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Date November 2016 Marks available 4 Reference code 16N.1.SL.TZ0.S_10
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

Let f ( x ) = cos x .

Let g ( x ) = x k , 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 ) .

[4]
a.

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

(ii)     Given that g ( 19 ) ( x ) = k ! ( k p ) ! ( x k 19 ) , find p .

[5]
b.

(i)     Find h ( x ) .

(ii)     Hence, show that h ( π ) = 21 ! 2 π 2 .

[7]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(i)     f ( x ) = sin x ,   f ( x ) = cos x ,   f ( 3 ) ( x ) = sin x ,   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 ,   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 ) ( k 18 ) × ( k 19 ) ! ( k 19 ) ! ,   k P 19

p = 19  (accept 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 ! ( k 2 ) ,   k ( k 1 ) ( k 2 ) = k ! ( k 3 ) ! ,   g ( 3 ) ( x ) = k P 3 ( x k 3 )

g ( 19 ) ( x ) = 19 ! ( k 19 ) ,   19 ! × k ! ( k 19 ) ! × 19 ! ,   k P 19

p = 19  (accept k ! ( k 19 ) ! x k 19 )     A1     N1

[5 marks]

b.

(i)     valid approach using product rule     (M1)

eg u v + v u ,   f ( 19 ) g ( 20 ) + f ( 20 ) g ( 19 )

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

eg g ( 20 ) ( x ) = 21 ! ( 21 20 ) ! x ,   f ( 20 ) ( x ) = cos x

h ( x ) = sin x ( 21 ! x ) + cos x ( 21 ! 2 x 2 )   ( accept  sin x ( 21 ! 1 ! x ) + cos x ( 21 ! 2 ! x 2 ) )    A1     N3

(ii)     substituting x = π (seen anywhere)     (A1)

eg f ( 19 ) ( π ) g ( 20 ) ( π ) + f ( 20 ) ( π ) g ( 19 ) ( π ) ,   sin π 21 ! 1 ! π + cos π 21 ! 2 ! π 2

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

eg sin π = 0 ,   cos π = 1

evidence of correct values substituted into h ( π )      A1

eg 21 ! ( π ) ( 0 π 2 ! ) ,   21 ! ( π ) ( π 2 ) ,   0 + ( 1 ) 21 ! 2 π 2

 

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

 

21 ! 2 π 2      AG     N0

[7 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
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