Solution — 2021 Question 4 Exponential & Logarithmic Functions| Graphs| Differentiation & Applications| Integration & Applications

(a)

When $y = 0,$

\begin{matrix} 2 - e^{1 - 3 x} & = 0 \\ e^{1 - 3 x} & = 2 \\ \ln e^{1 - 3 x} & = \ln 2 \\ 1 - 3 x & = \ln 2 \\ 3 x & = 1 - \ln 2 \\ x & = \frac{1 - \ln 2}{3} \end{matrix}

Let the $x$ -intercept be $B_{x} .$ $B_{x} = (\frac{1 - \ln 2}{3}, 0) ∎ .$

When $x = 0$ :

\begin{matrix} y & = 2 - e^{1 - 3 (0)} \\ = 2 - e \end{matrix}

Let the $y$ -intercept be $B_{y} .$ $B_{y} = (0, 2 - e) ∎$

(b)

\begin{matrix} y & = 2 - e^{1 - 3 x} \\ \frac{d y}{d x} & = 3 e^{1 - 3 x} \end{matrix}

When $x = \frac{1}{3},$

\begin{matrix} y & = 2 - e^{1 - 3 x} \\ = 2 - e^{1 - 3 (1)} \\ = 2 - e^{- 2} \end{matrix}

\begin{matrix} \frac{d y}{d x} & = 3 e^{1 - 3 (1)} \\ = 3 e^{- 2} \end{matrix}

Equation of tangent:

\begin{matrix} y - (2 - e^{- 2}) & = 3 e^{- 2} (x - 1) \\ y - 2 + e^{- 2} & = 3 e^{- 2} x - 3 e^{- 2} \\ y & = 3 e^{- 2} x - 4 e^{- 2} + 2 ∎ \end{matrix}

(c)

\begin{matrix} \int_{\frac{1}{3}}^{1} (2 - e^{1 - 3 x}) d x \\ = {[2 x + \frac{e^{1 - 3 x}}{3}]}_{\frac{1}{3}}^{1} \\ = 2 (1) + \frac{e^{1 - 3 (1)}}{3} - (2 (\frac{1}{3}) + \frac{e^{1 - 3 (\frac{1}{3})}}{3}) \\ = 1 + \frac{e^{- 2}}{3} \\ = 1 + \frac{1}{3 e^{2}} \end{matrix}

Hence $p = 1 ∎$ and $q = \frac{1}{3} ∎$