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AS & A Level · Practice questions · Locating roots by a change of sign

Trap the root between two numbers.

Six original Cambridge-style questions on locating roots by sketching graphs and by a change of sign, including proving a root to a given accuracy.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Rearrange to f(x) = 0, then test the sign.

01
Recall
[2 marks]

Show that the equation \(x^3 - 3x - 1 = 0\) has a root between 1 and 2.

  • \(f(x) = x^3 - 3x - 1\) is continuous. ✓
  • \(f(1) = 1 - 3 - 1 = -3\) and \(f(2) = 8 - 6 - 1 = 1\). ✓
  • Sign change, so there is a root between 1 and 2. ✓
02
Analysis
[4 marks]

By sketching the graphs of \(y = e^{-x}\) and \(y = 3 - x^2\) on the same diagram, state the number of real roots of the equation \(e^{-x} = 3 - x^2\). Show that one root lies between 1 and 2.

  • \(y = 3 - x^2\) is an inverted parabola with maximum \((0, 3)\); \(y = e^{-x}\) is decreasing through \((0, 1)\). ✓
  • At \(x = 0\) the parabola is above the curve; far to the left \(e^{-x}\) is huge, and to the right the parabola falls below the \(x\)-axis. So they cross twice: two real roots. ✓
  • \(f(x) = e^{-x} - 3 + x^2\): \(f(1) = 0.368 - 2 = -1.63\) and \(f(2) = 0.135 + 1 = 1.14\). ✓
  • Sign change and \(f\) continuous, so a root lies between 1 and 2. ✓
03
Calculation
[3 marks]

Show that the equation \(x = 2\sin x\) has a root between 1.8 and 2.

  • \(f(x) = x - 2\sin x\), continuous; radians. ✓
  • \(f(1.8) = 1.8 - 1.9477 = -0.148\) and \(f(2) = 2 - 1.8186 = 0.181\). ✓
  • Sign change, so a root lies between 1.8 and 2. ✓
04
Analysis
[3 marks]

A student says: "\(g(x) = \dfrac{1}{x-1}\) has \(g(0) = -1\) and \(g(2) = 1\), so the equation \(g(x) = 0\) has a root between 0 and 2." Explain why the student is wrong.

  • \(\dfrac{1}{x-1}\) can never be 0, because its numerator is never 0. ✓
  • \(g\) is not continuous on \([0, 2]\): it has a vertical asymptote at \(x = 1\). ✓
  • The sign change comes from the jump across the asymptote, so the test does not apply. ✓
05
Calculation
[3 marks]

The equation \(e^{x} = 3x\) has a root \(\beta\) between 0 and 1. Show that \(\beta = 0.62\) correct to 2 decimal places.

  • \(f(x) = e^{x} - 3x\); the bounds are 0.615 and 0.625. ✓
  • \(f(0.615) = 1.84966 - 1.845 = 0.00466\) and \(f(0.625) = 1.86825 - 1.875 = -0.00675\). ✓
  • Sign change, so \(0.615 < \beta < 0.625\), and \(\beta = 0.62\) to 2 d.p. ✓
06
Analysis
[6 marks]

A chord \(AB\) of a circle with centre \(O\) and radius \(r\) subtends an angle of \(\theta\) radians at \(O\). The area of the minor segment cut off by \(AB\) is one quarter of the area of the circle.

(a) Show that \(\theta = \sin\theta + \dfrac{\pi}{2}\).

(b) Show that this equation has a root between 2.3 and 2.4.

(c) Show that \(\theta = 2.31\) correct to 2 decimal places.

  • (a) Segment area \(= \dfrac{1}{2}r^2\theta - \dfrac{1}{2}r^2\sin\theta = \dfrac{1}{4}\pi r^2\). ✓
  • Divide by \(\dfrac{1}{2}r^2\): \(\theta - \sin\theta = \dfrac{\pi}{2}\), so \(\theta = \sin\theta + \dfrac{\pi}{2}\). ✓
  • (b) \(f(\theta) = \theta - \sin\theta - \dfrac{\pi}{2}\): \(f(2.3) = -0.0165\) and \(f(2.4) = 0.154\). Sign change, so a root lies between 2.3 and 2.4. ✓
  • (c) \(f(2.305) = -0.00816\) and \(f(2.315) = 0.00858\). Sign change, so \(2.305 < \theta < 2.315\) and \(\theta = 2.31\) to 2 d.p. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.