Location of rootsEdexcel International A Level Maths: Subtopic test

1. 1
   The function $f$ is defined by $f(x)=x^3+x-5$ for $x\in\mathbb{R}$. The equation $f(x)=0$ has exactly one real root $\alpha$.

   (a)

   Which of these intervals contains $\alpha$?

   [1 mark]

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

   (b)

   Find the value of $f(1.5)$.

   [1 mark]

   • A$0.125$
   • B$-3.125$
   • C$9.875$
   • D$-0.125$

   (c)

   Given that $f(1.6)=0.696$, explain how the values of $f(1.5)$ and $f(1.6)$ show that $1.5<\alpha<1.6$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The functions $h$ and $p$ are defined by $h(x)=\frac{1}{x-2}$ for $x\neq2$, and $p(x)=x^2-2$ for $x\in\mathbb{R}$.

   (a)

   Find the values of $h(1)$ and $h(3)$.

   [1 mark]

   • A$h(1)=-1$ and $h(3)=1$
   • B$h(1)=1$ and $h(3)=-1$
   • C$h(1)=1$ and $h(3)=1$
   • D$h(1)=-1$ and $h(3)=-1$

   (b)

   Which interval allows a change of sign of $p(x)$ to show that $p(x)=0$ has a root?

   [1 mark]

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

   (c)

   A student notes that $h(1)<0$ and $h(3)>0$, and concludes that $h(x)=0$ has a root in $[1,3]$. Explain why this conclusion is wrong.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The function $f$ is defined by $f(x)=e^x-3x$ for $x\in\mathbb{R}$. The equation $f(x)=0$ has two real roots, $\alpha$ and $\beta$, with $\alpha<\beta$.

   (a)

   Show that $\alpha$ lies in the interval $[0.5,0.7]$.

   [3 marks]

   (b)

   Show that $\beta$ lies in the interval $[1.5,1.6]$. Then, by evaluating $f(1.51)$ and $f(1.52)$, find an interval of width $0.01$ that contains $\beta$.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The function $g$ is defined by $g(x)=x^3-6x^2+9x-1$ for $x\in\mathbb{R}$. A calculator may be used.

   (a)

   Show that $g(x)=0$ has three distinct roots in the interval $[0,4]$.

   [6 marks]

   (b)

   (i) Show that $g(x)=0$ has a root $\alpha$ with $\alpha=3.5$ correct to 1 decimal place.
   (ii) A student evaluates $g(0)$ and $g(3.5)$, finds both are negative, and says that $g(x)=0$ has no roots in $[0,3.5]$. Use a suitable value of $g(1)$ to show that this is wrong, and explain the flaw in the reasoning.

   [6 marks]

   Total for question 4: 12 marks

End of questions