Question

Let $A\left(m\right)$ be the area bounded by the curves $y=x^2-3$ and $y=mx+2$, then

• A. The range of $A\left(m\right)$ is $[\frac{10\sqrt{5}}{3},\infty )$.
• B. The range of $A\left(m\right)$ is $[\frac{20\sqrt{5}}{3},\infty )$.
• C. $A\left(m\right)$ is many-one function for $m\in [-2,\infty )$
• D. The area is minimum when $m=1$.

Answer 1

Answer: (B), (C)

The correct options are
B The range of $A\left(m\right)$ is $[\frac{20\sqrt{5}}{3},\infty )$.
C $A\left(m\right)$ is many-one function for $m\in [-2,\infty )$


The line $y=mx+2$ is always passes through $(0,2).$

The point of intersection of the line and the parabola is,
$mx+2=x^2-3\Rightarrow x^2-mx-5=0$
Let the roots of the equation be $a,b$ where
$a>0;b<0$
Now,
$a+b=mab=-5(a-b)=\sqrt{(a+b{)}^2-4ab}=\sqrt{m^2+20}$
Now the area will be,
$A\left(m\right)=\underset{b}{\overset{a}{\int }}mx+2-x^2+3dx={[-\frac{x^3}{3}+m\frac{x^2}{2}+5x]}_b^a=(a-b)[-\left(\frac{a^2+b^2+ab}{3}\right)+\frac{m(a+b)}{2}+5]=\sqrt{m^2+20}[-\frac{1}{3}((a+b{)}^2-ab)+\frac{m^2}{2}+5]=\sqrt{m^2+20}[-\frac{1}{3}(m^2+5)+\frac{m^2}{2}+5]\Rightarrow A\left(m\right)=\frac{(m^2+20{)}^{3/2}}{6}$
Area will me minimum when $m=0$
So the minimum area will be,
$A\left(0\right)=\frac{{20}^{3/2}}{6}=\frac{20\sqrt{5}}{3}$
$A\left(m\right)$ will be many one function in the given domain.