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
B
g(x)=1x2
C
g(x)=1x2
D
g(x)=1+x2

Solution

## The correct option is A $$\displaystyle g(x)=\pm \sqrt{1-x^{2}}$$Since it is an equilateral triangle, thenSide $$a=\sqrt{(x-0)^{2}+(g(x)-0)^{2}}$$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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More