If the area bounded by \[y = {x^2} + 2x - 3\]and the line \[y = kx + 1\]is least, then its least area is
(a)$\dfrac{{64}}{3}sq.units$
(b) $\dfrac{{32}}{3}sq.units$
(c)$\dfrac{{25}}{3}sq.units$
(d)None of the above

Hint: Find the points of intersection of the given curves and integrate to find the area bounded by the curves. Then, substitute the least possible value for $k$to get the least area.
The equations given are
\[y = {x^2} + 2x - 3\] …(1)
\[y = kx + 1\] …(2)
Put (2) in (1),
\[kx + 1 = {x^2} + 2x - 3\]
\[{x^2} + \left( {2 - k} \right)x - 4 = 0\] …(3)
Let $x_1$and $x_2$be the roots of equation (3)
From (3),
$x_1 + x_2 = - \left( {2 - k} \right) = k - 2$ …(4)
$x_1 \times x_2 = - 4$ …(5)
Required area\[ = \int\limits_{x_1}^{x_2} {\left[ {\left( {kx + 1} \right) - \left( {{x^2} + 2x - 3} \right)} \right]dx} \]
$
   = \int\limits_{x_1}^{x_2} {\left[ {kx - 2x - {x^2} + 4} \right]dx} \\
   = \left[ {\left( {k - 2} \right)\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} + 4x} \right]_{x_1}^{x_2} \\
   = \left[ {\left( {k - 2} \right)\dfrac{{x_2{^2}}}{2} - \dfrac{{x_2{^3}}}{3} + 4\left( {x_2} \right) - \left\{ {\left( {k - 2} \right)\dfrac{{x_1{^2}}}{2} - \dfrac{{x_1{^3}}}{3} + 4\left( {x_1} \right)} \right\}} \right] \\
   = \left[ {\left( {k - 2} \right)\dfrac{{x_2{^2}}}{2} - \left( {k - 2} \right)\dfrac{{x_1{^2}}}{2} - \dfrac{{x_2{^3}}}{3} + \dfrac{{x_1{^3}}}{3} + 4x_2 + 4x_1} \right] \\
   = \left[ {\dfrac{{\left( {k - 2} \right)}}{2}\left( {x_2{^2} - x_1{^2}} \right) - \dfrac{1}{3}\left( {x_2{^3} - x_1{^3}} \right) + 4\left( {x_2 - x_1} \right)} \right] \\
$
We know that $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$and $\left( {{a^3} - {b^3}} \right) = {\left( {a - b} \right)^3} + 3ab\left( {a - b} \right)$
Using these, we get
\[ = \left[ {\dfrac{{\left( {k - 2} \right)}}{2}\left( {x_2 + x_1} \right)\left( {x_2 - x_1} \right) - \dfrac{1}{3}\left\{ {{{\left( {x_2 - x_1} \right)}^3} + 3x_1x_2\left( {x_2 - x_1} \right)} \right\} + 4\left( {x_2 - x_1} \right)} \right]\]
Using equations (4) and (5) and taking $\left( {x_2 - x_1} \right)$ common out, we get
$
   = \left[ {\dfrac{{\left( {k - 2} \right)}}{2}\left( {x_2 + x_1} \right)\left( {x_2 - x_1} \right) - \dfrac{1}{3}\left\{ {{{\left( {x_2 - x_1} \right)}^3} + 3x_1x_2\left( {x_2 - x_1} \right)} \right\} + 4\left( {x_2 - x_1} \right)} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{1}{3}\left\{ {{{\left( {x_2 - x_1} \right)}^2} + 3x_1x_2} \right\} + 4} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{1}{3}\left\{ {{{\left( {x_2 - x_1} \right)}^2} + 3x_1x_2} \right\} + 4} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{1}{3}\left\{ {{{\left( {x_2 - x_1} \right)}^2} - 4x_1x_2 + 3x_1x_2} \right\} + 4} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{1}{3}\left\{ {{{\left( {k - 2} \right)}^2} + 16 - 12} \right\} + 4} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{{{{\left( {k - 2} \right)}^2}}}{3} - \dfrac{4}{3} + 4} \right] \\
   = \left( {x_2 - x_1} \right)\left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{{{{\left( {k - 2} \right)}^2}}}{3} + \dfrac{8}{3}} \right] \\
$
We know that \[{\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab,\left( {a - b} \right) = \sqrt {{{\left( {a + b} \right)}^2} - 4ab} \]

$
   = \sqrt {{{(k - 2)}^2} + 16} \left[ {\dfrac{{{{\left( {k - 2} \right)}^2}}}{2} - \dfrac{{{{\left( {k - 2} \right)}^2}}}{3} + \dfrac{8}{3}} \right] \\
   = \sqrt {{{\left( {k - 2} \right)}^2} + 16} \left[ {\dfrac{{{{\left( {k - 2} \right)}^2} + 16}}{6}} \right] \\
   = \dfrac{1}{6}{\left[ {{{\left( {k - 2} \right)}^2} + 16} \right]^{\dfrac{1}{2} + 1}} \\
   = \dfrac{1}{6}{\left[ {{{\left( {k - 2} \right)}^2} + 16} \right]^{\dfrac{3}{2}}} \\
$
To get the least area, we must substitute $k = 2$in order for that term to get cancelled in the required area equation.
Required area when $k = 2$\[ = \dfrac{1}{6}{\left[ {{{\left( {2 - 2} \right)}^2} + 16} \right]^{\dfrac{3}{2}}}\]
$
   = \dfrac{1}{6}\left( {\sqrt {{{16}^3}} } \right) \\
   = \dfrac{{64}}{6} \\
   = \dfrac{{32}}{3}sq.units \\
$
Hence, the correct answer is option (b)

Note: From the two given equations, we get a single equation and find the sum of the roots and product of the roots in the form of two equations to be substituted in the later part of the solution. Then the integration is done to find the required bounded area and the unknown values are substituted. The least possible value for $k$ is substituted to get the least possible area.