By finding a suitable number of derivatives of \(f\), determine the Maclaurin series for \(f(x)\) as far as the term in \({x^3}\).

Hence, or otherwise, determine the exact value of \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}}\).

(i)     Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in this approximation.

(ii)     Deduce from the Lagrange error term whether the approximation will be greater than or less than the actual value of \(f(0.5)\).

attempt to use product rule     (M1)

\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\)    A1

\(f''(x) = 2{{\text{e}}^x}\cos x\)    A1

\(f''(x) = 2{{\text{e}}^x}\cos x - 2{{\text{e}}^x}\sin x\)    A1

\(f(0) = 0,{\text{ }}f'(0) = 1\)

\(f''(0) = 2,{\text{ }}f'''(0) = 2\)    (M1)

\({{\text{e}}^x}\sin x = x + {x^2} + \frac{{{x^3}}}{3} +  \ldots \)    (M1)A1

[7 marks]

METHOD 1

\(\frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}} = \frac{{x + {x^2} + \frac{{{x^3}}}{3} +  \ldots  - x - {x^2}}}{{{x^3}}}\)    M1A1

\( \to \frac{1}{3}\) as \(x \to 0\)     A1

METHOD 2

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - 1 - 2x}}{{3{x^2}}}\)    A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x}\cos x - 2}}{{6x}}\)    A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x}\cos x - 2{{\text{e}}^x}\sin x}}{6} = \frac{1}{3}\)    A1

[3 marks]

(i)     attempt to find \({{\text{4}}^{{\text{th}}}}\) derivative from the \({{\text{3}}^{{\text{rd}}}}\) derivative obtained in (a)     M1

\(f''''(x) =  - 4{{\text{e}}^x}\sin x\)    A1

Lagrange error term \( = \frac{{{f^{(n + 1)}}(c){x^{n + 1}}}}{{(n + 1)!}}\) (where c lies between 0 and \(x\))

\( =  - \frac{{4{{\text{e}}^c}\sin c \times {{0.5}^4}}}{{4!}}\)    (M1)

the maximum absolute value of this expression occurs when \(c = 0.5\)     (A1)

Note:     This A1 is independent of previous M marks.

therefore

upper bound \( = \frac{{4{{\text{e}}^{0.5}}\sin 0.5 \times {{0.5}^4}}}{{4!}}\)     (M1)

\( = 0.00823\)    A1

(ii)     the approximation is greater than the actual value because the Lagrange error term is negative     R1

[7 marks]

This part of the question was well answered by most candidates. In a few cases candidates failed to follow instructions and attempted to use known series; in a few cases mistakes in the determination of the derivatives prevented other candidates from achieving full marks; part (b) was also well answered using both the Maclaurin expansion or L’Hôpital rule; again in most cases that candidates failed to achieve full marks were due to mistakes in the determination of derivatives.

Part (a) of the question was well answered by most candidates. In a few cases candidates failed to follow instructions and attempted to use known series; in a few cases mistakes in the determination of the derivatives prevented other candidates from achieving full marks; part (b) was also well answered using both the Maclaurin expansion or L’Hôpital rule; again in most cases that candidates failed to achieve full marks were due to mistakes in the determination of derivatives.

Part (c) was poorly answered with few candidates showing familiarity with this part of the option. Most candidates quoted the formula and managed to find the \({4^{{\text{th}}}}\) derivative of \(f\) but then could not use it to obtain the required answer; in other cases candidates did obtain an answer but showed little understanding of its meaning when answering (c)(ii).