Question

If $g\left(x\right)$ be a function defined on $[-1,1]$. If the area of the equilateral triangle with two of its vertices at $(0,0)$ and $[x,g(x\left)\right]$ is $\frac{\sqrt{3}}{4}$, then the function is

• A. $g\left(x\right)=\pm \sqrt{1-x^2}$
• B. $g\left(x\right)=-\sqrt{1-x^2}$
• C. $g\left(x\right)=\sqrt{1-x^2}$
• D. $g\left(x\right)=\sqrt{1+x^2}$

Answer 1

The correct option is A $g\left(x\right)=\pm \sqrt{1-x^2}$

Since it is an equilateral triangle, then
Side $a=\sqrt{(x-0{)}^2+(g\left(x\right)-0{)}^2}$
Now, Area $=\frac{\sqrt{3}{a}^2}{4}$
Hence, $a^2=1$
Thus,
$1=x^2+g(x{)}^2$
$\Rightarrow g\left(x\right)=\pm \sqrt{1-x^2}$