Consider the following relations in the...

MathematicsDomain and Range of a RelationJEE Advanced 1979Moderate

Consider the following relations in the set of real numbers $R$ . $R = {(x, y) : x \in R, y \in R, x^{2} + y^{2} \leq 25}$ $R^{'} = {(x, y) : x \in R, y \in R, y \geq \frac{4}{9} x^{2}}$ Find the domain and range of $R \cap R^{'}$ . Is the relation $R \cap R^{'}$ a function?

Answer:Domain:

{x : x \in R, 16 x^{4} + 81 x^{2} - 2025 \leq 0}

, Range:

{y : y \in R, y \geq \frac{4 x ^{2}}{9}}

Visualized Solution (Hindi)

Visualizing Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Relation $R = {(x, y) : x^{2} + y^{2} \leq 25}$
This represents the interior and boundary of a circle.
Center: $(0, 0)$ , Radius: $5$ .

Visualizing Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R^{'}

Relation $R^{'} = {(x, y) : y \geq \frac{4}{9} x^{2}}$
This represents the region above and on the parabola $y = \frac{4}{9} x^{2}$ .
Vertex: $(0, 0)$ , Opening: Upwards.

Identifying the Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R \cap R^{'}

The intersection $R \cap R^{'}$ is the set of points $(x, y)$ satisfying:
1. $x^{2} + y^{2} \leq 25$ (Inside the circle)
2. $y \geq \frac{4}{9} x^{2}$ (Above the parabola)

Setting up the Equations

Boundary equations: $x^{2} + y^{2} = 25$ and $y = \frac{4}{9} x^{2}$ .
From the parabola: $x^{2} = \frac{9}{4} y$ .
Substitute into the circle equation: $\frac{9}{4} y + y^{2} = 25$ .

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Multiply by $4$ : $4 y^{2} + 9 y - 100 = 0$ .
Factoring: $(4 y + 25) (y - 4) = 0$ .
Possible values: $y = 4$ or $y = - \frac{25}{4}$ .
Since $y \geq 0$ , we take $y = 4$ .

Finding the Domain

Substitute $y = 4$ into $x^{2} = \frac{9}{4} y$ :
$x^{2} = \frac{9}{4} (4) = 9 ⟹ x = \pm 3$ .
The region exists for $x \in [- 3, 3]$ .
Domain: ${x : x \in R, 16 x^{4} + 81 x^{2} - 2025 \leq 0}$ (which simplifies to $∣ x ∣ \leq 3$ ).

Determining the Range

The lowest point is the vertex $(0, 0)$ , so $y_{min} = 0$ .
The highest point is the top of the circle $(0, 5)$ , so $y_{ma x} = 5$ .
Range: ${y : y \in R, 0 \leq y \leq 5}$ .
In terms of the relation: $y \geq \frac{4}{9} x^{2}$ restricted by $x^{2} + y^{2} \leq 25$ .

Is it a Function?

A relation is a function if each $x$ has exactly one $y$ .
Applying the Vertical Line Test: A vertical line (e.g., at $x = 0$ ) intersects the region at multiple points ( $y \in [0, 5]$ ).
Since one $x$ maps to multiple $y$ values, the relation is NOT a function.

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