_3 = 7.6\)
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Calculator 3 marks
Dev borrows £1800 from a bank.
Each month the amount he owes is multiplied by 1.01 (interest) and then Dev pays back £230.
\(A_n\) is the amount Dev owes, in pounds, at the end of month \(n\).
\(A_{n+1} = 1.01A_n - 230\) with \(A_0 = 1800\)
Calculate the amount Dev still owes at the end of the 2nd month.
Show answer
\(\text{£}1373.88\)Put \(A_0 = 1800\) into the formula to get the amount owed after month 1.
\(A_1 = 1.01 \times 1800 - 230 = 1818.00 - 230 = 1588.00\)
Feed that amount back in to get month 2.
\(A_2 = 1.01 \times 1588.00 - 230 = 1603.88 - 230 = 1373.88\)
Dev still owes £1373.88.
- M1month 1 balance \(1588.00\)
- M1multiplies by \(1.01\) and subtracts \(230\) again
- A1balance \(1373.88\)
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Either paper 2 marks
\(f(x) = x^3 + 4x - 37\)
Show that the equation \(f(x) = 0\) has a solution in the interval \(2 < x < 3\).
Show answer
\(f(2) = -21\) and \(f(3) = 2\). One is negative and the other is positive, so \(f(x) = 0\) has a solution between \(x = 2\) and \(x = 3\).Work out \(f(x)\) at each end of the interval.
\(f(2) = 2^3 + 4 \times 2 - 37 = -21\)
\(f(3) = 3^3 + 4 \times 3 - 37 = 2\)
One answer is negative and the other is positive, so the graph of \(f(x)\) crosses the \(x\)-axis somewhere between \(x = 2\) and \(x = 3\). That crossing point is a solution of \(f(x) = 0\).
Writing down both values and saying the words "change of sign" is what earns the second mark — the two evaluations on their own do not.
- M1both ends evaluated: \(f(2) = -21\) and \(f(3) = 2\)
- A1change of sign stated, so a solution lies between \(x = 2\) and \(x = 3\)
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Either paper 2 marks
Rearrange the equation \(x^3 + 9x - 49 = 0\) into the form \(x = \sqrt[3]{p + qx}\), where \(p\) and \(q\) are integers.
Show answer
\(x = \sqrt[3]{ 49 - 9x }\)Get the \(x^3\) term on its own on one side.
\(x^3 + 9x - 49 = 0\)
Add \(49\) to both sides and subtract \(9x\) from both sides:
\(x^3 = 49 - 9x\)
Now take the cube root of both sides:
\(x = \sqrt[3]{ 49 - 9x }\)
So \(p = 49\) and \(q = -9\).
- M1\(x^3\) isolated: \(x^3 = 49 - 9x\)
- A1cube root taken: \(x = \sqrt[3]{ 49 - 9x }\)
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Either paper 4 marks
At the start of year \(n\) the population of a species is \(p_n\) million.
At the start of the following year the population of the species is given by
\(p_{n+1} = kp_n\)
where \(k\) is a positive constant.
The population of the species at the start of year 1 is 32 million.
The population of the species at the start of year 2 is 30.4 million.
(a)Work out the population of the species at the start of year 3.
(b)At the start of year 5 the value of \(k\) is increased by 0.45 to a new constant value.
Aisha thinks that from the start of year 5 the population of the species would decrease year on year.
Is Aisha correct?
You must show how you get your answer.
Show answer
(a) \(28.88\ \text{million}\)
(b) No: the new value of \(k\) is \(1.4\), which is greater than 1, so the population would increase year on year.Each year the population is multiplied by \(k\), so \(k = \frac{ p_2 }{ p_1 }\):
\(k = \frac{ 30.4 }{ 32 } = 0.95\)
Year 3 is \(k\) times year 2:
\(p_3 = 0.95 \times 30.4 = 28.88\) million
The new value of \(k\) is the old value plus 0.45:
\(k = 0.95 + 0.45 = 1.4\)
Compare the new multiplier with 1:
\(1.4 > 1\)
Each year the population is multiplied by a number greater than 1, so it would increase year on year.
Aisha says it would decrease, so the answer is No.
- (a) M1\(k = 0.95\) found from \(30.4 \div 32\)
- (a) A1\(p_3 = 28.88\) million
- (b) M1new value of \(k\) = \(0.95 + 0.45 = 1.4\)
- (b) A1No, comparing the new \(k\) of \(1.4\) with 1
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Either paper 4 marks
A hot air balloon is descending.
The height of the balloon \(n\) minutes after it starts to descend is \(h_n\) metres.
The height of the balloon \((n + 1)\) minutes after it starts to descend, \(h_{n+1}\) metres, is given by
\(h_{n+1} = k h_n - 20\)
where \(k\) is a constant.
The balloon starts to descend from a height of 1600 metres at 09:31.
At 09:32 the height of the balloon is 1260 metres.
Work out the height of the balloon at 09:34.
Show answer
\(770.4\ \text{metres}\)The height after 1 minute comes from the starting height, so \(h_0 = 1600\) and \(h_1 = 1260\):
\(1260 = k \times 1600 - 20\)
Take the constant from both sides, then divide by 1600:
\(1600k = 1260 - (-20) = 1280\)
\(k = \frac{ 1280 }{ 1600 } = 0.8\)
The time asked for is 3 minutes after the start, so find \(h_3\) by applying the formula to \(h_1\) twice more.
\(h_2 = 0.8 \times 1260 - 20 = 1008 - 20 = 988\)
\(h_3 = 0.8 \times 988 - 20 = 790.4 - 20 = 770.4\)
The height at 09:34 is 770.4 metres.
- M1subtracting the constant, \(1600k = 1280\)
- A1\(k = 0.8\)
- M1\(h_2 = 988\) using \(h_1 = 1260\)
- A1\(h_3 = 770.4\) metres
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Calculator 2 marks
\(f(x) = x^3 - x - 40\)
Show that the equation \(f(x) = 0\) has a solution between \(x = 3.5\) and \(x = 3.6\).
Show answer
\(f(3.5) = -0.625\) and \(f(3.6) = 3.056\). The sign changes between them, so \(f(x) = 0\) has a solution between \(x = 3.5\) and \(x = 3.6\).Substitute each end of the interval into \(f(x)\).
\(f(3.5) = 3.5^3 - 1 \times 3.5 - 40 = -0.625\)
\(f(3.6) = 3.6^3 - 1 \times 3.6 - 40 = 3.056\)
The first is negative and the second is positive. A cubic has no breaks in it, so to get from a negative value to a positive one it must pass through zero somewhere between \(x = 3.5\) and \(x = 3.6\).
- M1both ends evaluated: \(f(3.5) = -0.625\) and \(f(3.6) = 3.056\)
- A1change of sign stated, so a solution lies between \(x = 3.5\) and \(x = 3.6\)
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Either paper 2 marks
A student uses the iteration formula
\(x_{n+1} = \sqrt[3]{ 4x_n + 8 }\)
The values it produces get closer and closer to one number.
Write down the equation, in the form \(x^3 + px + q = 0\), that this number is a solution of.
Show answer
\(x^3 - 4x - 8 = 0\)Once the iteration has settled, \(x_{n+1}\) and \(x_n\) are the same number. Call it \(x\), and the formula says
\(x = \sqrt[3]{ 4x + 8 }\)
Cube both sides to get rid of the root:
\(x^3 = 4x + 8\)
Take everything to the left-hand side:
\(x^3 - 4x - 8 = 0\)
So \(p = -4\) and \(q = -8\).
- M1cubing both sides: \(x^3 = 4x + 8\)
- A1all terms on one side: \(x^3 - 4x - 8 = 0\)
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Calculator 3 marks
\(x_{n+1} = \sqrt[3]{ 3x_n + 10 }\)
Starting with \(x_0 = 2\), work out the values of \(x_1\), \(x_2\) and \(x_3\).
Give your answers correct to 4 decimal places.
Show answer
\(x_1 \approx 2.5198\), \(x_2 \approx 2.5992\), \(x_3 \approx 2.6109\)Put \(x_0 = 2\) into the right-hand side to get \(x_1\).
\(x_1 = \sqrt[3]{ 3 \times 2 + 10 } = \sqrt[3]{ 16 } \approx 2.5198\)
Now feed that answer back in to get \(x_2\), and repeat.
\(x_2 = \sqrt[3]{ 3x_1 + 10 } \approx 2.5992\)
\(x_3 = \sqrt[3]{ 3x_2 + 10 } \approx 2.6109\)
Keep the full calculator value at every step and only round when you write the answer down. Rounding to 4 decimal places and then typing that back in will change the later digits.
- M1substitutes \(x_0 = 2\) to get \(x_1 = \sqrt[3]{ 16 }\)
- A1\(x_1 \approx 2.5198\) and uses it to find \(x_2 \approx 2.5992\)
- A1\(x_3 \approx 2.6109\)
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Calculator 2 marks
A student uses the iteration formula
\(x_{n+1} = \sqrt[3]{ 4x_n + b }\)
where \(b\) is an integer.
Starting with \(x_0 = 3\), the first iterate is \(x_1 = 4.0412\), correct to 4 decimal places.
Find the value of \(b\).
Show answer
\(54\)Cube both sides of the iteration formula to undo the cube root:
\(x_1^3 = 4x_0 + b\)
\(b = x_1^3 - 4x_0\)
Use the rounded value \(x_1 \approx 4.0412\):
\(b \approx 4.0412^3 - 12 \approx 53.9980\)
\(b\) is stated to be an integer, so round to the nearest whole number: \(b = 54\).
- M1cubing \(x_1\) and subtracting \(4x_0\): \(4.0412^3 - 12\)
- A1\(b = 54\), the whole number nearest \(53.9980\)