P9.1Locating roots by change of sign

(a) Answer: \(\tan x\) is not continuous on \([1, 2]\): it has an asymptote at \(x = \frac{\pi}{2} \approx 1.571\), where the sign changes. The roots of \(\tan x = 0\) are \(0, \pi, \ldots\), none in \((1, 2)\).

• B1 for \(\tan x\) has a discontinuity (vertical asymptote) at \(x = \frac{\pi}{2}\), which lies between 1 and 2
• B1 for the sign change is caused by the asymptote, not a root (the change of sign method needs a continuous function)

Worked solution: \(\tan x\) is not defined at \(x = \frac{\pi}{2} \approx 1.571\), which lies between 1 and 2. The function is not continuous there, and the sign changes across the asymptote. The roots of \(\tan x = 0\) are \(x = 0, \pi, 2\pi, \ldots\), so there is no root between 1 and 2.

(b) Answer: \(\mathrm{f}(2.1) = -0.239 \lt 0\), \(\mathrm{f}(2.2) = 0.648 \gt 0\); \(\mathrm{f}\) is continuous and changes sign.

• M1 for evaluating \(\mathrm{f}(2.1)\) and \(\mathrm{f}(2.2)\) where \(\mathrm{f}(x) = x^3 - 5x + 1\)
• A1 for \(-0.239\) and \(0.648\), with sign change, continuity and conclusion

Worked solution: Let \(\mathrm{f}(x) = x^3 - 5x + 1\). \(\mathrm{f}(2.1) = -0.239 \lt 0\) and \(\mathrm{f}(2.2) = 0.648 \gt 0\). \(\mathrm{f}\) is a polynomial, so it is continuous, and it changes sign, so there is a root between 2.1 and 2.2.

(c) Answer: \(\mathrm{g}\) has two roots, 1.4 and 1.6, in the interval; with an even number of roots the sign changes twice, so the end values have the same sign.

• B1 for there are two roots (1.4 and 1.6) in the interval, so the sign changes twice and there is no overall change of sign

Worked solution: \(\mathrm{g}(x) = 0\) at \(x = 1.4\) and \(x = 1.6\), both between 1 and 2. The graph crosses the axis twice, so \(\mathrm{g}\) is positive at both ends. No change of sign does not mean no roots.

(a) Answer: Shown

• M1 for attempting \(h(2.9)\) and \(h(3)\) with at least one value correct to 1 significant figure (rounded or truncated)
• A1 for both values correct to at least 1 s.f. (\(-0.135\) and \(0.0986\)), a reason (change of sign and \(h(x)\) continuous) and a conclusion

Worked solution: \(h(2.9) = -0.135 \lt 0\) and \(h(3) = 0.0986 \gt 0\).
There is a change of sign and \(h(x)\) is continuous on \([2.9, 3]\), so \(\alpha\) lies in the interval \([2.9, 3]\).

(b) Answer: Shown: \(h(2.955)\) and \(h(2.965)\) have opposite signs

• M1 for choosing the interval \([2.955, 2.965]\) (or a tighter interval that establishes the rounding) and evaluating \(h\) at both ends, with at least one value correct to 1 s.f.
• A1 for both values correct (\(-0.00650\) and \(0.0169\)), a change of sign, continuity and the conclusion \(\alpha = 2.96\) (2 d.p.)

Worked solution: \(h(2.955) = -0.00650 \lt 0\) and \(h(2.965) = 0.0169 \gt 0\).
There is a change of sign and \(h(x)\) is continuous, so \(2.955 \lt \alpha \lt 2.965\). Every number in this interval rounds to \(2.96\), so \(\alpha = 2.96\) to 2 decimal places.

(a) Answer: \(f(0) = 1.000, \ f(2.7) = 1.380\)

• B1 for both values: awrt \(1.000\) and awrt \(1.380\)

Worked solution: \(f(0) = \mathrm{e}^{0} - 5 \times 0 = 1.000\) and \(f(2.7) = \mathrm{e}^{2.7} - 5 \times 2.7 = 1.380\). Both are positive.

(b) Answer: No change of sign does not mean no root: there may be an even number of roots

• B1 for explaining that the absence of a change of sign does not rule out roots, because there could be an even number of roots (the curve could cross the axis twice) in the interval

Worked solution: If the curve crosses the \(x\)-axis an even number of times between the end points, the values at the ends have the same sign, so no change of sign does not mean that there is no root.

(c) Answer: Shown: sign changes on \([0, 1.6]\) and \([1.6, 2.7]\)

• M1 for \(f(1.6) = -3.05\) (correct to 1 s.f.)
• A1 for identifying the change of sign on both \([0, 1.6]\) and \([1.6, 2.7]\)
• A1 for continuity of \(f\) and the conclusion that there is a root in each sub-interval, so at least two roots

Worked solution: \(f(1.6) = -3.05 \lt 0\), while \(f(0) \gt 0\) and \(f(2.7) \gt 0\).
\(f\) is continuous, and it changes sign on \([0, 1.6]\) and on \([1.6, 2.7]\), so there is a root in each: at least two roots. (They are \(x = 0.259\) and \(x = 2.543\).)