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(A) Let the positive integer be $x = 3q + 1$.To find the square of this integer,we calculate $x^2 = (3q + 1)^2$.Using the algebraic identity $(a + b)^

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True

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(A) Let the positive integer be $x = 3q + 1$.To find the square of this integer,we calculate $x^2 = (3q + 1)^2$.Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$,we get:$x^2 = (3q)^2 + 2(3q)(1) + (1)^2$$x^2 = 9q^2 + 6q + 1$We can factor out $3$ from the first two terms:$x^2 = 3(3q^2 + 2q) + 1$Let $m = 3q^2 + 2q$,where $m$ is an integer.Thus,$x^2 = 3m + 1$.Therefore,the statement is true.

$Q.1.$ $HCF$ of $8$ and $6$	$A. 49$
$Q.2.$ $LCM$ of $(26, 91)$	$B. 182$
	$C. 2$

If a positive integer is of the form $3q + 1$,where $q$ is a natural number,then its square can be written in the form $3m + 1$. Is the statement true or false?

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