Functions 2 Question 14

15.

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: 15. ( b, c)

Solution:

  1. 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.



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