Solution — 2018 Question 4 Graphs| Differentiation & Applications| Integration & Applications

(a)

\begin{matrix} y & = x + e^{1 - 2 x} \\ \frac{d y}{d x} & = 1 + (- 2) e^{1 - 2 x} \\ = 1 - 2 e^{1 - 2 x} \end{matrix}

At the turning point of $C,$ $\frac{d y}{d x} = 0$

\begin{matrix} 1 - 2 e^{1 - 2 x} & = 0 \\ 2 e^{1 - 2 x} & = 1 \\ e^{1 - 2 x} & = \frac{1}{2} \\ 1 - 2 x & = \ln \frac{1}{2} \\ 2 x & = 1 - \ln \frac{1}{2} \\ x & = \frac{1 - \ln \frac{1}{2}}{2} \end{matrix}

When $x = \frac{1 - \ln \frac{1}{2}}{2},$

\begin{matrix} y & = \frac{1 - \ln \frac{1}{2}}{2} + e^{1 - 2 (\frac{1 - \ln \frac{1}{2}}{2})} \\ = \frac{1 - \ln \frac{1}{2}}{2} + e^{1 - (1 - \ln \frac{1}{2})} \\ = \frac{1 - \ln \frac{1}{2}}{2} + e^{\ln \frac{1}{2}} \\ = \frac{1 - \ln \frac{1}{2}}{2} + \frac{1}{2} \\ = \frac{1 - \ln \frac{1}{2} + 1}{2} \\ = \frac{2 - \ln \frac{1}{2}}{2} \end{matrix}

Hence the coordinates of the turning point are

$(\frac{1 - \ln \frac{1}{2}}{2}, \frac{2 - \ln \frac{1}{2}}{2}) ∎$

(b)

(c)

When $x = 2,$

\begin{matrix} y & = 2 + e^{1 - 2 (2)} \\ = 2.0498 \end{matrix}

\begin{matrix} \frac{d y}{d x} & = 1 - 2 e^{1 - 2 (2)} \\ = 0.900 43 \end{matrix}

Equation of tangent:

\begin{matrix} y - 2.0498 & = 0.900 43 (x - 2) \\ = 0.900 43 x - 1.8009 \\ y & = 0.900 43 x + 0.248 94 \\ = 0.900 x + 0.249 ∎ \end{matrix}

(d)

\begin{matrix} \int y d x \\ = \int (x + e^{1 - 2 x}) d x \\ = (\frac{x^{2}}{2}) + \frac{e^{1 - 2 x}}{- 2} + c \\ = \frac{x^{2}}{2} - \frac{e^{1 - 2 x}}{2} + c ∎ \end{matrix}

\begin{matrix} Area \\ = \int_{0}^{1} (x + e^{1 - 2 x}) d x \\ = {[\frac{x^{2} - e^{1 - 2 x}}{2}]}_{0}^{1} \\ = \frac{1^{2} - e^{1 - 2 (1)}}{2} - \frac{0^{2} - e^{1 - 2 (0)}}{2} \\ = \frac{1 - e^{- 1} + e}{2} ∎ \end{matrix}