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 (English)

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.

00:00 / 00:00

Conceptually Similar Problems

MathematicsDomain and Range of a FunctionJEE Advanced 1978Easy

View in:English Hindi

Find the domain and range of the function $f (x) = \frac{x ^{2}}{1 + x ^{2}}$ . Is the function one-to-one?

Answer:Domain: R, Range: [0, 1)

MathematicsTypes of RelationsJEE Main 2010Easy

View in:English Hindi

Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = {( $\frac{m}{n}, \frac{p}{q}$ ) | m, n, p and q are integers such that $n, q \neq = 0$ and qm = pn}. Then

(A)

Neither R nor S is an equivalence relation

(B)

S is an equivalence relation but R is not an equivalence relation

(C)

R and S both are equivalence relations

(D)

R is an equivalence relation but S is not an equivalence relation

Answer:B

MathematicsDomain and Range of a FunctionJEE Advanced 1985Easy

View in:English Hindi

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

Answer:(-2, 1), [-1, 1]

MathematicsTypes of RelationsJEE Main 2004Easy

View in:English Hindi

Let $R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}$ be a relation on the set $A = {1, 2, 3, 4}$ . The relation R is

(A)

reflexive

(B)

transitive

(C)

not symmetric

(D)

a function

Answer:C

MathematicsTypes of RelationsJEE Main 2008Easy

View in:English Hindi

Let $R$ be the real line. Consider the following subsets of the plane $R \times R$ : $S = {(x, y) : y = x + 1 and 0 < x < 2}$ , $T = {(x, y) : x - y is an integer}$ . Which one of the following is true?

(A)

Neither S nor T is an equivalence relation on R

(B)

Both S and T are equivalence relation on R

(C)

S is an equivalence relation on R but T is not

(D)

T is an equivalence relation on R but S is not

Answer:D

MathematicsDifferentiability of a FunctionJEE Advanced 1988Easy

View in:English Hindi

Let $R$ be the set of real numbers and $f : R \to R$ be such that for all $x$ and $y$ in $R$ $∣ f (x) - f (y) ∣ \leq ∣ x - y ∣^{2}$ . Prove that $f (x)$ is a constant.

Answer:

f (x) is constant

MathematicsClassification of FunctionsJEE Advanced 1979Easy

View in:English Hindi

Let $R$ be the set of real numbers. If $f : R \to R$ is a function defined by $f (x) = x^{2}$ , then $f$ is :

(A)

Injective but not surjective

(B)

Surjective but not injective

(C)

Bijective

(D)

None of these

Answer:D

MathematicsTypes of RelationsJEE Main 2011Easy

View in:English Hindi

Let R be the set of real numbers. Statement-1: $A = {(x, y) \in R \times R : y - x is an integer}$ is an equivalence relation on R. Statement-2: $B = {(x, y) \in R \times R : x = α y for some rational number α}$ is an equivalence relation on R.

(A)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B)

Statement-1 is true, Statement-2 is false.

(C)

Statement-1 is false, Statement-2 is true.

(D)

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Answer:B

MathematicsMonotonicityJEE Advanced 2011Difficult

View in:English Hindi

Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

List-I

(P)

The set

{Re (\frac{2 i z}{1 - z ^{2}}) : z is a complex number, ∣ z ∣ = 1, z \neq = \pm 1}

(Q)

The domain of the function

f (x) = sin^{- 1} (\frac{8 ( 3 ) ^{x - 2}}{1 - 3 ^{2 (x - 1)}})

(R)

f (θ) = 1 - tan θ - 1 tan θ 1 - tan θ 1 tan θ 1

, then the set

{f (θ) : 0 \leq θ < \frac{π}{2}}

(S)

f (x) = x^{3/2} (3 x - 10), x \geq 0

then

f (x)

is increasing in

List-II

(1)

(- \infty, - 1) \cup (1, \infty)

(2)

(- \infty, - 0) \cup (0, \infty)

(3)

[2, \infty)

(4)

(- \infty, - 1] \cup [1, \infty)

(5)

(0, 0] \cup [2, \infty)

Answer:P → 4, Q → 5, R → 3, S → 3

MathematicsClassification of FunctionsJEE Advanced 2014Moderate

View in:English Hindi

Let $f_{1} : R \to R, f_{2} : [0, \infty) \to R, f_{3} : R \to R$ and $f_{4} : R \to [0, \infty)$ be defined by $f_{1} (x) = {∣ x ∣ e^{x} if x < 0, if x \geq 0;$ $f_{2} (x) = x^{2}$ ; $f_{3} (x) = {sin x x if x < 0, if x \geq 0;$ and $f_{4} (x) = {f_{2} (f_{1} (x)) f_{2} (f_{1} (x)) - 1 if x < 0, if x \geq 0.$

List-I

(P)

f_{4}

(Q)

f_{3}

(R)

f_{2} \circ f_{1}

(S)

f_{2}

List-II

(1)

Onto but not one-one

(2)

Neither continuous nor one-one

(3)

Differentiable but not one-one

(4)

Continuous and one-one

Answer:P → 1, Q → 3, R → 2, S → 4