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1.Relation and Function
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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 :-
A
$\sqrt {1+x^2}$
B
$-\sqrt {1+x^2}$
C
$\sqrt {1-x^2}$ or $-\sqrt {1-x^2}$
D
None of these
Solution
since it is an equilateral triangle, then Side $a=\sqrt{(x-0)^{2}+(g(x)-0)^{2}}$
Now, Area $=\frac{\sqrt{3} a^{2}}{4}$
Hence, $a^{2}=1$
Thus,
$1=x^{2}+g(x)^{2}$
$\Rightarrow g(x)=\pm \sqrt{1-x^{2}}$
Standard 12
Mathematics
