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Date November 2016 Marks available 7 Reference code 16N.1.sl.TZ0.10
Level SL only Paper 1 Time zone TZ0
Command term Find and Hence Question number 10 Adapted from N/A

Question

Let \(f(x) = \cos x\).

Let \(g(x) = {x^k}\), where \(k \in {\mathbb{Z}^ + }\).

Let \(k = 21\) and \(h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)\).

(i)     Find the first four derivatives of \(f(x)\).

(ii)     Find \({f^{(19)}}(x)\).

[4]
a.

(i)     Find the first three derivatives of \(g(x)\).

(ii)     Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k - p)!}}({x^{k - 19}})\), find \(p\).

[5]
b.

(i)     Find \(h'(x)\).

(ii)     Hence, show that \(h'(\pi ) = \frac{{ - 21!}}{2}{\pi ^2}\).

[7]
c.

Markscheme

(i)     \(f'(x) =  - \sin x,{\text{ }}f''(x) =  - \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\)     A2     N2

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)

\({f^{(19)}}(x) = \sin x\)     A1     N2

[4 marks]

a.

(i)     \(g'(x) = k{x^{k - 1}}\)

\(g''(x) = k(k - 1){x^{k - 2}},{\text{ }}{g^{(3)}}(x) = k(k - 1)(k - 2){x^{k - 3}}\)     A1A1     N2

(ii)     METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative     A2

eg\(\,\,\,\,\,\)\(k(k - 1)(k - 2) \ldots (k - 18) \times \frac{{(k - 19)!}}{{(k - 19)!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}\))     A1     N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient     A2

eg\(\,\,\,\,\,\)\(g'' = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k - 1)(k - 2) = \frac{{k!}}{{(k - 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k - 3}})\)

\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k - 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}\))     A1     N1

[5 marks]

b.

(i)     valid approach using product rule     (M1)

eg\(\,\,\,\,\,\)\(uv' + vu',{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}\)

correct 20th derivatives (must be seen in product rule)     (A1)(A1)

eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 - 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)

\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\)    A1     N3

(ii)     substituting \(x = \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\({f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi  + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}\)

evidence of one correct value for \(\sin \pi \) or \(\cos \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\sin \pi  = 0,{\text{ }}\cos \pi  =  - 1\)

evidence of correct values substituted into \(h'(\pi )\)     A1

eg\(\,\,\,\,\,\)\(21!(\pi )\left( {0 - \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { - \frac{\pi }{2}} \right),{\text{ }}0 + ( - 1)\frac{{21!}}{2}{\pi ^2}\)

 

Note: If candidates write only the first line followed by the answer, award A1A0A0.

 

\(\frac{{ - 21!}}{2}{\pi ^2}\)     AG     N0

[7 marks]

c.

Examiners report

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c.

Syllabus sections

Topic 6 - Calculus » 6.2 » The product and quotient rules.
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