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Date May 2018 Marks available 3 Reference code 18M.2.hl.TZ1.9
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 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

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

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Syllabus sections

Topic 6 - Core: Calculus » 6.1 » Informal ideas of limit, continuity and convergence.
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