Q.Find the domain and range of

The minimum value of $x^2$ over all reals is 0 (at $x=0$), so the minimum of $f(x)=x^2+3$ is 3; $f$ can take any value from 3 upward.

For $f(x) = x^2 + 3$ with $x \in R$: since the square of any real number is never negative, $x^2 \ge 0$ for all $x$. Adding 3 to both sides, $f(x) = x^2 + 3 \ge 3$. The minimum value $3$ is attained at $x = 0$, and as $|x|$ grows without bound, $f(x)$ increases without bound. So $f(x)$ takes every value in $[3, \infty)$ and no value below 3 — this is the range.

$|x| \geq 0$ for every real $x$, and every value $y \geq 0$ is achieved (e.g. by $x = y$), so the range is $[0, \infty)$.

The modulus function is defined as:

$f(x) = |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$

In both cases the output is non-negative, so $f(x) \geq 0$ always -- the value $0$ is included (at $x=0$), and it's a closed lower bound, so the interval starts with a square bracket: $[0, \infty)$.

There is no upper bound since $|x|$ can be made arbitrarily large.

The literal stem $f(x)=\sqrt{x^2-4}$ has domain $(-\infty,-2]\cup[2,\infty)$, which is not one of the four choices — the choices match $f(x)=\sqrt{4-x^2}$ instead, so that is very likely the intended function, giving domain $[-2,2]$.

Working the literal stem first, honestly: for $f(x)=\sqrt{x^2-4}$ to be real-valued we need the expression under the root to be non-negative:

$x^2-4\ge 0 \implies x^2\ge 4 \implies x\le -2 \text{ or } x\ge 2.$

So the true domain of the function exactly as printed is $(-\infty,-2]\cup[2,\infty)$ — and none of options (a) [0,4], (b) [-2,2], (c) (-2,2), (d) (4,0) equal this set.

Because every one of the four options is a bounded interval centred on 0 (or close to it), and option (b) is exactly the domain of the closely related function $f(x)=\sqrt{4-x^2}$ (a very standard NCERT domain example), the most reasonable reading is that the paper has a sign slip and intended $4-x^2$ under the root, not $x^2-4$.

For $f(x)=\sqrt{4-x^2}$: need $4-x^2\ge 0 \implies x^2\le 4 \implies -2\le x\le 2$, i.e. domain $[-2,2]$.

$f(x) = |x-1|$ can never be negative and attains every non-negative value, so its range is $[0, \infty)$.

For $f: R \to R$, $f(x) = |x - 1|$.

By definition, the absolute value $|x-1| \ge 0$ for every real $x$, so $f(x)$ can never be negative.

As $x$ varies over all of $R$: when $x = 1$, $f(x) = 0$ (the minimum value); as $x \to \pm\infty$, $f(x) \to \infty$. Since $|x-1|$ is continuous and takes every value between $0$ and $\infty$ (e.g. $f(1+k) = k$ for any $k \ge 0$), the range is exactly $[0, \infty)$.

The expression under a square root must be non-negative, giving the domain $[-3, 3]$.

For $f(x) = \sqrt{9-x^2}$ to give a real value, the radicand must satisfy

$9 - x^2 \ge 0 \implies x^2 \le 9 \implies -3 \le x \le 3.$

So the domain is the closed interval $[-3, 3]$ (both endpoints included, since $9-x^2=0$ is allowed).

f(x) is undefined wherever the denominator 1 − x² = 0, so those points are excluded from the domain.

For $f(x) = 1/(1-x^2)$ to be defined we need the denominator to be non-zero:

$1 - x^2 \neq 0 \Rightarrow x^2 \neq 1 \Rightarrow x \neq 1 \text{ and } x \neq -1$

So f(x) is defined for every real number except x = 1 and x = −1.

The function is defined exactly where its denominator $|x|$ is nonzero.

$f(x)=\dfrac{x}{|x|}$ requires $|x|\ne0$, i.e. $x\ne0$.

So the domain is all real numbers except 0.