Numerical methods – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method (change of sign): if \(f\) is continuous on \([a,b]\) and \(f(a)\), \(f(b)\) have opposite signs, then \(f(x)=0\) has a root between \(a\) and \(b\). Always state both the change of sign and the continuity. Mark scheme as shown in brackets.

   1. \(f(1)=-4\lt 0\) and \(f(2)=2\gt 0\) (1); \(f\) is continuous so there is a root in \([1,2]\) (1).
   2. \(f(1.7)=-0.787\lt 0\) and \(f(1.8)=0.032\gt 0\) (1); continuous, change of sign, root in \([1.7,1.8]\) (1).

   Answer: the root of \(x^3-x-4=0\) lies between 1.7 and 1.8, and it is closer to 1.8 because \(|f(1.8)|\) is much smaller than \(|f(1.7)|\).

2. Answer to Question 2 (5 marks)

   Mark scheme as shown in brackets.

   1. rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|\lt 1\). Here \(x_1=\sqrt[3]{5.5}\). \(x_1=1.7652\), \(x_2=1.7931\), \(x_3=1.7960\) (2). The values increase towards the root in a staircase pattern.
   2. \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\). Each step follows the tangent at \(x_n\) down to the \(x\)-axis. Use the ANS key on the calculator and stop when two values agree to the required accuracy. Here \(f(1.8)=0.032\) and \(f'(1.8)=8.72\). \(f\prime(x)=3x^2-1\) (1). \(x_1=1.8-\dfrac{0.032}{8.72}=1.79633\) (1). \(x_2=1.79632\) (1). The two values agree to 4 decimal places, so \(\alpha=1.7963\) to 4 decimal places.

3. Answer to Question 3 (4 marks)

   Method (trapezium rule): \(\displaystyle\int_a^by\,dx\approx\dfrac h2\left[y_0+y_n+2(y_1+\dots+y_{n-1})\right]\) with strip width \(h=\dfrac{b-a}n\). The estimate is too large when the curve is convex and too small when it is concave. Mark scheme as shown in brackets.

   \(h=0.25\) and ordinates \(1,1.0308,1.1180,1.2500,1.4142\) (1). \(\dfrac{0.25}2\left[1+1.4142+2(1.0308+1.1180+1.2500)\right]\) (1) \(=0.125\times9.2118=1.1515\) (1). The curve is convex, so the estimate is too large (1).

   Answer: the integral is approximately 1.1515, which is a slight overestimate (the exact value is about 1.1478).

4. Answer to Question 4 (3 marks)

   Method (Newton-Raphson): \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\). Each step follows the tangent at \(x_n\) down to the \(x\)-axis. Use the ANS key on the calculator and stop when two values agree to the required accuracy. Mark scheme as shown in brackets.

   \(f\prime(x)=3x^2-2\) (1). \(x_1=0-\dfrac{2}{-2}=1\), \(x_2=1-\dfrac11=0\), \(x_3=1\) (1). The iteration cycles between 0 and 1 and never converges to the root near \(-1.77\) (1).

   This shows that Newton-Raphson can fail when the starting value is badly chosen; starting at \(x_0=-2\) instead would converge quickly to the root \(-1.769\).

5. Answer to Question 5 (4 marks)

   Method (fixed-point iteration): rearrange \(f(x)=0\) as \(x=g(x)\), then use \(x_{n+1}=g(x_n)\), keeping full calculator accuracy and rounding only when writing the values. The iteration converges near the root \(\alpha\) when \(|g'(\alpha)|\lt 1\). Mark scheme as shown in brackets.

   1. Write the cube root as a power \(\tfrac13\) and use the chain rule. \(g(x)=(x+4)^{1/3}\) so \(g\prime(x)=\dfrac13(x+4)^{-2/3}\) (2).
   2. Substitute \(\alpha=1.796\), so \(x+4=5.796\). \(g\prime(1.796)=\dfrac1{3(5.796)^{2/3}}=0.103\) (1). Since \(0\lt g\prime(\alpha)\lt 1\) the iteration converges, approaching \(\alpha\) from one side in a staircase (1). The error is multiplied by about 0.1 at each step, so the convergence is fast.

Total : 20 marks