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Date May 2022 Marks available 2 Reference code 22M.2.AHL.TZ1.6
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Show that and Hence Question number 6 Adapted from N/A

Question

Consider the curve y=x.

The shape of a piece of metal can be modelled by the region bounded by the functions f, g, the x-axis and the line segment [AB], as shown in the following diagram. The units on the x and y axes are measured in metres.

The piecewise function f is defined by

fx={x  0x0.161.25x+0.2  0.16<x0.5

The graph of g is obtained from the graph of f by:

Point A lies on the graph of f and has coordinates (0.5, 0.825). Point B is the image of A under the given transformations and has coordinates (p, q).

The piecewise function g is given by

gx={hx  0.2xa1.25x+b  a<xp

The area enclosed by y=g(x), the x-axis and the line x=p is 0.0627292m2 correct to six significant figures.

Find dydx.

[2]
a.i.

Hence show that the equation of the tangent to the curve at the point 0.16, 0.4 is y=1.25x+0.2.

[2]
a.ii.

Find the value of p and the value of q.

[2]
b.

Find an expression for h(x).

[2]
c.i.

Find the value of a.

[1]
c.ii.

Find the value of b.

[2]
c.iii.

Find the area enclosed by y=f(x), the x-axis and the line x=0.5.

[3]
d.i.

Find the area of the shaded region on the diagram.

[4]
d.ii.

Markscheme

y=x12           (M1)

dydx=12x-12          A1 

 

[2 marks]

a.i.

gradient at x=0.16 is 12×10.16          M1

=1.25


EITHER

y-0.4=1.25x-0.16          M1


OR

0.4=1.250.16+b          M1

 

Note: Do not allow working backwards from the given answer.

 

THEN

hence y=1.25x+0.2          AG

 

[2 marks]

a.ii.

p=0.45,  q=0.4125  (or 0.413)  (accept " (0.45, 0.4125) ")          A1A1

 

[2 marks]

b.

hx= 122x-0.2          A2


Note: Award A1 if only two correct transformations are seen. 

 

[2 marks]

c.i.

a= 0.28          A1


[1 mark]

c.ii.

EITHER

Correct substitution of their part (b) (or 0.28, 0.2) into the given expression         (M1)


OR

121.25×2x-0.2+0.2         (M1)


Note: Award M1 for transforming the equivalent expression for f correctly.


THEN

b= -0.15          A1


[2 marks]

c.iii.

recognizing need to add two integrals        (M1)

00.16xdx+0.160.51.25x+0.2dx         (A1)


Note: The second integral could be replaced by the formula for the area of a trapezoid 12×0.340.4+0.825.


0.251m2  0.250916          A1


[3 marks]

d.i.

EITHER

area of a trapezoid 12×0.050.4125+0.825=0.0309375        (M1)(A1)


OR

0.450.58.25x-3.3dx=0.0309375        (M1)(A1)


Note:
If the rounded answer of 0.413 from part (b) is used, the integral is 0.450.58.24x-3.295dx=0.03095 which would be awarded (M1)(A1).

 

THEN

shaded area =0.250916-0.0627292-0.0309375        (M1)


Note: Award (M1) for the subtraction of both 0.0627292 and their area for the trapezoid from their answer to (a)(i).

 

=0.157m2  0.15725          A1

 

[4 marks]

d.ii.

Examiners report

The differentiation using the power rule was well done. In part (ii) some candidates felt it was sufficient to refer to the equation being the same as the one generated by their calculator. Generally, for ‘show that’ questions an algebraic derivation is expected.

a.i.
[N/A]
a.ii.

The candidates were successful at applying transformations to points but very few were able to apply these transformations to derive the correct function h. In most cases it was due to not appreciating the effect the horizontal transformations have on x.

b.

The candidates were successful at applying transformations to points but very few were able to apply these transformations to derive the correct function h. In most cases it was due to not appreciating the effect the horizontal transformations have on x.

c.i.
[N/A]
c.ii.
[N/A]
c.iii.

Part (i) was frequently done well using the inbuilt functionality of the GDC. Part (ii) was less structured, and candidates needed to create a clear diagram so they could easily see which areas needed to be subtracted. Most of those who were successful used the formula for the trapezoid for the area they needed to find, though others were also successful through finding the equation of the line AB.

d.i.
[N/A]
d.ii.

Syllabus sections

Topic 5—Calculus » SL 5.4—Tangents and normal
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Topic 5—Calculus

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