Solution — 2017 Question 4 Graphs| Differentiation & Applications

(a)

(b)

\begin{matrix} y & = \ln (4 x - 5) \\ \frac{d y}{d x} & = \frac{4}{4 x - 5} \end{matrix}

When $x = 2.5,$

\begin{matrix} y & = \ln (4 x - 5) \\ = \ln (4 (2.5) - 5) \\ = \ln 5 \end{matrix}

\begin{matrix} \frac{d y}{d x} & = \frac{4}{4 (2.5) - 5} \\ = 0.8 \end{matrix}

Equation of tangent:

\begin{matrix} y - \ln 5 & = 0.8 (x - 2.5) \\ = 0.8 x - 2 \\ y - 0.8 x & = - 2 + \ln 5 \\ 5 y - 4 x & = - 10 + 5 \ln 5 ∎ \end{matrix}

(c)

Using GC, the coordinates of $P$ and $Q$ are

$P (0.488 20, 0) Q (0, - 0.390 56)$

Using Pythagoras’ Theorem/the coordinate geometry length formula,

\begin{matrix} Length of P Q \\ = \sqrt{{(0.488 20 - 0)}^{2} + {(- 0.390 56 - 0)}^{2}} \\ = 0.625 ∎ (to 3 s.f.) \end{matrix}