2023 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

3 marks

Curve $C$ : $x = 2t^{2} + 3$ , $y = 5t - 1$ , $t \geq \frac{1}{5}$ .

Approach

Before setting up the integral, we check where the curve starts. At $t = \frac{1}{5}$ , $y = 0$ , so $C$ begins on the $x$ -axis, and the required area is simply the area under $C$ from there to $x = 21$ .

At $t = \frac{1}{5}$ : $x = 2\left( \frac{1}{25} \right) + 3 = \frac{77}{25}$ , $y = 5\left( \frac{1}{5} \right) - 1 = 0$ .

At $x = 21$ : $2t^{2} + 3 = 21 \Rightarrow t^{2} = 9 \Rightarrow t = 3$ (since $t \geq \frac{1}{5}$ ).

Since the curve begins on the $x$ -axis, the required area is

\text{Area} = \int_{x = 77/25}^{21}y\, dx = \int_{1/5}^{3}y\,\frac{dx}{dt}\, dt

With $\frac{dx}{dt} = 4t$ ,

\text{Area} = \int_{1/5}^{3}(5t - 1)(4t)\, dt

= \int_{1/5}^{3}\left( 20t^{2} - 4t \right)\, dt

= \left\lbrack \frac{20}{3}t^{3} - 2t^{2} \right\rbrack_{1/5}^{3}

At $t = 3$ : $\frac{20}{3}(27) - 2(9) = 180 - 18 = 162$ . At $t = \frac{1}{5}$ : $\frac{20}{3}\left( \frac{1}{125} \right) - 2\left( \frac{1}{25} \right) = - \frac{2}{75}$ .

\text{Area} = 162 - \left( - \frac{2}{75} \right) = 162 + \frac{2}{75} = \frac{12\, 152}{75}

To check your answer:

\frac{12\, 152}{75} \approx 162.03

; since the region spans

x

from about

3.08

to

21

(width

\text{≈}17.9

) with

y

reaching up to

14

, an area around

162

is a reasonable order of magnitude ✓

$\therefore$ Area $\text{=}\frac{12\, 152}{75}$ .

Part (b)

5 marks

Curve $D$ : $x = 5u$ , $y = \frac{4}{u}$ , $u \neq 0$ .

Approach

At a point of intersection both parametrisations give the same $(x,y)$ . We eliminate $u$ using the $x$ -equations, substitute into the $y$ -equations, and solve for $t$ .

From $C$ : $x = 2t^{2} + 3$ . From $D$ : $x = 5u \Rightarrow u = \frac{2t^{2} + 3}{5}$ . Equating the $y$ -values,

5t - 1 = \frac{4}{u} = \frac{20}{2t^{2} + 3}

\Rightarrow (5t - 1)\left( 2t^{2} + 3 \right) = 20

\Rightarrow 10t^{3} - 2t^{2} + 15t - 23 = 0

Testing $t = 1$ : $10 - 2 + 15 - 23 = 0$ ✓, so $t = 1$ is a root. Dividing out the factor $(t - 1)$ ,

10t^{3} - 2t^{2} + 15t - 23 = (t - 1)\left( 10t^{2} + 8t + 23 \right)

To check your answer: expanding

(t - 1)\left( 10t^{2} + 8t + 23 \right) = 10t^{3} - 2t^{2} + 15t - 23

, matching the original cubic ✓

For the quadratic factor $10t^{2} + 8t + 23 = 0$ , the discriminant is

8^{2} - 4(10)(23) = 64 - 920 = - 856 < 0

Since the discriminant is negative, $10t^{2} + 8t + 23 = 0$ has no real roots, so $t = 1$ is the only real solution and there is only one point of intersection.

At $t = 1$ : $x = 2(1)^{2} + 3 = 5$ , $y = 5(1) - 1 = 4$ .

To check your answer: at

t = 1

,

u = \frac{2(1) + 3}{5} = 1

; on

D

this gives

x = 5(1) = 5

,

y = \frac{4}{1} = 4

, matching

A = (5,4)

from

C

✓

$\therefore$ $A = (5,4)$ , and this is the only point of intersection.

Part (c)

5 marks

For curve $C$ , $\frac{dy}{dt} = 5$ and $\frac{dx}{dt} = 4t$ , so $\frac{dy}{dx} = \frac{5}{4t}$ . At $A$ ( $t = 1$ ), $\frac{dy}{dx} = \frac{5}{4}$ .

The tangent to $C$ at $A(5,4)$ is

y - 4 = \frac{5}{4}(x - 5)

Substituting the parametrisation of $D$ , $x = 5u,\ y = \frac{4}{u}$ ,

\frac{4}{u} - 4 = \frac{5}{4}(5u - 5)

= \frac{25}{4}(u - 1)

Multiplying through by $4u$ ,

16 - 16u = 25u(u - 1)

\Rightarrow 16 - 16u = 25u^{2} - 25u

\Rightarrow 25u^{2} - 9u - 16 = 0

By the quadratic formula (discriminant $\text{=}( - 9)^{2} - 4(25)( - 16) = 1681 = 41^{2}$ ),

u = \frac{9 \pm 41}{50}

= 1\ \text{or}\ - \frac{16}{25}

To check your answer:

u = 1

corresponds to

A

itself (the tangent touches

C

at

A

, and

A

also lies on

D

, so

u = 1

is always a solution here); the “second time” the tangent meets

D

is therefore at the other root,

u = - \frac{16}{25}

✓

At $u = - \frac{16}{25}$ ,

x = 5u = - \frac{16}{5},\quad\quad y = \frac{4}{u} = - \frac{25}{4}

$\therefore$ the tangent meets $D$ a second time at $\left( - \frac{16}{5},\ - \frac{25}{4} \right)$ .

2023 H2 Maths Paper 2, Question 4

Part (a)

Part (b)

Part (c)

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