June 2024 Paper 1 Q3

Notes

M1: Attempts both f(3.6) and f(3.7) or a narrower interval that contains the root 3.672… (which may be implied by sight of f(3.6) = … and f(3.7) = … with at least one correct) and obtains at least one correct to 1 significant figure (rounded or truncated) for their interval and considers their signs. Use of degrees is M0.

Some examples for consideration of sign (which are also sufficient for the change of sign part of reasoning for the A1):

• \(\mathrm{f}(3.6) = -0.6 \lt 0\) and \(\mathrm{f}(3.7) = 0.2 \gt 0\)
• \(\mathrm{f}(3.6) \times \mathrm{f}(3.7) \lt 0\)
• \(\mathrm{f}(3.6) = -0.7,\ \mathrm{f}(3.7) = 0.2\) “change in sign”

For reference \(\mathrm{f}(3.6) = -0.68626\ldots\) and \(\mathrm{f}(3.7) = 0.21194\ldots\)

A1*: This mark requires:

• both f(3.6) and f(3.7) correct to 1 significant figure (rounded or truncated) (or their values correct to 1 significant figure if using a narrower interval)
• a reference to sign change
• a reference to continuity {of \(\mathrm{f}(x)\)}
• a (minimal) conclusion, e.g. “hence root”, “proved”, \(\checkmark\), #, QED, \(3.6 \lt \alpha \lt 3.7\)

Accept as a minimum, "change of sign, continuous, root".
Do not condone “change in sign therefore continuous” or other incorrect statements such as “\(x\) is continuous”, “the interval is continuous” – these score A0.
Condone “the graph is continuous”.
Condone reference to \(x\) in place of \(\alpha\) in their conclusion, e.g. “hence \(x\) lies in the interval”.
Condone statements such as “there is at least one root” in place of their conclusion.

Notes

Note: Their answer to (b) may be seen in part (c) provided that they have not clearly attempted part (b) incorrectly, e.g., an attempt at \(\mathrm{f}^{-1}(x)\) in (b).

M1: For \(\tan\left(\dfrac{1}{2}x\right) \rightarrow \ldots\sec^2\left(\dfrac{1}{2}x\right)\) o.e. The brackets are not required. You may see attempts at the quotient rule but the method should be correct and they should reach something equivalent to \(\ldots\sec^2\left(\dfrac{1}{2}x\right)\).

e.g. \(\tan\left(\dfrac{1}{2}x\right) = \dfrac{\sin\left(\dfrac{1}{2}x\right)}{\cos\left(\dfrac{1}{2}x\right)} \rightarrow \dfrac{k\cos\left(\dfrac{1}{2}x\right)\cos\left(\dfrac{1}{2}x\right) - -k\sin\left(\dfrac{1}{2}x\right)\sin\left(\dfrac{1}{2}x\right)}{\cos^2\left(\dfrac{1}{2}x\right)}\) where \(k\) is a positive constant scores M1. If the formula is seen it must be correct.

A1: \(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec^2\left(\dfrac{1}{2}x\right)\) o.e. which may be unsimplified and apply isw.

The brackets are not required. There is no need for \(\mathrm{f}^{\prime}(x) =\) just look for the expression.

Note that \(\{\mathrm{f}^{\prime}(x)=\}\ \dfrac{3}{2} + \dfrac{1}{2}\tan^2\left(\dfrac{1}{2}x\right)\) is correct and appears occasionally.

\(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec\dfrac{1}{2}x^2\) is condoned for M1A0 only but \(1 + \dfrac{1}{2}\left(\sec\dfrac{1}{2}x\right)^2\) scores M1A1.

Notes

M1: Attempts \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)}\) and obtains a value following through on their \(\mathrm{f}^{\prime}(x)\) as long as it is a “changed” function in terms of \(x\).

Just stating \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)} = \ldots\) without evidence of use of 3.7 in \(\mathrm{f}(x)\) (note that this evidence might come from part (a)) and in their \(\mathrm{f}^{\prime}(x)\) is M0 unless implied by a correct value for both \(\mathrm{f}(x)\) and \(\mathrm{f}^{\prime}(x)\) or by their final answer.
Must be a correct N-R formula used – you may need to check their values – accuracy of at least 3s.f. rounded or truncated required.
Allow if attempted in degrees. For reference in degrees \(\mathrm{f}(3.7) = 3.73\ldots\) and \(\mathrm{f}^{\prime}(3.7) = 1.50\ldots\) and gives \(\alpha = 1.21\ldots\)
Note that the full N-R accuracy is 3.672051617.
For reference, the value of \(\alpha\) is approximately 3.673194406… and scores M0A0 without other valid work.

A1: For awrt 3.672  Ignore any subsequent iterations.