2017 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

4 marks

Approach

The area of the plate is the region between the curve and the line, so we first find where they meet, then integrate the difference of the upper and lower expressions between those points.

Writing the line as $y = \frac{x - 1}{2}$ , we find where it meets the curve.

x^{2} - 6x + 5 = \frac{x - 1}{2}

\Rightarrow 2x^{2} - 12x + 10 = x - 1

\Rightarrow 2x^{2} - 13x + 11 = 0

\Rightarrow (x - 1)(2x - 11) = 0

\Rightarrow x = 1\ \text{or}\ x = \frac{11}{2}

At $x = 3$ (between the two roots), the line gives $y = 1$ and the curve gives $y = 9 - 18 + 5 = - 4$ , so the line lies above the curve on $\left\lbrack 1,\frac{11}{2} \right\rbrack$ . So

\text{Area} = \int_{1}^{11/2}\left\lbrack \frac{x - 1}{2} - \left( x^{2} - 6x + 5 \right) \right\rbrack dx

= \int_{1}^{11/2}\left( - x^{2} + \frac{13}{2}x - \frac{11}{2} \right)dx

= \left\lbrack - \frac{x^{3}}{3} + \frac{13}{4}x^{2} - \frac{11}{2}x \right\rbrack_{1}^{11/2}

= \frac{605}{48} - \left( - \frac{31}{12} \right)

= \frac{243}{16}

To check your answer: by GC, numerically integrating

\left\lbrack \frac{x - 1}{2} - \left( x^{2} - 6x + 5 \right) \right\rbrack

from

1

to

5.5

gives

15.1875 = \frac{243}{16}

. ✓

$\therefore$ the area of the plate is $\frac{243}{16}\ \text{units}^{2}$ .

Part (b)(i)

7 marks

Approach

The container is a solid of revolution about the $y$ -axis, so its volume is $\pi\int x^{2}\, dy$ . Since $x^{2} = \frac{y}{\left( a - y^{2} \right)^{2}}$ has a factor of $y$ multiplying a function of $a - y^{2}$ , the substitution $u = a - y^{2}$ turns it into a standard power of $u$ .

V = \pi\int_{0}^{1}x^{2}\, dy = \pi\int_{0}^{1}\frac{y}{\left( a - y^{2} \right)^{2}}\, dy

Let $u = a - y^{2}$ , so $du = - 2y\, dy$ .

\int\frac{y}{\left( a - y^{2} \right)^{2}}\, dy = - \frac{1}{2}\int\frac{du}{u^{2}}

= \frac{1}{2u} + C

= \frac{1}{2\left( a - y^{2} \right)} + C

V = \pi\left\lbrack \frac{1}{2\left( a - y^{2} \right)} \right\rbrack_{0}^{1}

= \pi\left( \frac{1}{2(a - 1)} - \frac{1}{2a} \right)

= \frac{\pi}{2}\left( \frac{a - (a - 1)}{a(a - 1)} \right)

= \frac{\pi}{2a(a - 1)}

$\therefore$ the volume of the container is $\frac{\pi}{2a(a - 1)}$ .

Part (b)(ii)

Approach

The volume formula from part (i) applies to the second container with $a$ replaced by $b$ , since it was formed the same way. We set this expression equal to $4$ times the first volume and solve for $b$ in terms of $a$ .

\frac{\pi}{2b(b - 1)} = 4\left( \frac{\pi}{2a(a - 1)} \right)

\Rightarrow \frac{1}{b(b - 1)} = \frac{4}{a(a - 1)}

\Rightarrow b(b - 1) = \frac{a(a - 1)}{4}

\Rightarrow b^{2} - b - \frac{a(a - 1)}{4} = 0

By the quadratic formula,

b = \frac{1 \pm \sqrt{1 + a(a - 1)}}{2} = \frac{1 \pm \sqrt{a^{2} - a + 1}}{2}

Since $a > 1$ , $a(a - 1) > 0$ , so $a^{2} - a + 1 > 1$ and $\sqrt{a^{2} - a + 1} > 1$ . So $b = \frac{1 - \sqrt{a^{2} - a + 1}}{2} < 0$ is rejected, since $b$ must be greater than $1$ .

b = \frac{1 - \sqrt{a^{2} - a + 1}}{2}\ \left( \text{rej. as}\ b > 1 \right),\quad b = \frac{1 + \sqrt{a^{2} - a + 1}}{2}

To check your answer: at

a = 2

, this gives

b = \frac{1 + \sqrt{3}}{2} \approx 1.37

; the volume with

a = 2

is

\frac{\pi}{4}

, and with this

b

the volume is

\frac{\pi}{2b(b - 1)} \approx \pi

, which is indeed

4 \times \frac{\pi}{4}

. ✓

$\therefore$ $b = \frac{1 + \sqrt{a^{2} - a + 1}}{2}$ .

2017 H2 Maths Paper 2, Question 4

Part (a)

Part (b)(i)

Part (b)(ii)

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