(A) Given the conditions for set $A$:$1$. $x^2 = 16 \implies x = 4$ or $x = -4$.$2$. $2x = 6 \implies x = 3$.For an element $x$ to be in set $A$,it mu

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(A) Given the conditions for set $A$:$1$. $x^2 = 16 \implies x = 4$ or $x = -4$.$2$. $2x = 6 \implies x = 3$.For an element $x$ to be in set $A$,it must satisfy both conditions simultaneously.Since there is no value of $x$ that satisfies both $x \in \{4, -4\}$ and $x = 3$,the set $A$ contains no elements.Therefore,$A = \phi$.

Class 11 Mathematics Set Theory Basic of Set theory

Easy

The set $A = \{ x : x \in \mathbb{R}, x^2 = 16 \text{ and } 2x = 6 \}$ is equal to

A
$\phi$
B
$\{4, -4, 3\}$
C
$\{3\}$
D
$\{4\}$

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Set Theory

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Basic of Set theory

Write the following set in the set-builder form: $\{ 2, 4, 8, 16, 32 \}$

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The remainder obtained when the polynomial $1 + x + x^3 + x^9 + x^{27} + x^{81} + x^{243}$ is divided by $x - 1$ is

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The set $A = \{ x : x \in \mathbb{R}, x^2 = 16 \text{ and } 2x = 6 \}$ is equal to

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List all the elements of the following set:
$E = \{ x : x \text{ is a month of a year not having } 31 \text{ days} \}$