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Then, (A) S Defines a Function From A To B (B) S0 Defines a Function From A To C (C) S0 Defines a Function From A To B (D) S Defines a Function From A To C - Mathematics

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Question

Let\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = \text{B and C} = \left\{ x \in R : x \geq 0 \right\} and\]\[S = \left\{ \left( x, y \right) \in A \times B : x^2 + y^2 = 1 \right\} \text{and } S_0 = \left\{ \left( x, y \right) \in A \times C : x^2 + y^2 = 1 \right\}\]

Then,


Options

  •  S defines a function from A to B

  •  `S_0` defines a function from A to C

  • S0 defines a function from A to B

  •  S defines a function from A to C

MCQ
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Solution

(a) S defines a function from A to B

\[\text{Let x} \in A\]
\[ \Rightarrow - 1 \leq x \leq 1\]
\[\text{Now}, x^2 + y^2 = 1\]
\[ \Rightarrow y^2 = 1 - x^2 \]
\[ \Rightarrow y = \pm \sqrt{1 - x^2}\]
\[ \Rightarrow - 1 \leq y \leq 1\]
\[ \therefore y \in B\]
\[\text{Thus, S defines a function from A to B} . \]

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Chapter 2: Functions - Exercise 2.6 [Page 75]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 1 | Page 75

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