Date | November 2016 | Marks available | 5 | 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)=cosx.
Let g(x)=xk, where k∈Z+.
Let k=21 and h(x)=(f(19)(x)×g(19)(x)).
(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)=k!(k−p)!(xk−19), find p.
(i) Find h′(x).
(ii) Hence, show that h′(π)=−21!2π2.
Markscheme
(i) f′(x)=−sinx, f″(x)=−cosx, f(3)(x)=sinx, f(4)(x)=cosx A2 N2
(ii) valid approach (M1)
egrecognizing that 19 is one less than a multiple of 4, f(19)(x)=f(3)(x)
f(19)(x)=sinx A1 N2
[4 marks]
(i) g′(x)=kxk−1
g″(x)=k(k−1)xk−2, g(3)(x)=k(k−1)(k−2)xk−3 A1A1 N2
(ii) METHOD 1
correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2
egk(k−1)(k−2)…(k−18)×(k−19)!(k−19)!, kP19
p=19 (accept k!(k−19)!xk−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
egg″=2!(k2), k(k−1)(k−2)=k!(k−3)!, g(3)(x)=kP3(xk−3)
g(19)(x)=19!(k19), 19!×k!(k−19)!×19!, kP19
p=19 (accept k!(k−19)!xk−19) A1 N1
[5 marks]
(i) valid approach using product rule (M1)
eguv′+vu′, f(19)g(20)+f(20)g(19)
correct 20th derivatives (must be seen in product rule) (A1)(A1)
egg(20)(x)=21!(21−20)!x, f(20)(x)=cosx
h′(x)=sinx(21!x)+cosx(21!2x2) (accept sinx(21!1!x)+cosx(21!2!x2)) A1 N3
(ii) substituting x=π (seen anywhere) (A1)
egf(19)(π)g(20)(π)+f(20)(π)g(19)(π), sinπ21!1!π+cosπ21!2!π2
evidence of one correct value for sinπ or cosπ (seen anywhere) (A1)
egsinπ=0, cosπ=−1
evidence of correct values substituted into h′(π) A1
eg21!(π)(0−π2!), 21!(π)(−π2), 0+(−1)21!2π2
Note: If candidates write only the first line followed by the answer, award A1A0A0.
−21!2π2 AG N0
[7 marks]