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Question
Find the value of \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right)\).
By using the series expansions for \({{\text{e}}^{{x^2}}}\) and cos x evaluate \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – {{\text{e}}^{{x^2}}}}}{{1 – \cos x}}} \right).\).
Answer/Explanation
Markscheme
Using l’Hopital’s rule,
\(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{\frac{1}{x}}}{{2\pi \cos 2\pi x}}} \right)\) M1A1
\( = \frac{1}{{2\pi }}\) A1
[3 marks]
\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – {{\text{e}}^{{x^2}}}}}{{1 – \cos x}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – \left( {1 + {x^2} + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + …} \right)}}{{1 – \left( {1 – \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} – …} \right)}}} \right)\) M1A1A1
Note: Award M1 for evidence of using the two series.
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { – {x^2} – \frac{{{x^4}}}{{2!}} – \frac{{{x^6}}}{{3!}} – …} \right)}}{{\left( {\frac{{{x^2}}}{{2!}} – \frac{{{x^4}}}{{4!}} + …} \right)}}} \right)\) A1
EITHER
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { – 1 – \frac{{{x^2}}}{{2!}} – \frac{{{x^4}}}{{3!}} – …} \right)}}{{\left( {\frac{1}{{2!}} – \frac{{{x^2}}}{{4!}} + …} \right)}}} \right)\) M1A1
\( = \frac{{ – 1}}{{\frac{1}{2}}} = – 2\) A1
OR
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { – 2x – \frac{{4{x^3}}}{{2!}} – \frac{{6{x^5}}}{{3!}} – …} \right)}}{{\left( {\frac{{2x}}{{2!}} – \frac{{4{x^3}}}{{4!}} + …} \right)}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { – 2 – \frac{{4{x^2}}}{{2!}} – \frac{{6{x^4}}}{{3!}} – …} \right)}}{{\left( {1 – \frac{{4{x^2}}}{{4!}} + …} \right)}}} \right)\)
\( = \frac{{ – 2}}{1} = – 2\) A1
[7 marks]
Examiners report
Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.
Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.
Question
The function f is defined by \(f(x) = {{\text{e}}^{({{\text{e}}^x} – 1)}}\) .
(a) Assuming the Maclaurin series for \({{\text{e}}^x}\) , show that the Maclaurin series for \(f(x)\)
is \(1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots {\text{ .}}\)
(b) Hence or otherwise find the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{f(x) – 1}}{{f'(x) – 1}}\) .
Answer/Explanation
Markscheme
(a) \({{\text{e}}^x} – 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^2}}}{6} + \ldots \) A1
\({{\text{e}}^{{{\text{e}}^x} – 1}} = 1 + \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right) + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^2}}}{2} + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^3}}}{6} + \ldots \) M1A1
\( = 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^3}}}{6} + \ldots \) M1A1
\( = 1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots \) AG
[5 marks]
(b) EITHER
\(f'(x) = 1 + 2x + \frac{{5{x^2}}}{2} + \ldots \) A1
\(\frac{{f(x) – 1}}{{f'(x) – 1}} = \frac{{x + {x^2} + 5{x^3}/6 + \ldots }}{{2x + 5{x^2}/2 + \ldots }}\) M1A1
\( = \frac{{1 + x + \ldots }}{{2 + 5x/2 + \ldots }}\) A1
\( \to \frac{1}{2}{\text{ as }}x \to 0\) A1
[5 marks]
OR
using l’Hopital’s rule, M1
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1′ – 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} – 1)}} – 1}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} – 1}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}}}}{{{{\text{e}}^{({{\text{e}}^x} + x – 1)}} \times ({{\text{e}}^x} + 1)}}\) A1
\( = \frac{1}{2}\) A1
[5 marks]
Total [10 marks]
Examiners report
Many candidates obtained the required series by finding the values of successive derivatives at x = 0 , failing to realise that the intention was to start with the exponential series and replace x by the series for \({{\text{e}}^x} – 1\). Candidates who did this were given partial credit for using this method. Part (b) was reasonably well answered using a variety of methods.
Question
(a) Using the Maclaurin series for \({(1 + x)^n}\), write down and simplify the Maclaurin series approximation for \({(1 – {x^2})^{ – \frac{1}{2}}}\) as far as the term in \({x^4}\)
(b) Use your result to show that a series approximation for arccos x is
\[\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}.\]
(c) Evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos ({x^2}) – {x^2}}}{{{x^6}}}\).
(d) Use the series approximation for \(\arccos x\) to find an approximate value for
\[\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,\]
giving your answer to 5 decimal places. Does your answer give the actual value of the integral to 5 decimal places?
Answer/Explanation
Markscheme
(a) using or obtaining \({(1 + x)^n} = 1 + nx + \frac{{n(n – 1)}}{2}{x^2} + \ldots \) (M1)
\({(1 – {n^2})^{ – \frac{1}{2}}} = 1 + ( – {x^2}) \times \left( { – \frac{1}{2}} \right) + \frac{{{{( – {x^2})}^2}}}{2} \times \left( { – \frac{1}{2}} \right) \times \left( { – \frac{3}{2}} \right) + \ldots \) (A1)
\( = 1 + \frac{1}{2}{x^2} + \frac{3}{8}{x^4} + \ldots \) A1
[3 marks]
(b) integrating, and changing sign
\(\arccos x = – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5} + C + \ldots \) M1A1
put x = 0,
\(\frac{\pi }{2} = C\) M1
\(\left( {\arccos x \approx \frac{\pi }{2} – x – \frac{1}{6}{x^3} – \frac{3}{{40}}{x^5}} \right)\) AG
[3 marks]
(c) EITHER
using \(\arccos {x^2} \approx \frac{\pi }{2} – {x^2} – \frac{1}{6}{x^6} – \frac{3}{{40}}{x^{10}}\) M1A1
\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} – \arccos {x^2} – {x^2}}}{{{x^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^6}}}{6} + {\text{higher powers}}}}{{{x^6}}}\) M1A1
\( = \frac{1}{6}\) A1
OR
using l’Hôpital’s Rule M1
\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} \times 2x – 2x}}{{6{x^5}}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 – {x^4}} }} – 1}}{{3{x^4}}}\) A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ – \frac{1}{2} \times \frac{1}{{{{(1 – {x^4})}^{3/2}}}} \times – 4{x^3}}}{{12{x^3}}}\) M1
\( = \frac{1}{6}\) A1
[5 marks]
(d) \(\int_0^{0.2} {\arccos \sqrt x {\text{d}}x \approx \int_0^{0.2} {\left( {\frac{\pi }{2} – {x^{\frac{1}{2}}} – \frac{1}{6}{x^{\frac{3}{2}}} – \frac{3}{{40}}{x^{\frac{5}{2}}}} \right){\text{d}}x} } \) M1
\( = \left[ {\frac{\pi }{2}x – \frac{2}{3}{x^{\frac{3}{2}}} – \frac{1}{{15}}{x^{\frac{5}{2}}} – \frac{3}{{140}}{x^{\frac{7}{2}}}} \right]_0^{0.2}\) (A1)
\( = \frac{\pi }{2} \times 0.2 – \frac{2}{3} \times {0.2^{\frac{3}{2}}} – \frac{1}{{15}} \times {0.2^{\frac{5}{2}}} – \frac{3}{{140}} \times {0.2^{\frac{7}{2}}}\) (A1)
= 0.25326 (to 5 decimal places) A1
Note: Accept integration of the series approximation using a GDC.
using a GDC, the actual value is 0.25325 A1
so the approximation is not correct to 5 decimal places R1
[6 marks]
Total [17 marks]
Examiners report
Many candidates ignored the instruction in the question to use the series for \({(1 + x)^n}\) to deduce the series for \({(1 – {x^2})^{ – 1/2}}\) and attempted instead to obtain it by successive differentiation. It was decided at the standardisation meeting to award full credit for this method although in the event the algebra proved to be too difficult for many. Many candidates used l’Hopital’s Rule in (c) – this was much more difficult algebraically than using the series and it usually ended unsuccessfully. Candidates should realise that if a question on evaluating an indeterminate limit follows the determination of a Maclaurin series then it is likely that the series will be helpful in evaluating the limit. Part (d) caused problems for many candidates with algebraic errors being common. Many candidates failed to realise that the best way to find the exact value of the integral was to use the calculator.
Question
Find \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – \cos {x^6}}}{{{x^{12}}}}} \right)\).
Answer/Explanation
Markscheme
METHOD 1
\(f(0) = \frac{0}{0}\), hence using l’Hôpital’s Rule, (M1)
\(g(x) = 1 – \cos ({x^6}),{\text{ }}h(x) = {x^{12}};{\text{ }}\frac{{g'(x)}}{{h'(x)}} = \frac{{6{x^5}\sin ({x^6})}}{{12{x^{11}}}} = \frac{{\sin ({x^6})}}{{2{x^6}}}\) A1A1
EITHER
\(\frac{{g'(0)}}{{h'(0)}} = \frac{0}{0}\), using l’Hôpital’s Rule again, (M1)
\(\frac{{g”(x)}}{{h”(x)}} = \frac{{6{x^5}\cos ({x^6})}}{{12{x^5}}} = \frac{{\cos ({x^6})}}{2}\) A1A1
\(\frac{{g”(0)}}{{h”(0)}} = \frac{1}{2}\), hence the limit is \(\frac{1}{2}\) A1
OR
So \(\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\) A1
\( = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\) A1
\( = \frac{1}{2}{\text{ since }}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}} = 1\) A1 (R1)
METHOD 2
substituting \({{x^6}}\) for x in the expansion \(\cos x = 1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} \ldots \) (M1)
\(\frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \frac{{1 – \left( {1 – \frac{{{x^{12}}}}{2} + \frac{{{x^{24}}}}{{24}}} \right) \ldots }}{{{x^{12}}}}\) M1A1
\( = \frac{1}{2} – \frac{{{x^{12}}}}{{24}} + …\) A1A1
\(\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos {x^6}}}{{{x^{12}}}} = \frac{1}{2}\) M1A1
Note: Accept solutions using Maclaurin expansions.
[7 marks]
Examiners report
Surprisingly, some weaker candidates were more successful in answering this question than stronger candidates. If candidates failed to simplify the expression after the first application of L’Hôpital’s rule, they generally were not successful in correctly differentiating the expression a \({2^{{\text{nd}}}}\) time, hence could not achieve the final three A marks.
Question
Find the first three terms of the Maclaurin series for \(\ln (1 + {{\text{e}}^x})\) .
Hence, or otherwise, determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) – x – \ln 4}}{{{x^2}}}\) .
Answer/Explanation
Markscheme
METHOD 1
\(f(x) = \ln (1 + {{\text{e}}^x});{\text{ }}f(0) = \ln 2\) A1
\(f'(x) = \frac{{{{\text{e}}^x}}}{{1 + {{\text{e}}^x}}};{\text{ }}f'(0) = \frac{1}{2}\) A1
Note: Award A0 for \(f'(x) = \frac{1}{{1 + {{\text{e}}^x}}};{\text{ }}f'(0) = \frac{1}{2}\)
\(f”(x) = \frac{{{{\text{e}}^x}(1 + {{\text{e}}^x}) – {{\text{e}}^{2x}}}}{{{{(1 + {{\text{e}}^x})}^2}}};{\text{ }}f”(0) = \frac{1}{4}\) M1A1
Note: Award M0A0 for \(f”(x){\text{ if }}f'(x) = \frac{1}{{1 + {{\text{e}}^x}}}\) is used
\(\ln (1 + {{\text{e}}^x}) = \ln 2 + \frac{1}{2}x + \frac{1}{8}{x^2} + …\) M1A1
[6 marks]
METHOD 2
\(\ln (1 + {{\text{e}}^x}) = \ln (1 + 1 + x + \frac{1}{2}{x^2} + …)\) M1A1
\( = \ln 2 + \ln (1 + \frac{1}{2}x + \frac{1}{4}{x^2} + …)\) A1
\( = \ln 2 + \left( {\frac{1}{2}x + \frac{1}{4}{x^2} + …} \right) – \frac{1}{2}{\left( {\frac{1}{2}x + \frac{1}{4}{x^2} + …} \right)^2} + …\) A1
\( = \ln 2 + \frac{1}{2}x + \frac{1}{4}{x^2} – \frac{1}{8}{x^2} + …\) A1
\( = \ln 2 + \frac{1}{2}x + \frac{1}{8}{x^2} + …\) A1
[6 marks]
METHOD 1
\(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) – x – \ln 4}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\ln 2 + x + \frac{{{x^2}}}{4} + {x^3}{\text{ terms & above}} – x – \ln 4}}{{{x^2}}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{4} + {\text{powers of }}x} \right) = \frac{1}{4}\) M1A1
Note: Accept + … as evidence of recognition of cubic and higher powers.
Note: Award M1AOM1A0 for a solution which omits the cubic and higher powers.
[4 marks]
METHOD 2
using l’Hôpital’s Rule
\(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) – x – \ln 4}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x} \div (1 + {{\text{e}}^x}) – 1}}{{2x}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x} \div {{(1 + {{\text{e}}^x})}^2}}}{2} = \frac{1}{4}\) M1A1
[4 marks]
Examiners report
In (a), candidates who found the series by successive differentiation were generally successful, the most common error being to state that the derivative of \(\ln (1 + {{\text{e}}^x})\) is \({(1 + {{\text{e}}^x})^{ – 1}}\). Some candidates assumed the series for \(\ln (1 + x)\) and \({{\text{e}}^x}\) attempted to combine them. This was accepted as an alternative solution but candidates using this method were often unable to obtain the required series.
In (b), candidates were equally split between using the series or using l’Hopital’s rule to find the limit. Both methods were fairly successful, but a number of candidates forgot that if a series was used, there had to be a recognition that it was not a finite series.
Question
Find \(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} – {x^2}} \right)}}{{\cot \pi x}}} \right)\).
Answer/Explanation
Markscheme
using l’Hôpital’s Rule (M1)
\(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} – {x^2}} \right)}}{{\cot \pi x}}} \right) = \mathop {\lim }\limits_{x \to \frac{1}{2}} \left[ {\frac{{ – 2}}{{ – \pi {\text{cose}}{{\text{c}}^2}\pi x}}} \right]\) A1A1
\( = \frac{{ – 1}}{{ – \pi {\text{cose}}{{\text{c}}^2}\frac{\pi }{2}}} = \frac{1}{\pi }\) (M1)A1
[5 marks]
Examiners report
This question was accessible to the vast majority of candidates, who recognised that L’Hôpital’s rule was required. However, some candidates omitted the factor \(\pi \) in the differentiation of \({\cot \pi x}\). Some candidates replaced \({\cot \pi x}\) by \(\cos \pi x{\text{/}}\sin \pi x\), which is a valid method but the extra algebra involved often led to an incorrect answer. Many fully correct solutions were seen.
Question
Use L’Hôpital’s Rule to find \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – 1 – x\cos x}}{{{{\sin }^2}x}}\) .
Answer/Explanation
Markscheme
apply l’Hôpital’s Rule to a \(0/0\) type limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – 1 – x\cos x}}{{{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} – \cos x + x\sin x}}{{2\sin x\cos x}}\) M1A1
noting this is also a \(0/0\) type limit, apply l’Hôpital’s Rule again (M1)
obtain \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} + \sin x + x\cos x + \sin x}}{{2\cos 2x}}\) A1
substitution of x = 0 (M1)
= 0.5 A1
[6 marks]
Examiners report
The vast majority of candidates were familiar with L’Hôpitals rule and were also able to apply the technique twice as required by the problem. The errors that occurred were mostly due to difficulty in applying the differentiation rules correctly or errors in algebra. A small minority of candidates tried to use the quotient rule but it seemed that most candidates had a good understanding of L’Hôpital’s rule and its application to finding a limit.
Question
The Taylor series of \(\sqrt x \) about x = 1 is given by
\[{a_0} + {a_1}(x – 1) + {a_2}{(x – 1)^2} + {a_3}{(x – 1)^3} + \ldots \]
Find the values of \({a_0},{\text{ }}{a_1},{\text{ }}{a_2}\) and \({a_3}\).
Hence, or otherwise, find the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}}\).
Answer/Explanation
Markscheme
let \(f(x) = \sqrt x ,{\text{ }}f(1) = 1\) (A1)
\(f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}},{\text{ }}f'(1) = \frac{1}{2}\) (A1)
\(f”(x) = – \frac{1}{4}{x^{ – \frac{3}{2}}},{\text{ }}f”(1) = – \frac{1}{4}\) (A1)
\(f”'(x) = \frac{3}{8}{x^{ – \frac{5}{2}}},{\text{ }}f”'(1) = \frac{3}{8}\) (A1)
\({a_1} = \frac{1}{2} \cdot \frac{1}{{1!}},{\text{ }}{a_2} = – \frac{1}{4} \cdot \frac{1}{{2!}},{\text{ }}{a_3} = \frac{3}{8} \cdot \frac{1}{{3!}}\) (M1)
\({a_0} = 1,{\text{ }}{a_1} = \frac{1}{2},{\text{ }}{a_2} = – \frac{1}{8},{\text{ }}{a_3} = \frac{1}{{16}}\) A1
Note: Accept \(y = 1 + \frac{1}{2}(x – 1) – \frac{1}{8}{(x – 1)^2} + \frac{1}{{16}}{(x – 1)^3} + \ldots \)
[6 marks]
METHOD 1
\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}(x – 1) – \frac{1}{8}{{(x – 1)}^2} + \ldots }}{{x – 1}}\) M1
\( = \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{2} – \frac{1}{8}(x – 1) + \ldots } \right)\) A1
\( = \frac{1}{2}\) A1
METHOD 2
using l’Hôpital’s rule, M1
\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x – 1}}{{x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}{x^{ – \frac{1}{2}}}}}{1}\) A1
\( = \frac{1}{2}\) A1
METHOD 3
\(\frac{{\sqrt x – 1}}{{x + 1}} = \frac{1}{{\sqrt x + 1}}\) M1A1
\(\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{2}\) A1
[3 marks]
Examiners report
Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.
Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.
Question
Let \(g(x) = \sin {x^2}\), where \(x \in \mathbb{R}\).
Using the result \(\mathop {{\text{lim}}}\limits_{t \to 0} \frac{{\sin t}}{t} = 1\), or otherwise, calculate \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{g(2x) – g(3x)}}{{4{x^2}}}\).
Use the Maclaurin series of \(\sin x\) to show that \(g(x) = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{{{x^{4n + 2}}}}{{(2n + 1)!}}} \)
Hence determine the minimum number of terms of the expansion of \(g(x)\) required to approximate the value of \(\int_0^1 {g(x){\text{d}}x} \) to four decimal places.
Answer/Explanation
Markscheme
METHOD 1
\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2} – \sin 9{x^2}}}{{4{x^2}}}\) M1
\( = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2}}}{{4{x^2}}} – \frac{9}{4}\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 9{x^2}}}{{9{x^2}}}\) A1A1
\( = 1 – \frac{9}{4} \times 1 = – \frac{5}{4}\) A1
METHOD 2
\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2} – \sin 9{x^2}}}{{4{x^2}}}\) M1
\( = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{8x\cos 4{x^2} – 18x\cos 9{x^2}}}{{8x}}\) M1A1
\( = \frac{{8 – 18}}{8} = – \frac{{10}}{8} = – \frac{5}{4}\) A1
[4 marks]
since \(\sin x = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{{{x^{(2n + 1)}}}}{{(2n + 1)!}}} \) \(\left( {{\text{or }}\sin x = \frac{x}{{1!}} – \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} – \ldots } \right)\) (M1)
\(\sin {x^2} = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{{{x^{2(2n + 1)}}}}{{(2n + 1)!}}} \) \(\left( {{\text{or }}\sin x = \frac{{{x^2}}}{{1!}} – \frac{{{x^6}}}{{3!}} + \frac{{{x^{10}}}}{{5!}} – \ldots } \right)\) A1
\(g(x) = \sin {x^2} = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{{{x^{4n + 2}}}}{{(2n + 1)!}}} \) AG
[2 marks]
let \(I = \int_0^1 {\sin {x^2}{\text{d}}x} \)
\( = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{1}{{(2n + 1)!}}} \int_0^1 {{x^{4n + 2}}{\text{d}}x{\text{ }}\left( {\int_0^1 {\frac{{{x^2}}}{{1!}}{\text{d}}x – } \int_0^1 {\frac{{{x^6}}}{{3!}}{\text{d}}x + } \int_0^1 {\frac{{{x^{10}}}}{{5!}}{\text{d}}x – \ldots } } \right)} \) M1
\( = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{1}{{(2n + 1)!}}} \frac{{[{x^{4n + 3}}]_0^1}}{{(4n + 3)}}{\text{ }}\left( {\left[ {\frac{{{x^3}}}{{3 \times 1!}}} \right]_0^1 – \left[ {\frac{{{x^7}}}{{7 \times 3!}}} \right]_0^1 + \left[ {\frac{{{x^{11}}}}{{11 \times 5!}}} \right]_0^1 – \ldots } \right)\) M1
\( = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{1}{{(2n + 1)!(4n + 3)}}} {\text{ }}\left( {\frac{1}{{3 \times 1!}} – \frac{1}{{7 \times 3!}} + \frac{1}{{11 \times 5!}} – \ldots } \right)\) A1
\( = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}{a_n}} \) where \({a_n} = \frac{1}{{(4n + 3)(2n + 1)!}} > 0\) for all \(n \in \mathbb{N}\)
as \(\{ {a_n}\} \) is decreasing the sum of the alternating series \(\sum\limits_{n = 0}^\infty {{{( – 1)}^n}{a_n}} \)
lies between \(\sum\limits_{n = 0}^N {{{( – 1)}^n}{a_n}} \) and \(\sum\limits_{n = 0}^N {{{( – 1)}^n}{a_n}} \pm {a_{N + 1}}\) R1
hence for four decimal place accuracy, we need \(\left| {{a_{N + 1}}} \right| < 0.00005\) M1
since \({a_{2 + 1}} < 0.00005\) R1
so \(N = 2\) (or 3 terms) A1
[7 marks]
Examiners report
Part (a) of this question was accessible to the vast majority of candidates, who recognised that L’Hôpital’s rule could be used. Most candidates were successful in finding the limit, with some making calculation errors. Candidates that attempted to use \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin x}}{x} = 1\) or a combination of this result and L’Hôpital’s rule were less successful.
In part (b) most candidates showed to be familiar with the substitution given and were successful in showing the result.
Very few candidates were able to do part (c) successfully. Most used trial and error to arrive at the answer.
Question
Consider the functions \(f\) and \(g\) given by \(f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2}{\text{ and }}g(x) = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2}\).
Show that \(f'(x) = g(x)\) and \(g'(x) = f(x)\).
Find the first three non-zero terms in the Maclaurin expansion of \(f(x)\).
Hence find the value of \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}}\).
Find the value of the improper integral \(\int_0^\infty {\frac{{g(x)}}{{{{\left[ {f(x)} \right]}^2}}}{\text{d}}x} \).
Answer/Explanation
Markscheme
any correct step before the given answer A1AG
eg, \(f'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } + {{\left( {{{\text{e}}^{ – x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2} = g(x)\)
any correct step before the given answer A1AG
eg, \(g'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } – {{\left( {{{\text{e}}^{ – x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2} = f(x)\)
[2 marks]
METHOD 1
statement and attempted use of the general Maclaurin expansion formula (M1)
\(f(0) = 1;{\text{ }}g(0) = 0\) (or equivalent in terms of derivative values) A1A1
\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\) A1A1
METHOD 2
\({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \) A1
\({{\text{e}}^{ – x}} = 1 – x + \frac{{{x^2}}}{{2!}} – \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \) A1
adding and dividing by 2 M1
\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\) A1A1
Notes: Accept 1, \(\frac{{{x^2}}}{2}\) and \(\frac{{{x^4}}}{{24}}\) or 1, \(\frac{{{x^2}}}{{2!}}\) and \(\frac{{{x^4}}}{{4!}}\).
Award A1 if two correct terms are seen.
[5 marks]
METHOD 1
attempted use of the Maclaurin expansion from (b) M1
\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – \left( {1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} + \ldots } \right)}}{{{x^2}}}\)
\(\mathop {{\text{lim}}}\limits_{x \to 0} \left( { – \frac{1}{2} – \frac{{{x^2}}}{{24}} – \ldots } \right)\) A1
\( = – \frac{1}{2}\) A1
METHOD 2
attempted use of L’Hôpital and result from (a) M1
\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ – g(x)}}{{2x}}\)
\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ – f(x)}}{2}\) A1
\( = – \frac{1}{2}\) A1
[3 marks]
METHOD 1
use of the substitution \(u = f(x)\) and \(\left( {{\text{d}}u = g(x){\text{d}}x} \right)\) (M1)(A1)
attempt to integrate \(\int_1^\infty {\frac{{{\text{d}}u}}{{{u^2}}}} \) (M1)
obtain \(\left[ { – \frac{1}{u}} \right]_1^\infty \) or \(\left[ { – \frac{1}{{f(x)}}} \right]_0^\infty \) A1
recognition of an improper integral by use of a limit or statement saying the integral converges R1
obtain 1 A1 N0
METHOD 2
\(\int_0^\infty {\frac{{\frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2}}}{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2}} \right)}^2}}}{\text{d}}x = \int_0^\infty {\frac{{2\left( {{{\text{e}}^x} – {{\text{e}}^{ – x}}} \right)}}{{{{\left( {{{\text{e}}^x} + {{\text{e}}^{ – x}}} \right)}^2}}}{\text{d}}x} } \) (M1)
use of the substitution \(u = {{\text{e}}^x} + {{\text{e}}^{ – x}}\) and \(\left( {{\text{d}}u = {{\text{e}}^x} – {{\text{e}}^{ – x}}{\text{d}}x} \right)\) (M1)
attempt to integrate \(\int_2^\infty {\frac{{2{\text{d}}u}}{{{u^2}}}} \) (M1)
obtain \(\left[ { – \frac{2}{u}} \right]_2^\infty \) A1
recognition of an improper integral by use of a limit or statement saying the integral converges R1
obtain 1 A1 N0
[6 marks]
Examiners report
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Question
Consider the infinite spiral of right angle triangles as shown in the following diagram.
The \(n{\text{th}}\) triangle in the spiral has central angle \({\theta _n}\), hypotenuse of length \({a_n}\) and opposite side of length 1, as shown in the diagram. The first right angle triangle is isosceles with the two equal sides being of length 1.
Consider the series \(\sum\limits_{n = 1}^\infty {{\theta _n}} \).
Using l’Hôpital’s rule, find \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\).
(i) Find \({a_1}\) and \({a_2}\) and hence write down an expression for \({a_n}\).
(ii) Show that \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\).
Using a suitable test, determine whether this series converges or diverges.
Answer/Explanation
Markscheme
\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\) is of the form \(\frac{0}{0}\)
and so will equal the limit of \(\frac{{\frac{{\frac{{ – 1}}{2}{{(x + 1)}^{ – \frac{3}{2}}}}}{{\sqrt {1 – \left( {\frac{1}{{x + 1}}} \right)} }}}}{{\frac{{ – 1}}{2}{x^{ – \frac{3}{2}}}}}\) M1M1A1A1
Note: M1 for attempting differentiation of the top and bottom, M1A1 for derivative of top (only award M1 if chain rule is used), A1 for derivative of bottom.
\( = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\frac{x}{{(x + 1)}}} \right)}^{\frac{3}{2}}}}}{{\sqrt {\frac{x}{{x + 1}}} }} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{x + 1}}} \right)\) M1
Note: Accept any intermediate tidying up of correct derivative for the method mark.
\( = 1\) A1
[6 marks]
(i) \({a_1} = \sqrt 2 ,{\text{ }}{a_2} = \sqrt 3 \) A1
\({a_n} = \sqrt {n + 1} \) A1
(ii) \(\sin {\theta _n} = \frac{1}{{{a_n}}} = \frac{1}{{\sqrt {n + 1} }}\) A1
Note: Allow \({\theta _n} = \arcsin \left( {\frac{1}{{{a_n}}}} \right)\) if \({a_n} = \sqrt {n + 1} \) in b(i).
so \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\) AG
[3 marks]
for \(\sum\limits_{n = 1}^\infty {\arcsin \frac{1}{{\sqrt {(n + 1)} }}} \) apply the limit comparison test (since both series of positive terms) M1
with \(\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}} \) A1
from (a) \(\mathop {\lim }\limits_{n \to \infty } \frac{{\arcsin \frac{1}{{\sqrt {(n + 1)} }}}}{{\frac{1}{{\sqrt n }}}} = 1\), so the two series either both converge or both diverge M1R1
\(\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt 2 }}} \) diverges (as is a \(p\)-series with \(p = \frac{1}{2}\)) A1
hence \(\sum\limits_{n = 1}^\infty {{\theta _n}} \) diverges A1
[6 marks]
Examiners report
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Question
The function f is defined on the domain \(\left] { – \frac{\pi }{2},\frac{\pi }{2}} \right[{\text{ by }}f(x) = \ln (1 + \sin x)\) .
Show that \(f”(x) = – \frac{1}{{(1 + \sin x)}}\) .
(i) Find the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
(ii) Explain briefly why your result shows that f is neither an even function nor an odd function.
Determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) – x}}{{{x^2}}}\).
Answer/Explanation
Markscheme
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) A1
\(f”(x) = \frac{{ – \sin x(1 + \sin x) – \cos x\cos x}}{{{{(1 + \sin x)}^2}}}\) M1A1
\( = \frac{{ – \sin x – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}\) A1
\( = – \frac{1}{{1 + \sin x}}\) AG
[4 marks]
(i) \(f”'(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) A1
\({f^{(4)}}(x) = \frac{{ – \sin x{{(1 + \sin x)}^2} – 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) M1A1
\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f”(0) = – 1\) M1
\(f”'(0) = 1,{\text{ }}{f^{(4)}}(0) = – 2\) A1
\(f(x) = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} – \frac{{{x^4}}}{{12}} + \ldots \) A1
(ii) the series contains even and odd powers of x R1
[7 marks]
\(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) – x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \ldots – x}}{{{x^2}}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ – 1}}{2} + \frac{x}{6} + \ldots }}{1}\) (A1)
\( = – \frac{1}{2}\) A1
Note: Use of l’Hopital’s Rule is also acceptable.
[3 marks]
Examiners report
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