Iteration and Newton-RaphsonAQA A-Level Maths: Subtopic test

1. 1
   The equation $x^3+x-3=0$ has a root $\alpha$ between 1 and 2. A student uses the iteration $x_{n+1}=\sqrt[3]{3-x_n}$ with $x_0=1$.

   (a)

   Find $x_1$, to 2 decimal places.

   [1 mark]

   • A$1.26$
   • B$2.00$
   • C$1.41$
   • D$8.00$

   (b)

   Which rearrangement of $x^3+x-3=0$ gives this iteration?

   [1 mark]

   • A$x^3=x+3$
   • B$x=3-x^3$
   • C$x^3=3-x$
   • D$x=\frac{3}{x^2+1}$

   (c)

   Find $x_2$ and $x_3$, giving each to 4 decimal places.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The Newton-Raphson method is used to approximate $\sqrt7$ by solving $f(x)=0$, where $f(x)=x^2-7$, with starting value $x_0=3$.

   (a)

   Which expression gives the Newton-Raphson iteration for $f(x)=x^2-7$?

   [1 mark]

   • A$x_{n+1}=x_n+\frac{x_n^2-7}{2x_n}$
   • B$x_{n+1}=x_n-(x_n^2-7)$
   • C$x_{n+1}=x_n-\frac{2x_n}{x_n^2-7}$
   • D$x_{n+1}=x_n-\frac{x_n^2-7}{2x_n}$

   (b)

   Find $x_1$.

   [1 mark]

   • A$\frac{10}{3}$
   • B$\frac83$
   • C$\frac13$
   • D$1$

   (c)

   Find $x_2$, giving your answer to 4 decimal places.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The equation $f(x)=0$, where $f(x)=x^3-2x-5$, has a root $\alpha$ close to 2. The Newton-Raphson method is used with $x_0=2$.

   (a)

   Show that the Newton-Raphson iteration is $x_{n+1}=x_n-\frac{x_n^3-2x_n-5}{3x_n^2-2}$, and hence show that $x_1=2.1$.

   [3 marks]

   (b)

   Find $x_2$ and $x_3$ to 6 decimal places, and hence write down $\alpha$ to 4 decimal places, justifying your answer.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The function $f(x)=x^3-3x+1$ has a root $\beta$ between 1 and 2.

   (a)

   (i) Show that $\beta$ lies between 1 and 2. (ii) Explain why the Newton-Raphson method fails with $x_0=1$. (iii) Use Newton-Raphson with $x_0=2$ to find $x_1$.

   [6 marks]

   (b)

   A student rearranges $f(x)=0$ to give $x_{n+1}=\frac{x_n^3+1}{3}$ and takes $x_0=1.6$, hoping to find $\beta=1.532$ (to 3 d.p.). Find $x_1$ and $x_2$ to 3 decimal places, and explain, using the gradient of $g(x)=\frac{x^3+1}{3}$, why the iteration does not converge to $\beta$.

   [6 marks]

   Total for question 4: 12 marks

End of questions