Functions Ques 41
Let $g(x)$ be a function defined on $[-1,1]$. If the area of the equilateral triangle with two of its vertices at $(0,0)$ and $[x, g(x)]$ is $\sqrt{3} / 4$, then the function $g(x)$ is
(1989, 2M)
(a) $g(x)= \pm \sqrt{1-x^{2}}$
(b) $g(x)=\sqrt{1-x^{2}}$
(c) $g(x)=-\sqrt{1-x^{2}}$
(d) $g(x)=\sqrt{1+x^{2}}$
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Answer:
Correct Answer: 41.( b, c)
Solution:
Formula:
- Since, area of equilateral triangle $=\frac{\sqrt{3}}{4}(B C)^{2}$
$\Rightarrow \quad \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4} \cdot\left[x^{2}+g^{2}(x)\right] \Rightarrow g^{2}(x)=1-x^{2}$
$\Rightarrow \quad g(x)=\sqrt{1-x^{2}}$ or $-\sqrt{1-x^{2}}$
Hence, (b) and (c) are the correct options.
BETA
