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Date November 2020 Marks available 7 Reference code 20N.2.SL.TZ0.S_10
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

Consider a function fx, for x0. The derivative of f is given by f'x=6xx2+4.

The graph of f is concave-down when x>n.

Show that f''x=24-6x2x2+42.

[4]
a.

Find the least value of n.

[2]
b.

Find 6xx2+4dx.

[3]
c.

Let R be the region enclosed by the graph of f, the x-axis and the lines x=1 and x=3. The area of R is 19.6, correct to three significant figures.

Find fx.

[7]
d.

Markscheme

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

METHOD 1

evidence of choosing the quotient rule        (M1)

eg     vu'-uv'v2

derivative of 6x is 6 (must be seen in rule)        (A1)

derivative of x2+4 is 2x (must be seen in rule)        (A1)

correct substitution into the quotient rule       A1

eg     6x2+4-6x2xx2+42

f''x=24-6x2x2+42       AG  N0

 

METHOD 2

evidence of choosing the product rule        (M1)

eg      vu'+uv'

derivative of 6x is 6 (must be seen in rule)        (A1)

derivative of x2+4-1 is -2xx2+4-2 (must be seen in rule)        (A1)

correct substitution into the product rule       A1

eg      6x2+4-1+-16x2xx2+4-2

f''x=24-6x2x2+42       AG  N0

 

[4 marks]

a.

METHOD 1 (2nd derivative)        (M1)

valid approach

eg     f''<0, 24-6x2<0 , n=±2, x=2

n=2 (exact)       A1  N2

 

METHOD 2 (1st derivative)

valid attempt to find local maximum on f'        (M1)

eg     sketch with max indicated, 2, 1.5, x=2

n=2 (exact)       A1  N2

 

[2 marks]

b.

evidence of valid approach using substitution or inspection      (M1)

eg     32x1udx , u=x2+4 , du=2xdx , 3×1udu

6xx2+4dx=3lnx2+4+c      A2  N3

[3 marks]

c.

recognizing that area =13fxdx  (seen anywhere)      (M1)

recognizing that their answer to (c) is their fx  (accept absence of c)      (M1)

eg     fx=3lnx2+4+c , fx=3lnx2+4

correct value for 133lnx2+4dx  (seen anywhere)      (A1)

eg     12.4859

correct integration for 13cdx  (seen anywhere)      (A1)

cx13 , 2c

adding their integrated expressions and equating to 19.6 (do not accept an expression which involves an integral)      (M1)

eg     12.4859+2c=19.6 , 2c=7.114

c=3.55700      (A1)

fx=3lnx2+4+3.56       A1  N4

[7 marks]

d.

Examiners report

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

Syllabus sections

Topic 5—Calculus » AHL 5.9—Differentiating standard functions and derivative rules
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