Let g(x) be a function defined on [-1, 1]. If the area of the equilateral triangle with two of its v

Q: Wait, why do we use the distance formula here? Can't we use vectors?

You absolutely can! Using vectors is a powerful alternative. The distance $s$ is simply the magnitude of the position vector $\vec{r} = x\hat{i} + g(x)\hat{j}$. The magnitude squared is $|\vec{r}|^2 = x^2 + [g(x)]^2$. It leads to the same result, just a different perspective. Both methods are perfectly valid in JEE.

Q: Why is the answer $\pm \sqrt{1-x^2}$ and not just $\sqrt{1-x^2}$?

Great catch. When you solve the equation $[g(x)]^2 = k$, the solution is $g(x) = \pm \sqrt{k}$. Geometrically, the point $(x, g(x))$ could be above or below the x-axis while maintaining the same distance from the origin. Both positions satisfy the condition of forming an equilateral triangle of the same area.

Q: Is there any restriction on $x$?

Yes, the problem defines $g(x)$ on $[-1, 1]$. For $g(x) = \pm \sqrt{1-x^2}$ to be a real-valued function, the term inside the square root must be non-negative, i.e., $1 - x^2 \ge 0$, which implies $x^2 \le 1$, or $x \in [-1, 1]$. This perfectly matches the given domain.

Question

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 $\frac{\sqrt{3}}{4}$, then the function $g(x)$ is

Options:

(A) $g(x) = \pm \sqrt{1 - x^2}$

(B) $g(x) = \sqrt{1 - x^2}$

(C) $g(x) = -\sqrt{1 - x^2}$

(D) $g(x) = \sqrt{1 + x^2}$

Sigma Percentile · Accepted Answer

### The Geometry of the Hidden Circle

Imagine you are standing on the Cartesian plane. You have a point fixed at the origin $(0,0)$, and another point, a wanderer, at $(x, g(x))$.

These two points are not just random coordinates; they are the anchors of an equilateral triangle. This is the beauty of coordinate geometry—it transforms a simple shape into a rigorous algebraic constraint.

### The Bridge: Distance and Side Length

Since this is an equilateral triangle, all its sides must be equal. Let us call this side length $s$.

The distance between our two points, the origin and $(x, g(x))$, is exactly $s$. We can express this distance using the fundamental distance formula:

$$s = \sqrt{(x-0)^2 + (g(x)-0)^2}$$

To make our lives easier, let us square both sides. This gives us the square of the side length:

$$s^2 = x^2 + [g(x)]^2$$

This equation is our bridge. It connects the geometric property of the triangle to the algebraic function $g(x)$ we are trying to uncover.

### The Constraint: Area as the Key

Now, we bring in the information provided by the problem. We are told the area of this equilateral triangle is $\frac{\sqrt{3}}{4}$.

The standard formula for the area of an equilateral triangle with side length $s$ is:

$$	ext{Area} = \frac{\sqrt{3}}{4} s^2$$

By equating our known area to this formula, we get:

$$\frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4}$$

This is the moment of clarity. The $\frac{\sqrt{3}}{4}$ on both sides cancels out perfectly, leaving us with the elegant result:

$$s^2 = 1$$

### The Final Resolution

Now, we simply substitute our expression for $s^2$ back into this result:

$$x^2 + [g(x)]^2 = 1$$

Our goal is to isolate $g(x)$. Subtracting $x^2$ from both sides, we find:

$$[g(x)]^2 = 1 - x^2$$

Finally, taking the square root of both sides, we arrive at the solution:

$$g(x) = \pm \sqrt{1 - x^2}$$

This result is more than just an algebraic expression; it is the equation of a **unit circle centered at the origin**. The point $(x, g(x))$ is constrained to move along the boundary of this circle, ensuring that its distance from the origin is always $1$.

This is the hidden geometric reality behind the function. You have successfully navigated the problem by linking geometry, algebra, and the fundamental properties of shapes.

Animated Solution for Mathematics - Functions: 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 $\frac{3}{4}$ , then the function $g (x)$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Side Length

Distance Formula

Area of Equilateral Triangle

Equating the Areas

Simplifying the Equation

Isolating $g (x)$

Solving for $g (x)$

Geometric Interpretation

The Sigma Insight: Classification of Functions

The Geometry of the Hidden Circle

The Bridge

The Constraint

The Final Resolution

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Select Answer:

Visualized Solution

Visualizing the Setup

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Distance Formula

Area of Equilateral Triangle

Equating the Areas

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Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

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