2024 H2 Mathematics Paper 1 Question 6

Differentiation I: Tangents and Normals, Parametric Curves

Answers

$p = 6,$ $q = - 4,$ $r = 1.$

$a = 4 e .$

$5.4 {units}^{2} .$

(a)

\begin{aligned} y & = 2 e^{x^{3}} + a \\ \frac{d y}{d x} & = 6 x^{2} e^{x^{3}} \end{aligned}

When $x = 1,$

\begin{aligned} y & = 2 e + a \\ \frac{d y}{d x} & = 6 e \end{aligned}

Equation of tangent:

\begin{matrix} y - (2 e + a) = 6 e (x - 1) \\ y = 6 e (x - 1) + 2 e + a \\ y = 6 e x - 4 e + a \\ y = e (6 x - 4) + a ∎ \end{matrix}

(i)

Since the tangent to $C$ at $T$ passes through the origin,

\begin{matrix} 0 = - 4 e + a \\ a = 4 e ∎ \end{matrix}

(ii)

Since $a = 4 e,$ coordinates of $T = (1, 6 e)$

graph

\begin{aligned} Area required \\ = \int_{0}^{1} (2 e^{x^{3}} + 4 e) d x - area of triangle \\ = \int_{0}^{1} (2 e^{x^{3}} + 4 e) d x - (\frac{1}{2} (1) (6 e)) \\ = 5.4 {units}^{2} ∎ \end{aligned}