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

$ ParseError: Can't use function '$' in math mode at position 1: $̲ P( 0.488,20, 0 ) \quad Q( 0, -0.390,56 )

Using Pythagoras’ Theorem/the coordinate geometry length formula,

\begin{align*} & \text{ Length of } PQ \\ &= \sqrt{\left( 0.488\,20 - 0 \right)^2 + \left( -0.390\,56 - 0 \right)^2} \\ &= 0.625\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.