If the area bounded by y = x² + 2x – 3 and the y = kx + 1 is the least and is equal to A, then

$\text{ Let }{x}_1\text{ and }{x}_2\text{ be the roots of the equation }$

$\begin{array}{r}{x}^2+2x-3=kx+1\\ \Rightarrow x^2+(2-k)x-4=0\\ \Rightarrow x_1+x_2=k-2,x_1{x}_2=-4\end{array}$

$$\begin{gathered} \text{ Now, }A={\int }_{x_1}^{x_2} \left[(kx+1)-\left(x^2+2x-3\right)\right]dx \\ \begin{array}{c}={\left[(\mathrm{k}-2)\frac{x^2}{2}-\frac{x^3}{3}+4x\right]}_{x_1}^{x_2}\\ =\left[(\mathrm{k}-2)\frac{x_2^2-x_1^2}{2}-\frac{1}{3}\left(x_2^3-x_1^3\right)+4\left(x_2-x_1\right)\right]\\ =\left(x_2-x_1\right)\left[\frac{(k-2{)}^2}{2}-\frac{1}{3}\left({\left(x_2+x_1\right)}^2-x_1{x}_2\right)+4\right]\\ =\sqrt{{\left(x_2+x_1\right)}^2-4x_1{x}_2}\left[\frac{(k-2{)}^2}{2}-\frac{1}{3}\left((k-2{)}^2+4\right)+4\right]\\ =\frac{\sqrt{(k-2{)}^2+16}}{6}\left[\frac{1}{6}(k-2{)}^2+\frac{16}{3}\right]\\ =\frac{{\left[(\mathrm{k}-2{)}^2+16\right]}^{3/2}}{6}.\end{array} \end{gathered}$$

$\Rightarrow \mathrm{A}\text{ is minimum if }\mathrm{k}=2\text{ and }{\mathrm{A}}_{min}=\frac{32}{3}\text{. }$