Numerical methods in contextAQA A-Level Maths: Subtopic test

1. 1
   The depth $d$ metres of water in a tank satisfies the equation $d+\ln d=4$, where $d>0$. Let $f(d)=d+\ln d-4$.

   (a)

   Why is a numerical method needed to solve this equation?

   [1 mark]

   • AThe equation has no real solution.
   • B$d$ must be a whole number.
   • C$\ln d$ is not defined for $d>0$.
   • D$d$ appears both on its own and inside $\ln d$, so it cannot be made the subject by algebraic rearrangement.

   (b)

   Which interval must contain a solution?

   [1 mark]

   • A$1<d<2$
   • B$2.9<d<3$
   • C$2<d<2.9$
   • D$3<d<4$

   (c)

   Use the iteration $d_{n+1}=4-\ln d_n$ with $d_0=3$ to find $d_1$ and $d_2$, giving your answers to 3 decimal places.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The angle $\theta$ (in radians) in a model of an orbit satisfies $\theta-0.5\sin\theta=1$. Let $f(\theta)=\theta-0.5\sin\theta-1$. The Newton-Raphson method is used with $\theta_0=1.5$.

   (a)

   Which expression is $f'(\theta)$?

   [1 mark]

   • A$1-0.5\cos\theta$
   • B$1+0.5\cos\theta$
   • C$1-0.5\sin\theta$
   • D$-0.5\cos\theta$

   (b)

   Find $\theta_1$, to 4 decimal places.

   [1 mark]

   • A$1.5013$
   • B$0.5265$
   • C$1.4987$
   • D$1.5354$

   (c)

   Find $\theta_2$ to 6 decimal places and hence state the value of $\theta$ to 3 decimal places, with a reason.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The integral $I=\int_0^1e^{-x^2}\,dx$ cannot be found by integrating $e^{-x^2}$ using the standard methods in the specification. It is estimated using the trapezium rule with 4 strips of equal width.

   (a)

   Find the trapezium rule estimate of $I$, giving your answer to 3 significant figures.

   [3 marks]

   (b)

   A calculator gives $I=0.7468$ (4 d.p.). Find the percentage error in your estimate, to 2 significant figures. Explain why a numerical method is needed here, and how the estimate could be made more accurate.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The width $x$ metres of a support in a design satisfies $x^3-4x-5=0$, which has a root $\alpha$ between 2 and 3.

   (a)

   (i) Show that $\alpha$ lies between 2 and 3. (ii) Show that $x^3-4x-5=0$ can be rearranged to give $x_{n+1}=\sqrt[3]{4x_n+5}$. (iii) Use this iteration with $x_0=2.5$ to find $x_1$, $x_2$ and $x_3$ to 4 decimal places, and write down $\alpha$ to 2 decimal places.

   [6 marks]

   (b)

   A student instead rearranges the equation as $x_{n+1}=\frac{x_n^3-5}{4}$ and uses $x_0=2.5$. Find $x_1$ and $x_2$ to 3 decimal places, and explain, by considering the gradient of $g(x)=\frac{x^3-5}{4}$ near $\alpha$, why this iteration fails to find $\alpha$.

   [6 marks]

   Total for question 4: 12 marks

End of questions