Chapter 2 - Relations and Functions

Domain:
For $f(x)$ to be defined, $9 - x^2 \ge 0$
$\implies x^2 \le 9$
$\implies -3 \le x \le 3$
$\therefore$ Domain = $[-3, 3]$.

Range:
Let $y = \sqrt{9-x^2}$. Since square root is non-negative, $y \ge 0$.
Squaring both sides: $y^2 = 9 - x^2 \implies x^2 = 9 - y^2$.
Since $x^2 \ge 0$, we have $9 - y^2 \ge 0 \implies y^2 \le 9 \implies -3 \le y \le 3$.
Combining with $y \ge 0$, we get $0 \le y \le 3$.
$\therefore$ Range = $[0, 3]$.

Domain:
The denominator $1+x^2$ is never zero for any real $x$ (since $x^2 \ge 0$).
$\therefore$ Domain = $\mathbb{R}$.

Range:
Let $y = \frac{x}{1+x^2}$
$y(1+x^2) = x \implies yx^2 - x + y = 0$.
This is a quadratic in $x$. For $x$ to be real, Discriminant $D \ge 0$.
$D = b^2 - 4ac = (-1)^2 - 4(y)(y) = 1 - 4y^2$.
$1 - 4y^2 \ge 0 \implies 4y^2 \le 1 \implies y^2 \le \frac{1}{4}$
$\implies -\frac{1}{2} \le y \le \frac{1}{2}$.
$\therefore$ Range = $[-\frac{1}{2}, \frac{1}{2}]$.

We can redefine the function based on critical points $x=1$ and $x=2$:

Case 1: $x < 1$
$f(x) = -(x-1) - (x-2) = -2x + 3$

Case 2: $1 \le x < 2$
$f(x) = (x-1) - (x-2) = 1$

Case 3: $x \ge 2$
$f(x) = (x-1) + (x-2) = 2x - 3$

The graph consists of a ray with slope -2 for $x<1$, a horizontal line segment $y=1$ between $x=1$ and $x=2$, and a ray with slope 2 for $x \ge 2$.

For the function to be defined, the term inside the square root must be positive (since it's in the denominator).
$x - |x| > 0 \implies x > |x|$.
We know that $|x| \ge x$ for all real numbers.
Specifically, if $x \ge 0$, $|x| = x$, so $x > x$ is false.
If $x < 0$, $|x| = -x$, so $x > -x \implies 2x > 0 \implies x > 0$, which contradicts $x < 0$.
Thus, there is no real $x$ satisfying the condition.
$\therefore$ Domain = $\phi$ (Empty Set).

Given $y = f(x) = \frac{ax-b}{cx-a}$.
We need to find $f(y)$.
$f(y) = \frac{ay-b}{cy-a}$
Substitute $y = \frac{ax-b}{cx-a}$:
$f(y) = \frac{a(\frac{ax-b}{cx-a}) - b}{c(\frac{ax-b}{cx-a}) - a}$
$= \frac{a(ax-b) - b(cx-a)}{c(ax-b) - a(cx-a)}$
$= \frac{a^2x - ab - bcx + ab}{acx - bc - acx + a^2}$
$= \frac{a^2x - bcx}{a^2 - bc} = \frac{x(a^2 - bc)}{a^2 - bc} = x$.
Hence Proved.

Domain:
Since $-1 \le \sin 3x \le 1$, the denominator $2 - \sin 3x$ lies between $2-1=1$ and $2-(-1)=3$.
The denominator is never zero.
$\therefore$ Domain = $\mathbb{R}$.

Range:
We know $-1 \le \sin 3x \le 1$.
Multiply by -1: $1 \ge -\sin 3x \ge -1$.
Add 2: $3 \ge 2 - \sin 3x \ge 1$.
Take reciprocal (inequality reverses): $\frac{1}{3} \le \frac{1}{2 - \sin 3x} \le 1$.
$\therefore$ Range = $[\frac{1}{3}, 1]$.