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Date	May 2018	Marks available	3	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 $\pi$ 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

$2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}$ A1A1

Note: Award A1 for each side.

if $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ then either $x = 0$ or $y = 0$ M1A1

$x = 0 \Rightarrow$ two solutions for $y\left( {y = \pm \sqrt[4]{5}} \right)$ 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) $4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 4\frac{{{\text{d}}y}}{{{\text{d}}x}}$ M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{1}{2}$ (A1)

gradient of normal is 2 M1

1 = 4 + c (M1)

equation of normal is $y = 2x - 3$ A1

[5 marks]

b.

substituting (M1)

${x^2}\left( {2x - 3} \right) = 5 - {\left( {2x - 3} \right)^4}$ or ${\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 - {y^4}$ (A1)

$x = 0.724$ A1

[3 marks]

c.

recognition of two volumes (M1)

volume $1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 - {y^4}}}{y}} {\text{d}}y\left( { = 101\pi = 3.178 \ldots } \right)$ M1A1A1

Note: Award M1 for attempt to use $\pi \int {{x^2}} {\text{d}}y$ , A1 for limits, A1 for ${\frac{{5 - {y^4}}}{y}}$ Condone omission of $\pi$ at this stage.

volume 2

EITHER

$= \frac{1}{3}\pi \times {2^2} \times 4\left( { = 16.75 \ldots } \right)$ (M1)(A1)

OR

$= \pi \int_{ - 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)$ (M1)(A1)

THEN

total volume = 19.9 A1

[7 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

Syllabus sections

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

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