Interval bisection and linear interpolationEdexcel International A Level Further Maths: Subtopic test

1. 1
   Let $f(x)=x^3+2x-7$ for real $x$.

   (a)

   Find the value of $f(1.5)$.

   [1 mark]

   • A$0.625$
   • B$-0.625$
   • C$-1.625$
   • D$-3.625$

   (b)

   Interval bisection is applied twice, starting with the interval $[1,2]$ that contains a root of $f(x)=0$. Which interval contains the root after the second bisection?

   [1 mark]

   • A$[1.5,1.75]$
   • B$[1.75,2]$
   • C$[1,1.5]$
   • D$[1.5,2]$

   (c)

   Show that the equation $f(x)=0$ has a root in the interval $[1,2]$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Two functions are defined for real $x$: $f(x)=\frac{1}{x-2}$ (for $x\neq2$) and $g(x)=(x-2)^2$.

   (a)

   A student notes that $f(1)=-1$ and $f(3)=1$. Which statement is correct?

   [1 mark]

   • A$f(x)=0$ has a root in $[1,3]$ because $f(1)<0<f(3)$
   • B$f(x)=0$ has no root in $[1,3]$ because $f(1)$ and $f(3)$ have the same sign
   • C$f(x)=0$ has no root in $[1,3]$; the sign change occurs because $f$ is not continuous at $x=2$
   • D$f(x)=0$ has exactly two roots in $[1,3]$

   (b)

   For $g$, $g(1)=1$ and $g(3)=1$. Which statement is correct?

   [1 mark]

   • A$g(x)=0$ has no root in $[1,3]$ because there is no change of sign
   • B$g(x)=0$ has two distinct roots in $[1,3]$
   • C$g(x)=0$ has no root in $[1,3]$ because $g$ is not continuous
   • D$g(x)=0$ has a root in $[1,3]$ even though there is no change of sign

   (c)

   Explain why the change of sign of $f$ between $x=1$ and $x=3$ does not show that $f(x)=0$ has a root in $[1,3]$.

   [2 marks]

   Total for question 2: 4 marks

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

   (a)

   Use linear interpolation on the interval $[1,2]$ to find a first approximation to $\alpha$.

   [3 marks]

   (b)

   Taking $1.375$ as the first approximation, use linear interpolation again, on a suitable interval, to find a second approximation to $\alpha$. Give your answer to 3 significant figures.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The equation $\ln x+x-3=0$ has a single root $\alpha$. Let $f(x)=\ln x+x-3$ for $x>0$.

   (a)

   (i) Show that $\alpha$ lies in the interval $[2,3]$.
   (ii) Use interval bisection twice, starting with $[2,3]$, to find an interval of width $0.25$ that contains $\alpha$.

   [6 marks]

   (b)

   (i) Use linear interpolation on the interval $[2,2.25]$ to find an estimate for $\alpha$ to 3 decimal places.
   (ii) By considering the signs of $f(2.2075)$ and $f(2.2085)$, determine $\alpha$ correct to 3 decimal places, and comment on your answer to (i).

   [6 marks]

   Total for question 4: 12 marks

End of questions