Numerical methods (Newton–Raphson, iteration) – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

$\mathrm{f}(x)=x^{3}-4x+1$

(a) Show that the equation $\mathrm{f}(x)=0$ has a root $\alpha$ in the interval $[1.8,\,1.9]$. [2]

(b) The function $\mathrm{g}$ is defined by $\mathrm{g}(x)=\dfrac{1}{x-2}$, $x\neq2$. Explain why $\mathrm{g}(1)$ and $\mathrm{g}(3)$ have different signs but the equation $\mathrm{g}(x)=0$ has no root in $[1,3]$. [1]

(a) Evaluates $\mathrm{f}(1.8)$ and $\mathrm{f}(1.9)$ with at least one correct to 1 s.f. (rounded or truncated): $\mathrm{f}(1.8)=-0.368$, $\mathrm{f}(1.9)=0.259$M1
Both values correct to at least 1 s.f., rounded or truncated (e.g. $-0.4$ and $0.3$, or $-0.3$ and $0.2$, or better), states change of sign and that f is continuous, and concludes there is a root in $[1.8,1.9]$A1
(b) g is not continuous on $[1,3]$: it has a vertical asymptote (discontinuity) at $x=2$, so the change of sign does not indicate a root (accept 'the sign change is caused by the asymptote/discontinuity, not a root'; an answer that only says $\mathrm{g}(x)$ is never 0 without referring to the discontinuity scores B0)B1
Answer: (a) $\mathrm{f}(1.8)=-0.368<0$, $\mathrm{f}(1.9)=0.259>0$, sign change and f continuous so root (b) discontinuity at $x=2$
Examiner tip: The A mark needs both values, 'change of sign', 'f is continuous' and a conclusion – miss any one and it is lost.

2StandardCalculator6 marks

$\mathrm{f}(x)=x^{3}-4x+1$. The equation $\mathrm{f}(x)=0$ has a root $\alpha$ between 1.8 and 1.9.

(a) Show that $\mathrm{f}(x)=0$ can be rearranged into the form $x=\sqrt[3]{4x-1}$. [1]

(b) Using the iterative formula $x_{n+1}=\sqrt[3]{4x_{n}-1}$ with $x_{0}=1.8$, find the values of $x_{1}$, $x_{2}$ and $x_{3}$, giving your answers to 4 decimal places. [3]

(a) $x^{3}=4x-1\Rightarrow x=\sqrt[3]{4x-1}$ *B1
(b) Attempts to use the formula: $x_{1}=\sqrt[3]{4(1.8)-1}$M1
$x_{1}=1.8371$A1
$x_{2}=1.8516,\ x_{3}=1.8573$A1
(c) Chooses the interval $[1.8605,1.8615]$ (or tighter containing the root) and evaluates f at both endsM1
$\mathrm{f}(1.8605)=-0.00195\ldots<0$, $\mathrm{f}(1.8615)=0.00444\ldots>0$; sign change, f continuous, so $\alpha=1.861$ to 3 d.p.A1
Answer: (b) $x_1=1.8371,\ x_2=1.8516,\ x_3=1.8573$ (c) see mark scheme
Examiner tip: Use the bounds $1.8605$ and $1.8615$ – checking $1.860$ and $1.862$ does not prove the 3 d.p. value.

3StandardCalculator5 marks

$\mathrm{f}(x)=\mathrm{e}^{x}-3x$. The equation $\mathrm{f}(x)=0$ has a root $\beta$ close to 1.5.

(a) Find $\mathrm{f}'(x)$. [1]

(b) Taking $x_{0}=1.5$ as a first approximation, apply the Newton–Raphson procedure once to find a second approximation for $\beta$, giving your answer to 3 decimal places. [3]

(a) $\mathrm{f}'(x)=\mathrm{e}^{x}-3$B1
(b) Uses $x_{1}=1.5-\dfrac{\mathrm{f}(1.5)}{\mathrm{f}'(1.5)}$ with their $\mathrm{f}'$M1
$\mathrm{f}(1.5)=-0.0183\ldots$ and $\mathrm{f}'(1.5)=1.4816\ldots$ (may be implied)A1
$x_{1}=1.512$A1
(c) $\mathrm{f}'(\ln3)=\mathrm{e}^{\ln3}-3=0$, so the tangent is horizontal / division by zero in the formulaB1
Answer: (a) $\mathrm{e}^{x}-3$ (b) $1.512$ (c) $\mathrm{f}'(\ln3)=0$
Examiner tip: Show $\mathrm{f}(1.5)$ and $\mathrm{f}'(1.5)$ as numbers – a bare final answer with an error earns nothing.

4HardCalculator7 marks

(a) Show that the equation $x^{2}-2x-1=0$ can be written in the form $x=2+\dfrac{1}{x}$. [1]

The iterative formula $x_{n+1}=2+\dfrac{1}{x_{n}}$ is used with $x_{1}=1$.

(b) Find the values of $x_{2}$, $x_{3}$ and $x_{4}$, giving your answers to 4 decimal places where appropriate. [2]

The diagram shows the graphs of $y=x$ and $y=2+\dfrac{1}{x}$ for $x>0$, together with part of the diagram for this iteration.

(d) Find the exact value of the limit of the sequence $x_{1},x_{2},x_{3},\ldots$, giving a reason for your choice. [2]

(a) Divides $x^{2}=2x+1$ by $x$ ($x\neq0$) to give $x=2+\dfrac1x$ *B1
(b) Uses the formula at least once: $x_{2}=3$M1
$x_{3}=2.3333$ (or $\tfrac73$), $x_{4}=2.4286$ (or $\tfrac{17}{7}$)A1
(c) Cobweb (diagram)B1
Converges to the root, oscillating either side of it (terms alternately above and below)B1
(d) Solves $x^{2}-2x-1=0$: $x=1\pm\sqrt2$M1
$1+\sqrt2$, rejecting $1-\sqrt2$ because it is negative and all terms are positiveA1
Answer: (b) $3,\ 2.3333,\ 2.4286$ (c) cobweb; converges, oscillating (d) $1+\sqrt2$
Examiner tip: A staircase diagram approaches from one side; a cobweb spirals in – name it from what you see.

5HardCalculator7 marks

The equation $x^{3}+2x-5=0$ has a single real root $\alpha$, where $\alpha\approx1.3$.

A student uses the iterative formula $$x_{n+1}=\frac{5-x_{n}^{3}}{2}$$ with $x_{0}=1.3$.

(a) Find $x_{1}$, $x_{2}$ and $x_{3}$, giving your answers to 4 decimal places (use the unrounded values from your calculator in each step). [2]

(b) By considering the gradient of $y=\dfrac{5-x^{3}}{2}$ near $\alpha$, explain why this iteration fails to converge to $\alpha$. [2]

(c) Use one iteration of the Newton–Raphson method, starting with $x_{0}=1.3$, to find a better approximation to $\alpha$, giving your answer to 4 decimal places. [3]

(a) Uses the formula: $x_{1}=\dfrac{5-1.3^{3}}{2}=1.4015$M1
$x_{1}=1.4015,\ x_{2}=1.1236,\ x_{3}=1.7908$ (awrt; allow $x_3=1.7907$ from using rounded $x_2$)A1
(b) Differentiates: with $\mathrm{g}(x)=\dfrac{5-x^{3}}{2}$, $\mathrm{g}'(x)=-\tfrac32x^{2}$ and evaluates near $\alpha$, e.g. $\mathrm{g}'(1.3)=-2.535$ (allow evaluation at any value from $1.3$ to $1.33$)M1
$|\mathrm{g}'(x)|>1$ near $\alpha$ (gradient steeper than $-1$), so successive values move away from $\alpha$ (diverge / oscillate with increasing distance from $\alpha$)A1
(c) $\mathrm{f}'(x)=3x^{2}+2$ and uses $x_{1}=1.3-\dfrac{\mathrm{f}(1.3)}{\mathrm{f}'(1.3)}$M1
$\mathrm{f}(1.3)=-0.203$, $\mathrm{f}'(1.3)=7.07$A1
$1.3287$A1
Answer: (a) $1.4015,\ 1.1236,\ 1.7908$ (b) gradient $-\tfrac32x^{2}\approx-2.6$ near $\alpha$, $|\,\cdot\,|>1$ so diverges (c) $1.3287$
Examiner tip: Quote the convergence condition precisely: the iteration converges only if $|\mathrm{g}'(x)|<1$ near the root.

6StretchCalculator7 marks

The curve $C$ has equation $y=x^{2}\ln x-3x$, $x>0$.

(a) Show that the $x$-coordinate of the stationary point of $C$ is a solution of $$2x\ln x+x-3=0$$ [2]

(b) Show that this stationary point has $x$-coordinate between 1.5 and 1.6. [2]

(a) Product rule on $x^{2}\ln x$: $2x\ln x+x^{2}\cdot\dfrac1x$M1
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x\ln x+x-3$, set $=0$ for stationary point csoA1*
(b) Evaluates $h(x)=2x\ln x+x-3$ at 1.5 and 1.6: $h(1.5)=-0.284$, $h(1.6)=0.104$M1
Both correct (1 s.f.), sign change, $h$ continuous, so the stationary point lies between 1.5 and 1.6A1
(c) Differentiates $h$: $h'(x)=2\ln x+3$ (product rule seen)M1
Uses $x_{1}=1.6-\dfrac{h(1.6)}{h'(1.6)}$ with their $h'$M1
$x_{1}=1.5736$A1
Answer: (c) $x_1=1.5736$
Examiner tip: In (c) you are applying Newton–Raphson to $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, so you need the second derivative of $y$ – not $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ again.

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