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Question: If the area bounded by \[y = {x^2} + 2x - 3\]and the line \[y = kx + 1\]is least, then its least are...

If the area bounded by y=x2+2x−3y = {x^2} + 2x - 3and the line y=kx+1y = kx + 1is least, then its least area is
(a)643sq.units\dfrac{{64}}{3}sq.units
(b) 323sq.units\dfrac{{32}}{3}sq.units
(c)253sq.units\dfrac{{25}}{3}sq.units
(d)None of the above

Explanation

Solution

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 kkto get the least area.
The equations given are
y=x2+2x−3y = {x^2} + 2x - 3 …(1)
y=kx+1y = kx + 1 …(2)
Put (2) in (1),
kx+1=x2+2x−3kx + 1 = {x^2} + 2x - 3
x2+(2−k)x−4=0{x^2} + \left( {2 - k} \right)x - 4 = 0 …(3)
Let x1x_1and x2x_2be the roots of equation (3)
From (3),
x1+x2=−(2−k)=k−2x_1 + x_2 = - \left( {2 - k} \right) = k - 2 …(4)
x1×x2=−4x_1 \times x_2 = - 4 …(5)
Required area=∫x1x2[(kx+1)−(x2+2x−3)]dx= \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 (a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)and (a3−b3)=(a−b)3+3ab(a−b)\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 (x2−x1)\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 (a−b)2=(a+b)2−4ab,(a−b)=(a+b)2−4ab{\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab,\left( {a - b} \right) = \sqrt {{{\left( {a + b} \right)}^2} - 4ab}

=(k−2)2+16[(k−2)22−(k−2)23+83] =(k−2)2+16[(k−2)2+166] =16[(k−2)2+16]12+1 =16[(k−2)2+16]32  = \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=2k = 2in 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 kk is substituted to get the least possible area.