Solution — 2023 Question 4 Graphs| Integration & Applications

(a)

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

When $x = 3,$

\begin{matrix} y & = \ln (1 + x^{2}) \\ = \ln (1 + 3^{2}) \\ = \ln 10 \end{matrix}

\begin{matrix} \frac{d y}{d x} & = \frac{2 (3)}{1 + 3^{2}} \\ = \frac{3}{5} \end{matrix}

Equation of tangent:

\begin{matrix} y - \ln 10 & = \frac{3}{5} (x - 3) \\ = \frac{3}{5} x - \frac{9}{5} \\ \frac{3}{5} x - y & = - \ln 10 + \frac{9}{5} \\ 3 x - 5 y & = - 5 \ln 10 + 9 ∎ \end{matrix}

(b)

Using GC, the tangent meets $D$ again at

$(- 0.485, 0.211) ∎$

(c)

where $A = (3 - \frac{5 \ln 10}{3}, 0)$ and $B = (0, \ln 10 - \frac{9}{5}) .$

(d)

\begin{matrix} Area \\ = \int_{0}^{3} \ln (1 + x^{2}) d x \\ = 3.41 ∎ (to 3 s.f.) \end{matrix}