Question 14

We will simplify each algebraic expression. Then we will check if the simplified form matches the given condition.

Step 1 — Checking statement (i)

Statement (i) is about an algebraic expression. Let us expand the given expression. The expression is $(k + 1) (k + 2) - (k + 3)$.

$(k + 1) (k + 2) - (k + 3)$

$= (k^2 + 2k + k + 2) - (k + 3)$

$= (k^2 + 3k + 2) - k - 3$

$= k^2 + 3k + 2 - k - 3$

$= k^2 + 2k - 1$

This expression is $k^2 + 2k - 1$. The statement says it is always 2. Let us test with a value for $k$. If $k = 0$, the expression is $0^2 + 2(0) - 1 = \mathbf{-1}$. Since $\mathbf{-1}$ is not equal to 2, the statement is false.

$\boxed{\text{Statement (i) is False}}$

Step 2 — Checking statement (ii)

Statement (ii) is about a multiple of 4. Let us expand the given expression. The expression is $(2q + 1) (2q - 3)$.

$(2q + 1) (2q - 3)$

$= (2q)(2q) + (2q)(-3) + (1)(2q) + (1)(-3)$

$= 4q^2 - 6q + 2q - 3$

$= 4q^2 - 4q - 3$

We can factor out 4 from the first two terms. $= 4(q^2 - q) - 3$

Let $m = q^2 - q$. $m$ is an integer if $q$ is an integer. The expression is $4m - 3$. This means the expression is 3 less than a multiple of 4. A multiple of 4 must be of the form $4m$. Since $4m - 3$ is not of the form $4m$, it is never a multiple of 4. Let us test with a value for $q$. If $q = 1$, the expression is $(2(1) + 1)(2(1) - 3) = (3)(-1) = \mathbf{-3}$. $\mathbf{-3}$ is not a multiple of 4. So, the statement is false.

$\boxed{\text{Statement (ii) is False}}$

Step 3 — Checking statement (iii)

Statement (iii) has two parts. Part 1: Squares of even numbers are multiples of 4. Let an even number be $2n$. Here, $n$ is an integer. Let us find its square.

$(2n)^2$

$= 4n^2$

Since $n^2$ is an integer, $4n^2$ is a multiple of 4. So, the first part of the statement is true.

Part 2: Squares of odd numbers are 1 more than multiples of 8. Let an odd number be $2n + 1$. Here, $n$ is an integer. Let us find its square.

$(2n + 1)^2$

$= (2n)^2 + 2(2n)(1) + 1^2$

$= 4n^2 + 4n + 1$

We can factor out $4n$ from the first two terms. $= 4n(n + 1) + 1$

Now, consider the term $n(n + 1)$. This is the product of two consecutive integers. One of these integers must be even. So, $n(n + 1)$ is always an even number. Let $n(n + 1) = 2m$. Here, $m$ is an integer. Substitute $2m$ into the expression.

$4n(n + 1) + 1$

$= 4(2m) + 1$

$= 8m + 1$

This means the square of an odd number is 1 more than a multiple of 8. So, the second part of the statement is also true. Since both parts are true, the entire statement (iii) is true.

$\boxed{\text{Statement (iii) is True}}$

Step 4 — Checking statement (iv)

Statement (iv) compares an expression to a square number. Let us simplify the given expression. The expression is $(6n + 2)^2 - (4n + 3)^2$. We use the identity $a^2 - b^2 = (a - b)(a + b)$. Let $a = 6n + 2$ and $b = 4n + 3$.

$(6n + 2)^2 - (4n + 3)^2$

$= ((6n + 2) - (4n + 3)) ((6n + 2) + (4n + 3))$

$= (6n + 2 - 4n - 3) (6n + 2 + 4n + 3)$

$= (2n - 1) (10n + 5)$

We can factor out 5 from the second bracket. $= (2n - 1) \cdot 5(2n + 1)$

$= 5 (2n - 1)(2n + 1)$

Now, we use the identity $(x - y)(x + y) = x^2 - y^2$. Here, $x = 2n$ and $y = 1$. $= 5 ((2n)^2 - 1^2)$

$= 5 (4n^2 - 1)$

$= 20n^2 - 5$

The statement says this expression is 5 less than a square number. This means $20n^2 - 5 = X^2 - 5$ for some integer $X$. If this is true, then $20n^2 = X^2$. For $X^2$ to be equal to $20n^2$, $20n^2$ must be a perfect square. $20n^2 = (4 \times 5)n^2 = 4 \times 5 \times n^2$. For this to be a perfect square, $5 \times n^2$ must be a perfect square. This is only true if $n$ is a multiple of $\sqrt{5}$ (not an integer) or if $n = 0$. Let us test with $n = 1$. The expression is $20(1)^2 - 5 = \mathbf{15}$. Is $\mathbf{15}$ five less than a square number? If $X^2 - 5 = 15$, then $X^2 = 20$. $\mathbf{20}$ is not a perfect square. So, the statement is not always true. It is false.

$\boxed{\text{Statement (iv) is False}}$

Answer

(i) False (ii) False (iii) True (iv) False