Locating roots and iterationEdexcel A-Level Maths: Subtopic test

1. 1
   Let $f(x)=x^3-3x-5$, a continuous function. The equation $f(x)=0$ has a single real root $\alpha$.

   (a)

   Which interval of length 1 contains $\alpha$?

   [1 mark]

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

   (b)

   Which statement correctly explains why a sign change of $f$ on an interval shows that it contains a root?

   [1 mark]

   • AAny function that changes sign must have a root.
   • BA cubic function always has exactly one root.
   • C$f$ is closer to zero at one end of the interval.
   • D$f$ is continuous and changes sign, so it passes through zero.

   (c)

   Show that $\alpha$ lies in the interval $[2.2,\,2.3]$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The root $\alpha$ of $x^3-3x-5=0$ is estimated using the iteration $x_{n+1}=\sqrt[3]{3x_n+5}$ with $x_0=2$.

   (a)

   Find $x_1$, correct to 3 decimal places.

   [1 mark]

   • A$2.224$
   • B$3.317$
   • C$11.000$
   • D$1.000$

   (b)

   Find $x_2$, correct to 3 decimal places.

   [1 mark]

   • A$2.224$
   • B$6.883$
   • C$2.268$
   • D$1.187$

   (c)

   Find $x_3$ and $x_4$, and hence write down the value of $\alpha$ correct to 2 decimal places.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Let $f(x)=\frac{1}{x-2}$ for $x\ne2$, and $g(x)=x^2-5x+6$.

   (a)

   Show that $f(1)$ and $f(3)$ have opposite signs, and explain why this does not show that $f(x)=0$ has a root in $[1,3]$.

   [3 marks]

   (b)

   A student calculates $g(1.5)$ and $g(3.5)$, finds both are positive and concludes there is no root in $[1.5,\,3.5]$. Show that this conclusion is wrong and explain how a root can be located.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The equation $\mathrm{e}^x=4x$ has two roots, $\alpha$ and $\beta$, where $\alpha<\beta$.

   (a)

   (i) Show that $\mathrm{e}^x=4x$ can be rearranged as $x=\ln(4x)$.
   (ii) Use the iteration $x_{n+1}=\ln(4x_n)$ with $x_0=2$ to find $x_1$, $x_2$ and $x_3$ to 4 decimal places.
   (iii) By considering $\mathrm{e}^x-4x$ at $x=2.145$ and $x=2.155$, show that $\beta=2.15$ to 2 decimal places.

   [6 marks]

   (b)

   The iteration $x_{n+1}=\frac{\mathrm{e}^{x_n}}{4}$ with $x_0=1$ is also used. Find $x_1$ and $x_2$, and explain, by considering the gradient of $y=\frac{\mathrm{e}^x}{4}$ at each root, which root the iteration finds and why it cannot find the other.

   [6 marks]

   Total for question 4: 12 marks

End of questions