# IB DP Maths Topic 9.6 Maclaurin series for ex , sinx , cosx , ln(1+x) , (1+x)p , P∈Q . HL Paper 3

## Question

The function f is defined by

$f(x) = \ln \left( {\frac{1}{{1 – x}}} \right).$

(a)     Write down the value of the constant term in the Maclaurin series for $$f(x)$$ .

(b)     Find the first three derivatives of $$f(x)$$ and hence show that the Maclaurin series for $$f(x)$$ up to and including the $${x^3}$$ term is $$x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$$.

(c)     Use this series to find an approximate value for ln 2 .

(d)     Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.

(e)     How good is this upper bound as an estimate for the actual error?

## Markscheme

(a)     Constant term = 0     A1

[1 mark]

(b)     $$f'(x) = \frac{1}{{1 – x}}$$     A1

$$f”(x) = \frac{1}{{{{(1 – x)}^2}}}$$     A1

$$f”'(x) = \frac{2}{{{{(1 – x)}^3}}}$$     A1

$$f'(0) = 1;{\text{ }}f”(0) = 1;{\text{ }}f”'(0) = 2$$     A1

Note: Allow FT on their derivatives.

$$f(x) = 0 + \frac{{1 \times x}}{{1!}} + \frac{{1 \times {x^2}}}{{2!}} + \frac{{2 \times {x^3}}}{{3!}} + …$$     M1A1

$$= x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$$     AG

[6 marks]

(c)     $$\frac{1}{{1 – x}} = 2 \Rightarrow x = \frac{1}{2}$$     (A1)

$$\ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}}$$     M1

$$= \frac{2}{3}{\text{ (0.667)}}$$     A1

[3 marks]

(d)     Lagrange error $${\text{ = }}\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}} \times {\left( {\frac{1}{2}} \right)^{n + 1}}$$     (M1)

$$= \frac{6}{{{{(1 – c)}^4}}} \times \frac{1}{{24}} \times {\left( {\frac{1}{2}} \right)^4}$$     A1

$$< \frac{6}{{{{\left( {1 – \frac{1}{2}} \right)}^4}}} \times \frac{1}{{24}} \times \frac{1}{{16}}$$     A2

giving an upper bound of 0.25.     A1

[5 marks]

(e)     Actual error $$= \ln 2 – \frac{2}{3} = 0.0265$$     A1

The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate.     R1

[2 marks]

Total [17 marks]

## Examiners report

In (a), some candidates appeared not to understand the term ‘constant term’. In (b), many candidates found the differentiation beyond them with only a handful realising that the best way to proceed was to rewrite the function as $$f(x) = – \ln (1 – x)$$. In (d), many candidates were unable to use the Lagrange formula for the upper bound so that (e) became inaccessible.

## Question

(a)     Given that $$y = \ln \cos x$$ , show that the first two non-zero terms of the Maclaurin series for y are $$– \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}}$$.

(b)     Use this series to find an approximation in terms of $$\pi {\text{ for }}\ln 2$$ .

## Markscheme

(a)     $$f(x) = \ln \cos x$$

$$f'(x) = \frac{{ – \sin x}}{{\cos x}} = – \tan x$$     M1A1

$$f”(x) = – {\sec ^2}x$$     M1

$$f”'(x) = – 2\sec x\sec x\tan x$$     A1

$${f^{iv}}(x) = – 2{\sec ^2}x({\sec ^2}x) – 2\tan x(2{\sec ^2}x\tan x)$$

$$= – 2{\sec ^4}x – 4{\sec ^2}x{\tan ^2}x$$     A1

$$f(x) = f(0) + xf'(0) + \frac{{{x^2}}}{{2!}}f”(0) + \frac{{{x^3}}}{{3!}}f”'(0) + \frac{{{x^4}}}{{4!}}{f^{iv}}(0) + …$$

$$f(0) = 0,$$     M1

$$f'(0) = 0,$$

$$f”(0) = – 1,$$

$$f”'(0) = 0,$$

$${f^{iv}}(0) = – 2,$$     A1

Notes: Award the A1 if all the substitutions are correct.

Allow FT from their derivatives.

$$\ln (\cos x) \approx – \frac{{{x^2}}}{{2!}} – \frac{{2{x^4}}}{{4!}}$$     A1

$$= – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}}$$     AG

[8 marks]

(b)     Some consideration of the manipulation of ln 2     (M1)

Attempt to find an angle     (M1)

EITHER

Taking $$x = \frac{\pi }{3}$$     A1

$$\ln \frac{1}{2} \approx – \frac{{{{\left( {\frac{\pi }{3}} \right)}^2}}}{{2!}} – \frac{{2{{\left( {\frac{\pi }{3}} \right)}^4}}}{{4!}}$$     A1

$$– \ln 2 \approx – \frac{{\frac{{{\pi ^2}}}{9}}}{{2!}} – \frac{{2\frac{{{\pi ^4}}}{{81}}}}{{4!}}$$     A1

$$\ln 2 \approx \frac{{{\pi ^2}}}{{18}} + \frac{{{\pi ^4}}}{{972}} = \frac{{{\pi ^2}}}{9}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{108}}} \right)$$     A1

OR

Taking $$x = \frac{\pi }{4}$$     A1

$$\ln \frac{1}{{\sqrt 2 }} \approx – \frac{{{{\left( {\frac{\pi }{4}} \right)}^2}}}{{2!}} – \frac{{2{{\left( {\frac{\pi }{4}} \right)}^4}}}{{4!}}$$     A1

$$– \frac{1}{2}\ln 2 \approx – \frac{{\frac{{{\pi ^2}}}{{16}}}}{{2!}} – \frac{{2\frac{{{\pi ^4}}}{{256}}}}{{4!}}$$     A1

$$\ln 2 \approx \frac{{{\pi ^2}}}{{16}} + \frac{{{\pi ^4}}}{{1536}} = \frac{{{\pi ^2}}}{8}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{192}}} \right)$$     A1

[6 marks]

Total [14 marks]

## Examiners report

Some candidates had difficulty organizing the derivatives but most were successful in getting the series. Using the series to find the approximation for $$\ln 2$$ in terms of $$\pi$$ was another story and it was rare to see a good solution.

## Question

The variables x and y are related by $$\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x = \cos x$$ .

(a)     Find the Maclaurin series for y up to and including the term in $${x^2}$$ given that

$$y = – \frac{\pi }{2}$$ when x = 0 .

(b)     Solve the differential equation given that y = 0 when $$x = \pi$$ . Give the solution in the form $$y = f(x)$$ .

## Markscheme

(a)     from $$\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x + \cos x$$ , $$f'(0) = 1$$     A1

now $$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = y{\sec ^2}x + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x – \sin x$$     M1A1A1A1

Note: Award A1 for each term on RHS.

$$\Rightarrow f”(0) = – \frac{\pi }{2}$$     A1

$$\Rightarrow y = – \frac{\pi }{2} + x – \frac{{\pi {x^2}}}{4}$$     A1

[7 marks]

(b)     recognition of integrating factor     (M1)

integrating factor is $${{\text{e}}^{\int { – \tan x{\text{d}}x} }}$$

$$= {{\text{e}}^{\ln \cos x}}$$     (A1)

$$= \cos x$$     (A1)

$$\Rightarrow y\cos x = \int {{{\cos }^2}x{\text{d}}x}$$     M1

$$\Rightarrow y\cos x = \frac{1}{2}\int {(1 + \cos 2x){\text{d}}x}$$     A1

$$\Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} + k$$     A1

when $$x = \pi ,{\text{ }}y = 0 \Rightarrow k = – \frac{\pi }{2}$$     M1A1

$$\Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}$$     (A1)

$$\Rightarrow y = \sec x\left( {\frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}} \right)$$     A1

[10 marks]

Total [17 marks]

## Examiners report

Part (a) of the question was set up in an unusual way, which caused a problem for a number of candidates as they tried to do part (b) first and then find the Maclaurin series by a standard method. Few were successful as they were usually weaker candidates and made errors in finding the solution $$y = f(x)$$ . The majority of candidates knew how to start part (b) and recognised the need to use an integrating factor, but a number failed because they missed out the negative sign on the integrating factor, did not realise that $${{\text{e}}^{\ln \cos x}} = \cos x$$ or were unable to integrate $${{{\cos }^2}x}$$ . Having said this, a number of candidates succeeded in gaining full marks on this question.

## 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 $$\mathop {\lim }\limits_{x \to 0} \frac{{f(x) – 1}}{{f'(x) – 1}}$$ .

## 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

$\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,$

## 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

(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.

## Question

Find the first three terms of the Maclaurin series for $$\ln (1 + {{\text{e}}^x})$$ .

[6]
a.

Hence, or otherwise, determine the value of $$\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) – x – \ln 4}}{{{x^2}}}$$ .

[4]
b.

## 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]

a.

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]

b.

## 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.

a.

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.

b.

## Question

Use the limit comparison test to prove that $$\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}}$$ converges.

[5]
a.

Using the Maclaurin series for $$\ln (1 + x)$$ , show that the Maclaurin series for $$\left( {1 + x} \right)\ln \left( {1 + x} \right)$$ is $$x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}}$$.

[3]
c.

## Markscheme

apply the limit comparison test with $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$     M1

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{n(n + 1)}}}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{n(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1$$     M1A1

(since the limit is finite and $$\ne 0$$ ) both series do the same     R1

we know that $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ converges and hence $$\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}}$$ also converges     R1AG

[5 marks]

a.

$$(1 + x)\ln (1 + x) = (1 + x)\left( {x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4}…} \right)$$     A1

$$= \left( {x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4}…} \right) + \left( {{x^2} – \frac{{x3}}{2} + \frac{{{x^4}}}{3} – \frac{{{x^5}}}{4}…} \right)$$

EITHER

$$= x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^n}{x^{n + 1}}}}{{n + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{n}} }$$     A1

$$= x + \sum\limits_{n = 1}^\infty {{{( – 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{{ – 1}}{{n + 1}} + \frac{1}{n}} \right)}$$     M1

OR

$$x + \left( {1 – \frac{1}{2}} \right){x^2} – \left( {\frac{1}{2} – \frac{1}{3}} \right){x^3} + \left( {\frac{1}{3} – \frac{1}{4}} \right){x^4} – …$$     A1

$$= x + \sum\limits_{n = 1}^\infty {{{( – 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right)}$$     M1

$$= x + \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}}$$     AG

[3 marks]

c.

## Examiners report

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ converges.

a.

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ converges.

In part (b) finding the partial fractions was well done. The second part involving the use of telescoping series was less well done, and students were clearly not as familiar with this technique as with some others.

Part (c) was the least well done of all the questions. It was expected that students would use explicitly the result from the first part of 4(b) or show it once again in order to give a complete answer to this question, rather than just assuming that a pattern spotted in the first few terms would continue.

Candidates need to be informed that unless specifically told otherwise they may use without proof any of the Maclaurin expansions given in the Information Booklet. There were many candidates who lost time and gained no marks by trying to derive the expansion for $$\ln (1 + x)$$.

c.

## 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 $$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{g(2x) – g(3x)}}{{4{x^2}}}$$.

[4]
a.

Use the Maclaurin series of $$\sin x$$ to show that $$g(x) = \sum\limits_{n = 0}^\infty {{{( – 1)}^n}\frac{{{x^{4n + 2}}}}{{(2n + 1)!}}}$$

[2]
b.

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.

[7]
c.

## 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]

a.

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]

b.

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]

c.

## 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.

a.

In part (b) most candidates showed to be familiar with the substitution given and were successful in showing the result.

b.

Very few candidates were able to do part (c) successfully. Most used trial and error to arrive at the answer.

c.

## 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 $$g'(x) = f(x)$$.

[2]
a.

Find the first three non-zero terms in the Maclaurin expansion of $$f(x)$$.

[5]
b.

Hence find the value of $$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}}$$.

[3]
c.

Find the value of the improper integral $$\int_0^\infty {\frac{{g(x)}}{{{{\left[ {f(x)} \right]}^2}}}{\text{d}}x}$$.

[6]
d.

## 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]

a.

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]

b.

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]

c.

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]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

Given that $$f(x) = \ln x$$, use the mean value theorem to show that, for $$0 < a < b$$, $$\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}$$.

[7]
a.

Hence show that $$\ln (1.2)$$ lies between $$\frac{1}{m}$$ and $$\frac{1}{n}$$, where $$m$$, $$n$$ are consecutive positive integers to be determined.

[2]
b.

## Markscheme

$$f'(x) = \frac{1}{x}$$    (A1)

using the MVT $$f'(c) = \frac{{f(b) – f(a)}}{{b – a}}$$ (where $$c$$ lies between $$a$$ and $$b$$)     (M1)

$$f'(c) = \frac{{\ln b – \ln a}}{{b – a}}$$    A1

$$\ln \frac{b}{a} = \ln b – \ln a$$    (M1)

$$f'(c) = \frac{{\ln \frac{b}{a}}}{{b – a}}$$

since $$f'(x)$$ is a decreasing function or $$a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a}$$     R1

$$f'(b) < f'(c) < f'(a)$$    (M1)

$$\frac{1}{b} < \frac{{\ln \frac{b}{a}}}{{b – a}} < \frac{1}{a}$$    A1

$$\frac{{b – a}}{b} < \ln \frac{b}{a} < \frac{{b – a}}{a}$$    AG

[7 marks]

a.

putting $$b = 1.2,{\text{ }}a = 1$$, or equivalent     M1

$$\frac{1}{6} < \ln 1.2 < \frac{1}{5}$$    A1

$$(m = 6,{\text{ }}n = 5)$$

[2 marks]

b.

## Examiners report

Although many candidates achieved at least a few marks in this question, the answers revealed difficulties in setting up a proof. The Mean value theorem was poorly quoted and steps were often skipped. The conditions under which the Mean value theorem is valid were largely ignored, as were the reasoned steps towards the answer.

a.

There were inequalities everywhere, without a great deal of meaning or showing progress. A number of candidates attempted to work backwards and presented the work in a way that made it difficult to follow their reasoning; in part (b) many candidates ignored the instruction ‘hence’ and just used GDC to find the required values; candidates that did notice the link to part a) answered this question well in general. A number of candidates guessed the answer and did not present an analytical derivation as required.

b.

## Question

The function $$f$$ is defined by $$f(x){\text{ }}={\text{ }}{(\arcsin{\text{ }}x)^2},{\text{ }} – 1 \leqslant x \leqslant 1$$.

The function $$f$$ satisfies the equation $$\left( {1 – {x^2}} \right)f”\left( x \right) – xf’\left( x \right) – 2 = 0$$.

Show that $$f’\left( 0 \right) = 0$$.

[2]
a.

By differentiating the above equation twice, show that

$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 5x{f^{\left( 3 \right)}}\left( x \right) – 4f”\left( x \right) = 0$

where $${f^{\left( 3 \right)}}\left( x \right)$$ and $${f^{\left( 4 \right)}}\left( x \right)$$ denote the 3rd and 4th derivative of $$f\left( x \right)$$ respectively.

[4]
b.

Hence show that the Maclaurin series for $$f\left( x \right)$$ up to and including the term in $${x^4}$$ is $${x^2} + \frac{1}{3}{x^4}$$.

[3]
c.

Use this series approximation for $$f\left( x \right)$$ with $$x = \frac{1}{2}$$ to find an approximate value for $${\pi ^2}$$.

[2]
d.

## Markscheme

$$f’\left( x \right) = \frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 – {x^2}} }}$$     M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for $$f’\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)$$.

$$f’\left( 0 \right) = 0$$     AG

[2 marks]

a.

differentiating gives $$\left( {1 – {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) – 2xf”\left( x \right) – f’\left( x \right) – xf”\left( x \right)\left( { = 0} \right)$$      M1A1

differentiating again gives $$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 2x{f^{\left( 3 \right)}}\left( x \right) – 3f”\left( x \right) – 3x{f^{\left( 3 \right)}}\left( x \right) – f”\left( x \right)\left( { = 0} \right)$$     M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

$$\left( {1 – {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) – 5x{f^{\left( 3 \right)}}\left( x \right) – 4f”\left( x \right) = 0$$      AG

[4 marks]

b.

attempting to find one of $$f”\left( 0 \right)$$, $${f^{\left( 3 \right)}}\left( 0 \right)$$ or $${f^{\left( 4 \right)}}\left( 0 \right)$$ by substituting $$x = 0$$ into relevant differential equation(s)       (M1)

Note: Condone $$f”\left( 0 \right)$$ found by calculating $$\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 – {x^2}} }}} \right)$$ at $$x = 0$$.

$$\left( {f\left( 0 \right) = 0,\,f’\left( 0 \right) = 0} \right)$$

$$f”\left( 0 \right) = 2$$ and $${f^{\left( 4 \right)}}\left( 0 \right) – 4f”\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8$$      A1

$${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ and so $$\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}$$     A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to $${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ stated or $${f^{\left( 3 \right)}}\left( 0 \right) = 0$$ seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if $${f^{\left( 4 \right)}}\left( 0 \right) = 8$$ is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for $$f\left( x \right)$$ up to and including the term in $${x^4}$$ is $${x^2} + \frac{1}{3}{x^4}$$     AG

[3 marks]

c.

substituting $$x = \frac{1}{2}$$ into $${x^2} + \frac{1}{3}{x^4}$$      M1

the series approximation gives a value of $$\frac{{13}}{{48}}$$

so $${\pi ^2} \simeq \frac{{13}}{{48}} \times 36$$

$$\simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)$$     A1

Note: Accept 9.76.

[2 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## 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 $$f”(x) = – \frac{1}{{(1 + \sin x)}}$$ .

[4]
a.

(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.

[7]
b.

Determine the value of $$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) – x}}{{{x^2}}}$$.

[3]
c.

## 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]

a.

(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]

b.

$$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) – x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \ldots – x}}{{{x^2}}}$$     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]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.