Find the pairs of equal sets,if any,give reasons:
$A = \{ 0 \}$
$B = \{ x : x > 15 \text{ and } x < 5 \}$
$C = \{ x : x - 5 = 0 \}$
$D = \{ x : x^2 = 25 \}$
$E = \{ x : x \text{ is an integral positive root of the equation } x^2 - 2x - 15 = 0 \}$
$A = \{ 0 \}$
$B = \{ x : x > 15 \text{ and } x < 5 \}$
$C = \{ x : x - 5 = 0 \}$
$D = \{ x : x^2 = 25 \}$
$E = \{ x : x \text{ is an integral positive root of the equation } x^2 - 2x - 15 = 0 \}$
(C) First,we determine the elements of each set:
$A = \{ 0 \}$
$B = \phi$ (since no number is both greater than $15$ and less than $5$)
$C = \{ 5 \}$ (since $x - 5 = 0 \implies x = 5$)
$D = \{ -5, 5 \}$ (since $x^2 = 25 \implies x = \pm 5$)
$E = \{ 5 \}$ (since $x^2 - 2x - 15 = 0 \implies (x - 5)(x + 3) = 0$,so $x = 5$ or $x = -3$. The positive integral root is $5$)
Comparing the sets:
$A = \{ 0 \}, B = \phi, C = \{ 5 \}, D = \{ -5, 5 \}, E = \{ 5 \}$.
We observe that $C = E$ because they contain the same elements.
Thus,the only pair of equal sets is $(C, E)$.
$A = \{ 0 \}$
$B = \phi$ (since no number is both greater than $15$ and less than $5$)
$C = \{ 5 \}$ (since $x - 5 = 0 \implies x = 5$)
$D = \{ -5, 5 \}$ (since $x^2 = 25 \implies x = \pm 5$)
$E = \{ 5 \}$ (since $x^2 - 2x - 15 = 0 \implies (x - 5)(x + 3) = 0$,so $x = 5$ or $x = -3$. The positive integral root is $5$)
Comparing the sets:
$A = \{ 0 \}, B = \phi, C = \{ 5 \}, D = \{ -5, 5 \}, E = \{ 5 \}$.
We observe that $C = E$ because they contain the same elements.
Thus,the only pair of equal sets is $(C, E)$.
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