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Date May 2018 Marks available 5 Reference code 18M.2.AHL.TZ1.H_9
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number H_9 Adapted from N/A

Question

The following graph shows the two parts of the curve defined by the equation x 2 y = 5 y 4 , and the normal to the curve at the point P(2 , 1).

 

Show that there are exactly two points on the curve where the gradient is zero.

[7]
a.

Find the equation of the normal to the curve at the point P.

[5]
b.

The normal at P cuts the curve again at the point Q. Find the x -coordinate of Q.

[3]
c.

The shaded region is rotated by 2 π about the y -axis. Find the volume of the solid formed.

[7]
d.

Markscheme

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

differentiating implicitly:       M1

2 x y + x 2 d y d x = 4 y 3 d y d x      A1A1

Note: Award A1 for each side.

if  d y d x = 0  then either  x = 0 or  y = 0        M1A1

x = 0  two solutions for  y ( y = ± 5 4 )       R1

y = 0  not possible (as 0 ≠ 5)     R1

hence exactly two points      AG

Note: For a solution that only refers to the graph giving two solutions at   x = 0  and no solutions for  y = 0 award R1 only.

[7 marks]

a.

at (2, 1)   4 + 4 d y d x = 4 d y d x      M1

d y d x = 1 2      (A1)

gradient of normal is 2       M1

1 = 4 + c       (M1)

equation of normal is  y = 2 x 3      A1

[5 marks]

b.

substituting      (M1)

x 2 ( 2 x 3 ) = 5 ( 2 x 3 ) 4 or  ( y + 3 2 ) 2 y = 5 y 4        (A1)

x = 0.724       A1

[3 marks]

c.

recognition of two volumes      (M1)

volume  1 = π 1 5 4 5 y 4 y d y ( = 101 π = 3.178 )       M1A1A1

Note: Award M1 for attempt to use  π x 2 d y A1 for limits, A1 for  5 y 4 y  Condone omission of π at this stage.

volume 2

EITHER

= 1 3 π × 2 2 × 4 ( = 16.75 )      (M1)(A1)

OR

= π 3 1 ( y + 3 2 ) 2 d y ( = 16 π 3 = 16.75 )      (M1)(A1)

THEN

total volume = 19.9      A1

[7 marks]

d.

Examiners report

[N/A]
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d.

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus
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