P3: Numerical methodsEdexcel International A Level Maths: Topic test

1. 1
   The function $f$ is defined by $f(x)=x^{3}-7x+4$ for $x\in\mathbb{R}$. The equation $f(x)=0$ has a root $\alpha$ in the interval $[0,1]$.

   (a)

   Which of the following intervals must contain a root of $f(x)=0$?

   [1 mark]

   • A$[1,2]$
   • B$[0,1]$
   • C$[3,4]$
   • D$[-1,0]$

   (b)

   It is given that $f(0.6)=0.016$, $f(0.65)=-0.275$ and $f(0.7)=-0.557$, each to 3 decimal places. Which of the following is a correct deduction about $\alpha$?

   [1 mark]

   • A$0.65<\alpha<0.7$, so $\alpha=0.7$ to 1 decimal place
   • B$0.6<\alpha<0.65$, so $\alpha=0.7$ to 1 decimal place
   • Cnothing can be deduced, because $f(0.65)$ is negative
   • D$0.6<\alpha<0.65$, so $\alpha=0.6$ to 1 decimal place

   (c)

   Show that $f(x)=0$ has another root $\beta$ in the interval $[2.2,2.3]$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The function $f$ is defined by $f(x)=\tan x+x-4$ for $0\le x\le2$ with $x\ne\frac{\pi}{2}$, where $x$ is in radians. The equation $f(x)=0$ has a root $\alpha$ in the interval $[1.2,1.3]$.

   (a)

   A student notes that $f(1.5)=11.6$ and $f(1.6)=-36.6$ and concludes that $f(x)=0$ has a root in $[1.5,1.6]$. Which statement explains the error?

   [1 mark]

   • A$f$ is not continuous at $x=\frac{\pi}{2}\approx1.571$, which lies in the interval, so the change of sign does not guarantee a root
   • B$f(1.5)$ is too far from zero for a root to be close to it
   • Ca root is only guaranteed when $f$ is positive at both ends of the interval
   • Dthe interval $[1.5,1.6]$ is too short to contain a root

   (b)

   It is given that $f(1.2)=-0.228$, $f(1.25)=0.260$ and $f(1.3)=0.902$, each to 3 decimal places. What is the narrowest interval from the following that must contain $\alpha$?

   [1 mark]

   • A$1.25<\alpha<1.3$
   • B$1.2<\alpha<1.3$
   • C$1.2<\alpha<1.25$
   • D$\alpha=1.25$

   (c)

   Show that the equation $f(x)=0$ can be rearranged into the form $x=\arctan(4-x)$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The equation $\tan x+x-4=0$ has a root $\alpha$ in the interval $[1.2,1.3]$, where $x$ is in radians. The iteration $x_{n+1}=\arctan(4-x_n)$ with $x_0=1.2$ is used to approximate $\alpha$.

   (a)

   Find $x_1$, $x_2$ and $x_3$, giving each answer to 4 decimal places.

   [3 marks]

   (b)

   By considering the sign of $f(x)=\tan x+x-4$ at $x=1.215$ and $x=1.225$, show that $\alpha=1.22$ to 2 decimal places.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The equation $xe^{x}=2$ has a single positive root $\alpha$. Let $f(x)=xe^{x}-2$.

   (a)

   (i) Show that $\alpha$ lies in the interval $[0.8,0.9]$.
   (ii) Show that the equation can be rearranged into $x=2e^{-x}$.
   (iii) Use the iteration $x_{n+1}=2e^{-x_n}$ with $x_0=1$ to find $x_1$, $x_2$ and $x_3$, giving each answer to 4 decimal places.

   [6 marks]

   (b)

   (i) Show that the equation can be rearranged into $x=\ln\dfrac{2}{x}$.
   (ii) Use the iteration $x_{n+1}=\ln\dfrac{2}{x_n}$ with $x_0=0.8$ to find $x_1$, $x_2$ and $x_3$, giving each answer to 4 decimal places.
   (iii) Explain why this iteration is not suitable for finding $\alpha$.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   The sequence $x_0=3$, $x_{n+1}=\sqrt{3x_n+2}$ converges to a limit $\alpha$.

   (a)

   Find $x_2$ to 3 significant figures.

   [1 mark]

   • A$3.32$
   • B$3.56$
   • C$3.46$
   • D$11.9$

   (b)

   Which equation is satisfied by $\alpha$?

   [1 mark]

   • A$x^{2}-3x-2=0$
   • B$x^{2}-3x+2=0$
   • C$x^{2}+3x-2=0$
   • D$x^{2}-2x-3=0$

   (c)

   Find the exact value of $\alpha$.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   The function $g$ is defined by $g(x)=\ln(x+2)-x^{2}$ for $x>-2$.

   (a)

   Which of the following intervals must contain a root of $g(x)=0$?

   [1 mark]

   • A$[0,1]$
   • B$[1.2,2]$
   • C$[-0.5,0]$
   • D$[1,1.2]$

   (b)

   The equation $g(x)=0$ has a root $\beta$ in $[1,1.2]$. Which pair of values of $g$ is enough to show that $\beta=1.1$ to 1 decimal place?

   [1 mark]

   • A$g(1.1)$ and $g(1.2)$
   • B$g(1.05)$ and $g(1.15)$
   • C$g(1.0)$ and $g(1.1)$
   • D$g(1.1)$ alone

   (c)

   Show that $g(x)=0$ has a root in the interval $[-1,-0.5]$.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   The equation $x^{3}+2x^{2}-6=0$ has a single positive root $\alpha$, where $1.3<\alpha<1.4$.

   (a)

   Show that the equation can be rearranged into $x=\sqrt{\dfrac{6}{x+2}}$.

   [3 marks]

   (b)

   Use the iteration $x_{n+1}=\sqrt{\dfrac{6}{x_n+2}}$ with $x_0=1.5$ to find $x_1$, $x_2$ and $x_3$, giving each answer to 4 decimal places.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   The equation $\cos x+x-2=0$, where $x$ is in radians, has a root $\alpha$ in the interval $[2.9,3]$. Let $f(x)=\cos x+x-2$.

   (a)

   (i) Show that $f(x)=0$ has a root in the interval $[2.9,3]$.
   (ii) Use the iteration $x_{n+1}=2-\cos x_n$ with $x_0=3$ to find $x_1$, $x_2$ and $x_3$, giving each answer to 4 decimal places.

   [6 marks]

   (b)

   (i) By considering the sign of $f(x)$ at $x=2.985$ and $x=2.995$, show that $\alpha=2.99$ to 2 decimal places.
   (ii) A student rearranges the equation as $x_{n+1}=\arccos(2-x_n)$ and uses $x_0=3$. Show that this iteration fails at $x_2$.

   [6 marks]

   Total for question 8: 12 marks

End of questions