| Marks available | 9 |
| Reference code | 10N.3ca.hl.TZ0.3 |
Question
(a) Using the Maclaurin series for the function \({{\text{e}}^x}\), write down the first four terms of the Maclaurin series for \({{\text{e}}^{ – \frac{{{x^2}}}{2}}}\).
(b) Hence find the first four terms of the series for \(\int_0^x {{{\text{e}}^{ – \frac{{{u^2}}}{2}}}} {\text{d}}u\).
(c) Use the result from part (b) to find an approximate value for \(\frac{1}{{\sqrt {2\pi } }}\int_0^1 {{{\text{e}}^{ – \frac{{{x^2}}}{2}}}{\text{d}}x} \).
Markscheme
(a) \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
putting \(x = \frac{{ – {x^2}}}{2}\) (M1)
\({{\text{e}}^{ – \frac{{{x^2}}}{2}}} \approx 1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{{2^2} \times 2!}} – \frac{{{x^6}}}{{{2^3} \times 3!}} \approx \left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}}} \right)\) A2
[3 marks]
(b) \(\int_0^x {{{\text{e}}^{ – \frac{{{u^2}}}{2}}}{\text{d}}u \approx \left[ {u – \frac{{{u^3}}}{{3 \times 2}} + \frac{{{u^5}}}{{5 \times {2^2} \times 2!}} – \frac{{{u^7}}}{{7 \times {2^3} \times 3!}}} \right]_0^x} \) M1(A1)
\( = x – \frac{{{x^3}}}{{3 \times 2}} + \frac{{{x^5}}}{{5 \times {2^2} \times 2!}} – \frac{{{x^7}}}{{7 \times {2^3} \times 3!}}\) A1
\(\left( { = x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}}} \right)\)
[3 marks]
(c) putting x = 1 in part (b) gives \(\int_0^1 {{{\text{e}}^{ – \frac{{{x^2}}}{2}}}{\text{d}}x \approx } 0.85535 \ldots \) (M1)(A1)
\(\frac{1}{{\sqrt {2\pi } }}\int_0^1 {{{\text{e}}^{ – \frac{{{x^2}}}{2}}}{\text{d}}x \approx } 0.341\) A1
[3 marks]
Total [9 marks]
Examiners report
This was one of the most successfully answered questions. Some candidates however failed to use the data booklet for the expansion of the series, thereby wasting valuable time.
| Marks available | 2 |
| Reference code | 13N.3ca.hl.TZ0.4 |
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.
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.
| Marks available | 7 |
| Reference code | 13N.3ca.hl.TZ0.4 |
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.
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.
| Marks available | 4 |
| Reference code | 14N.3ca.hl.TZ0.4 |
Question
In this question you may assume that \(\arctan x\) is continuous and differentiable for \(x \in \mathbb{R}\).
Consider the infinite geometric series
\[1 – {x^2} + {x^4} – {x^6} + \ldots \;\;\;\left| x \right| < 1.\]
Show that the sum of the series is \(\frac{1}{{1 + {x^2}}}\).
Hence show that an expansion of \(\arctan x\) is \(\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \)
\(f\) is a continuous function defined on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) with \(f'(x) > 0\) on \(]a,{\text{ }}b[\).
Use the mean value theorem to prove that for any \(x,{\text{ }}y \in [a,{\text{ }}b]\), if \(y > x\) then \(f(y) > f(x)\).
(i) Given \(g(x) = x – \arctan x\), prove that \(g'(x) > 0\), for \(x > 0\).
(ii) Use the result from part (c) to prove that \(\arctan x < x\), for \(x > 0\).
Use the result from part (c) to prove that \(\arctan x > x – \frac{{{x^3}}}{3}\), for \(x > 0\).
Hence show that \(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\).
Markscheme
\(r = – {x^2},\;\;\;S = \frac{1}{{1 + {x^2}}}\) A1AG
[1 mark]
\(\frac{1}{{1 + {x^2}}} = 1 – {x^2} + {x^4} – {x^6} + \ldots \)
EITHER
\(\int {\frac{1}{{1 + {x^2}}}{\text{d}}x} = \int {1 – {x^2} + {x^4} – {x^6} + \ldots } {\text{d}}x\) M1
\(\arctan x = c + x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \) A1
Note: Do not penalize the absence of \(c\) at this stage.
when \(x = 0\) we have \(\arctan 0 = c\) hence \(c = 0\) M1A1
OR
\(\int_0^x {\frac{1}{{1 + {t^2}}}{\text{d}}t = } \int_0^x {1 – {t^2} + {t^4}} – {t^6} + \ldots {\text{d}}t\) M1A1A1
Note: Allow \(x\) as the variable as well as the limit.
M1 for knowing to integrate, A1 for each of the limits.
\([\arctan t]_0^x = \left[ {t – \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} – \frac{{{t^7}}}{7} + \ldots } \right]_0^x\) A1
hence \(\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \) AG
[4 marks]
applying the \(MVT\) to the function \(f\) on the interval \([x,{\text{ }}y]\) M1
\(\frac{{f(y) – f(x)}}{{y – x}} = f'(c)\;\;\;({\text{for some }}c \in ]x,{\text{ }}y[)\) A1
\(\frac{{f(y) – f(x)}}{{y – x}} > 0\;\;\;({\text{as }}f'(c) > 0)\) R1
\(f(y) – f(x) > 0{\text{ as }}y > x\) R1
\( \Rightarrow f(y) > f(x)\) AG
Note: If they use \(x\) rather than \(c\) they should be awarded M1A0R0, but could get the next R1.
[4 marks]
(i) \(g(x) = x – \arctan x \Rightarrow g'(x) = 1 – \frac{1}{{1 + {x^2}}}\) A1
this is greater than zero because \(\frac{1}{{1 + {x^2}}} < 1\) R1
so \(g'(x) > 0\) AG
(ii) (\(g\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(g'(x) > 0\) on \(]0,{\text{ }}b[\) for all \(b \in \mathbb{R}\))
(If \(x \in [0,{\text{ }}b]\) then) from part (c) \(g(x) > g(0)\) M1
\(x – \arctan x > 0 \Rightarrow \arctan x < x\) M1
(as \(b\) can take any positive value it is true for all \(x > 0\)) AG
[4 marks]
let \(h(x) = \arctan x – \left( {x – \frac{{{x^3}}}{3}} \right)\) M1
(\(h\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(h'(x) > 0\) on \(]0,{\text{ }}b[\))
\(h'(x) = \frac{1}{{1 + {x^2}}} – (1 – {x^2})\) A1
\( = \frac{{1 – (1 – {x^2})(1 + {x^2})}}{{1 + {x^2}}} = \frac{{{x^4}}}{{1 + {x^2}}}\) M1A1
\(h'(x) > 0\) hence \(({\text{for }}x \in [0,{\text{ }}b]){\text{ }}h(x) > h(0)( = 0)\) R1
\( \Rightarrow \arctan x > x – \frac{{{x^3}}}{3}\) AG
Note: Allow correct working with \(h(x) = x – \frac{{{x^3}}}{3} – \arctan x\).
[5 marks]
use of \(x – \frac{{{x^3}}}{3} < \arctan x < x\) M1
choice of \(x = \frac{1}{{\sqrt 3 }}\) A1
\(\frac{1}{{\sqrt 3 }} – \frac{1}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) M1
\(\frac{8}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) A1
Note: Award final A1 for a correct inequality with a single fraction on each side that leads to the final answer.
\(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\) AG
[4 marks]
Total [22 marks]
Examiners report
Most candidates picked up this mark for realizing the common ratio was \( – {x^2}\).
Quite a few candidates did not recognize the importance of ‘hence’ in this question, losing a lot of time by trying to work out the terms from first principles.
Of those who integrated the formula from part (a) only a handful remembered to include the ‘\( + c\)’ term, and to verify that this must be equal to zero.
Most candidates were able to achieve some marks on this question. The most commonly lost mark was through not stating that the inequality was unchanged when multiplying by \(y – x{\text{ as }}y > x\).
The first part of this question proved to be very straightforward for the majority of candidates.
In (ii) very few realized that they had to replace the lower variable in the formula from part (c) by zero.
Candidates found this part difficult, failing to spot which function was required.
Many candidates, even those who did not successfully complete (d) (ii) or (e), realized that these parts gave them the necessary inequality.
| Marks available | 7 |
| Reference code | 15M.3ca.hl.TZ0.1 |
Question
The function \(f\) is defined by \(f(x) = {{\text{e}}^{ – x}}\cos x + x – 1\).
By finding a suitable number of derivatives of \(f\), determine the first non-zero term in its Maclaurin series.
Markscheme
\(f(0) = 0\) A1
\(f'(x) = – {{\text{e}}^{ – x}}\cos x – {{\text{e}}^{ – x}}\sin x + 1\) M1A1
\(f'(0) = 0\) (M1)
\(f”(x) = 2{{\text{e}}^{ – x}}\sin x\) A1
\(f”(0) = 0\)
\({f^{(3)}}(x) = – 2{{\text{e}}^{ – x}}\sin x + 2{{\text{e}}^{ – x}}\cos x\) A1
\({f^{(3)}}(0) = 2\)
the first non-zero term is \(\frac{{2{x^3}}}{{3!}}\;\;\;\left( { = \frac{{{x^3}}}{3}} \right)\) A1
Note: Award no marks for using known series.
[7 marks]
Examiners report
Most students had a good understanding of the techniques involved with this question. A surprising number forgot to show \(f(0) = 0\). Some candidates did not simplify the second derivative which created extra work and increased the chance of errors being made.
| Marks available | 4 |
| Reference code | 15N.3ca.hl.TZ0.4 |
Question
Consider the function \(f(x) = \frac{1}{{1 + {x^2}}},{\text{ }}x \in \mathbb{R}\).
Illustrate graphically the inequality, \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \).
Use the inequality in part (a) to find a lower and upper bound for \(\pi \).
Show that \(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}} \).
Hence show that \(\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\).
Markscheme
A1A1A1
A1 for upper rectangles, A1 for lower rectangles, A1 for curve in between with \(0 \le x \le 1\)
hence \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \) AG
[3 marks]
attempting to integrate from \(0\) to \(1\) (M1)
\(\int_0^1 {f(x){\text{d}}x = [\arctan x]_0^1} \)
\( = \frac{\pi }{4}\) A1
attempt to evaluate either summation (M1)
\(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \frac{\pi }{4} < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)
hence \(\frac{4}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \pi < \frac{4}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)
so \(2.93 < \pi < 3.33\) A1A1
Note: Accept any answers that round to \(2.9\) and \(3.3\).
[5 marks]
EITHER
recognise \(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}}} \) as a geometric series with \(r = – {x^2}\) M1
sum of \(n\) terms is \(\frac{{1 – {{( – {x^2})}^n}}}{{1 – – {x^2}}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}\) M1AG
OR
\(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}(1 + {x^2}){x^{2r}} = (1 + {x^2}){x^0} – (1 + {x^2}){x^2} + (1 + {x^2}){x^4} + \ldots } \)
\( + {( – 1)^{n – 1}}(1 + {x^2}){x^{2n – 2}}\) M1
cancelling out middle terms M1
\( = 1 + {( – 1)^{n – 1}}{x^{2n}}\) AG
[2 marks]
\(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{1}{{1 + {x^2}}} + {{( – 1)}^{n – 1}}\frac{{{x^{2n}}}}{{1 + {x^2}}}} \)
integrating from \(0\) to \(1\) M1
\(\left[ {\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} } \right]_0^1 = \int_0^1 {f(x){\text{d}}x + {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } \) A1A1
\(\int_0^1 {f(x){\text{d}}x = \frac{\pi }{4}} \) A1
so \(\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\) AG
[4 marks]
Total [14 marks]
Examiners report
| Marks available | 7 |
| Reference code | SPNone.3ca.hl.TZ0.1 |
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 \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) – x}}{{{x^2}}}\).
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]