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Date	May 2022	Marks available	6	Reference code	22M.2.AHL.TZ2.7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Show that	Question number	7	Adapted from	N/A

Question

Consider $\lim_{x \to 0} \frac{arctan (\cos x) - k}{x^{2}}$ , where $k \in ℝ$ .

Show that a finite limit only exists for $k = \frac{π}{4}$ .

[2]

Using l’Hôpital’s rule, show algebraically that the value of the limit is $- \frac{1}{4}$ .

[6]

Markscheme

(as $\lim_{x \to 0} x^{2} = 0$ , the indeterminate form $\frac{0}{0}$ is required for the limit to exist)

$\Rightarrow \lim_{x \to 0} (arctan (\cos x) - k) = 0$ M1

$arctan 1 - k = 0 (k = arctan 1)$ A1

so $k = \frac{π}{4}$ AG

Note: Award M1A0 for using $k = \frac{π}{4}$ to show the limit is $\frac{0}{0}$ .

[2 marks]

$\lim_{x \to 0} \frac{arctan (\cos x) - \frac{π}{4}}{x^{2}} (= \frac{0}{0})$

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x}$ A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

recognises to apply l’Hôpital’s rule again (M1)

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x} (= \frac{0}{0})$

Note: Award M0 if their limit is not the indeterminate form $\frac{0}{0}$ .

EITHER

$= \lim_{x \to 0} \frac{\frac{- \cos x (1 + \cos^{2} x) - 2 \sin^{2} x \cos x}{{(1 + \cos^{2} x)}^{2}}}{2}$ A1A1

Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.

$\lim_{x \to 0} \frac{- \cos x}{2 (1 + \cos^{2} x) - 4 x \sin x \cos x}$ A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

THEN

substitutes $x = 0$ into the correct expression to evaluate the limit A1

Note: The final A1 is dependent on all previous marks.

$= - \frac{1}{4}$ AG

[6 marks]

Examiners report

Part (a) Many candidates recognised the indeterminate form and provided a nice algebraic proof. Some verified by substituting the given value. Therefore, there is a need to teach the candidates the difference between proof and verification. Only a few candidates were able to give a complete 'show that' proof.

Part (b) Many candidates realised that they needed to apply the L'Hôpital's rule twice. There were many mistakes in differentiation using the chain rule. Not all candidates clearly showed the final substitution.

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.5—Unit circle definitions of sin, cos, tan. Exact trig ratios, ambiguous case of sine rule

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