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Question

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


A
g(x)=±1x2
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B
g(x)=1x2
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C
g(x)=1x2
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D
g(x)=1+x2
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Solution

The correct option is A $$\displaystyle g(x)=\pm \sqrt{1-x^{2}}$$
Since it is an equilateral triangle, then
Side $$a=\sqrt{(x-0)^{2}+(g(x)-0)^{2}}$$
Now, Area $$=\dfrac{\sqrt{3}a^{2}}{4}$$
Hence, $$a^{2}=1$$
Thus,
$$1=x^{2}+g(x)^{2}$$
$$\Rightarrow g(x)=\pm\sqrt{1-x^{2}}$$

Mathematics

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