Date | November 2016 | Marks available | 4 | Reference code | 16N.1.sl.TZ0.10 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | 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 \({f^{(19)}}(x)\).
(i) Find the first three derivatives of \(g(x)\).
(ii) Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k - p)!}}({x^{k - 19}})\), find \(p\).
(i) Find \(h'(x)\).
(ii) Hence, show that \(h'(\pi ) = \frac{{ - 21!}}{2}{\pi ^2}\).
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]
(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]
(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]