Q.Let $f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}$ be a linear function from $\mathbb{Z}$ into $\mathbb{Z}$. Find $f(x)$.

The key idea is simple: a pair $(x,y)$ belongs to $R$ if and only if $y = 1 - |x|$ and both $x$ and $y$ are rational numbers. Since all given coordinates are rational, we only need to check the equation. The absolute value means we treat $x$ as non‑negative or negative, but the formula $1 - |x|$ is straightforward: it’s $1 - x$ when $x \ge 0$, and $1 + x$ when $x < 0$.

Let’s test each ordered pair one by one.

1. $(1,0)$

   Here $x = 1$, so $|x| = 1$. Then $1 - |x| = 1 - 1 = 0$. The given $y$ is $0$, so it matches.

   $\Rightarrow$ This pair is in $R$.

2. $(-2,-1)$

   $x = -2$, so $|x| = 2$. Then $1 - |x| = 1 - 2 = -1$. The given $y$ is $-1$, so it matches.

   $\Rightarrow$ This pair is in $R$.

3. $(7,-6)$

   $x = 7$, so $|x| = 7$. Then $1 - |x| = 1 - 7 = -6$. The given $y$ is $-6$, so it matches.

   $\Rightarrow$ This pair is in $R$.

4. $(-3,4)$

   $x = -3$, so $|x| = 3$. Then $1 - |x| = 1 - 3 = -2$. But the given $y$ is $4$, not $-2$.

   $\Rightarrow$ This pair is not in $R$.

5. $(0,2)$

   $x = 0$, so $|x| = 0$. Then $1 - |x| = 1 - 0 = 1$. The given $y$ is $2$, not $1$.

   $\Rightarrow$ This pair is not in $R$.

So the only pairs not in $R$ are $(-3,4)$ and $(0,2)$.

Now look at the options: …

The problem tests your understanding of composite functions and, more importantly, the domain restrictions that come with them. When you write $fog$, you mean $f(g(x))$. For this to make sense, $x$ must first be in the domain of $g$, and then $g(x)$ must be in the domain of $f$. Here, $g(x) = \sqrt{x}$ is only defined for $x \ge 0$. So $fog$ is only defined for $x \ge 0$, no matter what $f$ does afterwards.

Let’s check each option carefully.

1. Option (A): $(fog)(-4) = 4$

   $fog$ means $f(g(x))$. For $x = -4$, we first need $g(-4) = \sqrt{-4}$. But $\sqrt{-4}$ is not a real number — $g$ is not defined for negative inputs. So $fog(-4)$ simply does not exist as a real-valued function. The statement claims it equals $4$, which is false. This is the one that is not true.

2. Option (B): $(fog)(2) = 2$

   Here $x = 2$ is in the domain of $g$. Compute $g(2) = \sqrt{2}$. Then $f(g(2)) = (\sqrt{2})^2 = 2$. This is true.

3. Option (C): $(gof)(-2) = 2$ …

Step 1 — The condition. For $f(x) = \sqrt{x^2 - 7x + 12}$ to be a real number we need

Step 2 — Factorise.

The roots are $x = 3$ and $x = 4$.

Step 3 — Sign analysis. The parabola opens upwards (coefficient of $x^2$ is $+1$), so the expression is negative strictly between the roots and non-negative outside them:

We are dealing with two functions: $f(x)=x^2$ defined for all real numbers, and $g(x)=\sqrt{x}$ defined only for $x \ge 0$ (since we are working in real numbers). The composition $f \circ g$ means "first apply $g$, then apply $f$"; $g \circ f$ means "first apply $f$, then apply $g$". The critical point is that $g$ cannot accept a negative input — if the inner function's output is negative, the composition is not defined in the reals.

Let us check each option step by step.

1. Option (A): $gof(4)$

   First compute $f(4) = 4^2 = 16$.

   Then $g(16) = \sqrt{16} = 4$.

   So $gof(4) = 4$. This is true.

2. Option (B): $fog(-4)$

   First compute $g(-4)$. But $g(x) = \sqrt{x}$ is defined only for $x \ge 0$. Since $-4 < 0$, $g(-4)$ is not defined in the real numbers.

   Therefore $fog(-4)$ does not exist as a real number. The statement claims it equals $4$, which is false. …