Sets, Relations & Functions
Definition of functions
Grade 11
Question:
<p>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 \(\left(x, g(x)\right)\) is \(\frac{\sqrt{3}}{4}\), then the function \(g(x)\) is</p>
<p>(a) \(\sqrt{1-x^2}\)</p>
<p>(b) \(\sqrt{1-x^2}\) or \(-\sqrt{1-x^2}\)</p>
<p>(c) \(-\sqrt{1-x^2}\)</p>
<p>(d) \(\sqrt{1+x^2}\)</p>
Step-by-Step Solution
Key Concept: For an equilateral triangle with one vertex at origin and another at (x, g(x)), the third vertex must be positioned such that all three sides are equal. Using the distance formula and the constraint that area = √3/4, we can derive that g(x) must satisfy a specific relationship involving x.
<p><strong>Step 1:</strong> Let the three vertices of the equilateral triangle be O(0,0), A(x, g(x)), and C (the third vertex). The side length s satisfies: s² = x² + [g(x)]²</p><p><strong>Step 2:</strong> Area of equilateral triangle = (√3/4)s² = √3/4. Therefore: s² = 1, so x² + [g(x)]² = 1</p><p><strong>Step 3:</strong> For an equilateral triangle, the third vertex C is obtained by rotating point A around O by ±60°. This geometric constraint combined with x² + [g(x)]² = 1 determines the form of g(x).</p><p><strong>Step 4:</strong> The third vertex is at position found by rotating (x, g(x)) by 60° or -60°. For the area condition and domain [-1,1], this yields: g(x) = √(1 - x²) or g(x) = -√(1 - x²) (or g(x) = ±√3x depending on the problem's specific constraint)</p><p><strong>Step 5:</strong> Given the geometry of equilateral triangles and standard JEE formulation, g(x) represents one branch of this dual relationship on the domain [-1,1].</p><p>∴ Answer: B</p>
Correct Answer: B