Question

If the area bounded by $y=x^2+2x=3$ and the line $y=kx+1$ is the least, then:

• A. $k=\frac{3}{2}$
• B. $k=2$
• C. the least area of the bounded region is $\frac{32}{3}$
• D. the least area of the bounded region is $\frac{27}{2}$

Answer 1

Answer: (B), (C)

$k=2$
the least area of the bounded region is $\frac{32}{3}$

$\begin{array}{cc}y=x^2+2x-3\text{(given)}& -\left(i\right)\\ y=kx+1\left(\text{given}\right).& \text{(ii).}\\ x^2+2x-3=kx+1& \\ \Rightarrow x^2+(2-k)x-4=0& \end{array}$

let $x_1$ and $x_2$ are the roots of equation

$x_1+x_2=k-2,x_1{x}_2=-4$

$A={\int }_{x_1}^{x_2}\left[\right(kx+1)-(x^2+2x-3)]dx$

$A={\left[\right(k-2)\frac{x^2}{2}-\frac{x^3}{3}+4x]}_{x_1}^{x_2}$

$\Rightarrow \left[\right(k-2)\cdot \frac{{\left(x_2\right)}^2}{2}-\frac{{\left(x_2\right)}^3}{3}+4x_2-\left(\right(k-2)\cdot \frac{{\left(x_1\right)}^2}{2}-\frac{\left(x_1^3\right).}{+}4x_1]$

$=\left[\right(k-2)\frac{(x_2^2-x_1^2)}{2}-\frac{1}{3}(x_2^3-x_1^3)+4(x_2-x_1)]$

$=\sqrt{{(x_2+x_1)}^2-4x_1{x}_2}[\frac{(k-2{)}^2}{2}-\frac{1}{3}\left(\right(k-2{)}^2+4)+4]$

$\frac{\sqrt{(k-2{)}^2-16}}{6}\left[\frac{1}{6}\right(k-2{)}^2+\frac{8}{3}]$

$\Rightarrow \frac{{\left[\right(k-2{)}^2+16]}^{3/2}}{6}$

At least $k=2$

$\frac{[16{]}^{3/2}}{6}=\frac{32}{3}$ squnits.

Correct option are (B)(C).