Locating roots by change of signAQA A-Level Maths: Subtopic test

1. 1
   The function $f(x)=x^3-5x+3$ is continuous for all real $x$.

   (a)

   Which interval must contain a root of $f(x)=0$?

   [1 mark]

   • A$-1\le x\le0$
   • B$2\le x\le3$
   • C$0.2\le x\le0.5$
   • D$1\le x\le2$

   (b)

   Given that $f(1.8)=-0.168$ and $f(1.9)=0.359$, which conclusion is justified?

   [1 mark]

   • AThere is at least one root between 1.8 and 1.9.
   • BThe root is exactly 1.85.
   • CThere is no root, because neither value is zero.
   • DThere is exactly one root between 1.8 and 1.9.

   (c)

   Show that $f(x)=0$ has a root between $x=0.6$ and $x=0.7$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Two functions are defined for $x\ne1$ by $f(x)=\frac{1}{x-1}$ and for all real $x$ by $g(x)=(x-1)^2$.

   (a)

   Given that $f(0)=-1$ and $f(2)=1$, which statement is correct?

   [1 mark]

   • A$f$ has a root in the interval because it changes sign.
   • B$f$ changes sign between 0 and 2 but has no root there, because $f$ is not continuous at $x=1$.
   • C$f$ has a root at $x=1$, where it changes sign.
   • D$f$ has two roots in the interval.

   (b)

   Given that $g(0)=1$ and $g(2)=1$, which statement is correct?

   [1 mark]

   • A$g$ has no root between 0 and 2, because $g(0)$ and $g(2)$ are both positive.
   • B$g$ has two different roots between 0 and 2.
   • CThere is no sign change, but $g(x)=0$ has a root at $x=1$.
   • D$g$ has a root only if $g(0)<0$.

   (c)

   Explain why the change of sign of $f$ between $x=0$ and $x=2$ does not show that $f(x)=0$ has a root in that interval.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The equation $e^x=3x$ is written as $f(x)=0$, where $f(x)=e^x-3x$.

   (a)

   Show that $f(x)=0$ has a root $\alpha$ in the interval $0.6<x<0.7$.

   [3 marks]

   (b)

   Use interval bisection twice, starting with the interval $[0.6,0.7]$, to find an interval of width 0.025 that contains $\alpha$.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The function $f(x)=x^3-2x^2-4x+8$ is continuous for all real $x$.

   (a)

   A student finds that $f(1)=3$ and $f(3)=5$ and concludes that $f(x)=0$ has no root between $x=1$ and $x=3$. (i) Show that $f(x)=0$ has a root between $x=-3$ and $x=-1$. (ii) Show that the student's conclusion is wrong, and explain why the change-of-sign test did not detect the root.

   [6 marks]

   (b)

   By considering $k(x)=f(x)-1$, show that $f(x)=1$ has three roots, one in each of the intervals $[-3,-1]$, $[1,2]$ and $[2,3]$. Explain how this is related to the repeated root of $f(x)=0$.

   [6 marks]

   Total for question 4: 12 marks

End of questions