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(C) Let the vertices of the equilateral triangle be $A(0, 0)$,$B(x, g(x))$,and $C$. Since the triangle is equilateral,the side length $a$ is the dista

Vedclass · Accepted Answer

$\sqrt{1-x^2}$ or $-\sqrt{1-x^2}$

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(C) Let the vertices of the equilateral triangle be $A(0, 0)$,$B(x, g(x))$,and $C$. Since the triangle is equilateral,the side length $a$ is the distance between $A$ and $B$. $a = \sqrt{(x-0)^2 + (g(x)-0)^2} = \sqrt{x^2 + g(x)^2}$. The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$. Given the area is $\frac{\sqrt{3}}{4}$,we have $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4}$. This implies $a^2 = 1$. Substituting the expression for $a^2$,we get $x^2 + g(x)^2 = 1$. Solving for $g(x)$,we get $g(x)^2 = 1 - x^2$. Therefore,$g(x) = \pm \sqrt{1 - x^2}$.

If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt{3}}{4}$,then $g(x)$ equals :-

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