\begin{align*} y & = e^{2x} - 3 \\ {dy \over dx} & = 2 e^{2x} \phantom{000000} \left[ {d \over dx} [ e^{f(x)}] = f'(x) e^{f(x)} \right] \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = e^{2(1)} - 3 \\ y & = e^2 - 3 \\ \\ \therefore & \phantom{.} A(1, e^2 - 3) \\ \\ \text{Substitute } & x = 1 \text{ into } {dy \over dx}, \\ {dy \over dx} & = 2 e^{2(1)} \\ & = 2 e^2 \\ \\ y & = mx + c \\ y & = 2e^2 x + c \\ \\ \text{Using } & A(1, e^2 - 3), \\ e^2 - 3 & = 2 e^2 (1) + c \\ e^2 - 3 & = 2 e^2 + c \\ - e^2 - 3 & = c \\ \\ \therefore y & = 2e^2 x - e^2 - 3 \end{align*}

\begin{align*} \text{Gradient of normal} & = -1 \div 2e^2 \phantom{000000} [m_1 \times m_2 = -1] \\ & = - {1 \over 2e^2} \\ \\ y & = mx + c \\ y & = - {1 \over 2e^2} x + c \\ \\ \text{Using } & A(1, e^2 - 3), \\ e^2 - 3 & = -{1 \over 2e^2} (1) + c \\ e^2 - 3 & = - {1 \over 2e^2} + c \\ e^2 + {1 \over 2e^2} - 3 & = c \\ \\ \therefore y & = -{1 \over 2e^2}x + e^2 + {1 \over 2e^2} - 3 \end{align*}

\begin{align*} y & = \ln (x^2 + 1) \\ \\ {dy \over dx} & = { 2x \over x^2 + 1} \phantom{000000} \left[ {d \over dx} [ \ln f(x) ] = {f'(x) \over f(x)} \right] \\ \\ \text{Line: } y - x & = 2 \\ y & = x + 2 \phantom{000000000} [y = mx + c] \\ \\ \text{Gradient of line} & = 1 \\ \\ \text{Let } & {dy \over dx} = 1, \\ 1 & = {2x \over x^2 + 1} \\ x^2 + 1 & = 2x \\ x^2 - 2x + 1 & = 0 \\ (x - 1)(x - 1) & = 0 \\ (x - 1)^2 & = 0 \\ x - 1 & = 0 \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = \ln (1^2 + 1) \\ y & = \ln 2 \\ \\ \therefore & \phantom{.} (1, \ln 2) \end{align*}

\begin{align*} x \text{-axis is a } & \text{horizontal line with gradient} = 0 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {2x \over x^2 + 1} \\ 0 & = 2x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = \ln (0^2 + 1) \\ y & = 0 \\ \\ \therefore & \phantom{.} (0, 0) \end{align*}

\begin{align*} \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = -(2)^2 + 3(2) - 5 \\ y & = -3 \\ \\ \therefore & \phantom{.} P(2, -3) \\ \\ {dy \over dx} & = -2x + 3 \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = -2(2) + 3 \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & P(2, -3), \\ -3 & = -2 + c \\ -1 & = c \\ \\ \text{Eqn of tangent at } P : & \phantom{0} y = -x - 1 \\ \\ \implies & \phantom{.} B(0, -1) \\ \\ \text{Let } & y = 0, \\ 0 & = -x - 1 \\ x & = -1 \\ \\ \therefore & \phantom{.} A(-1, 0) \\ \\ \text{Area of triangle } AOB & = {1 \over 2} \times OA \times OB \\ & = {1 \over 2} \times 1 \times 1 \\ & = 0.5 \text{ unit}^2 \end{align*}