Date | May 2022 | Marks available | 2 | 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 limx→0arctan(cos x)-kx2, where k∈ℝ.
Show that a finite limit only exists for k=π4.
Using l’Hôpital’s rule, show algebraically that the value of the limit is -14.
Markscheme
(as limx→0x2=0, the indeterminate form 00 is required for the limit to exist)
⇒limx→0(arctan(cos x)-k)=0 M1
arctan 1-k=0 (k=arctan 1) A1
so k=π4 AG
Note: Award M1A0 for using k=π4 to show the limit is 00.
[2 marks]
limx→0arctan(cos x)-π4x2(=00)
=limx→0-sin x1+cos2 x2x A1A1
Note: Award A1 for a correct numerator and A1 for a correct denominator.
recognises to apply l’Hôpital’s rule again (M1)
=limx→0-sin x1+cos2 x2x (=00)
Note: Award M0 if their limit is not the indeterminate form 00.
EITHER
=limx→0-cos x(1+cos2 x)-2 sin2 x cos x(1+cos2 x)22 A1A1
Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.
OR
limx→0-cos x2(1+cos2 x)-4x 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.
=-14 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.