\(\text { The domain is }[0, \infty)\)
\(\text { The codomain is also }[0, \infty)\)

\(\mathrm{f}(x)=\left\lfloor\begin{array}{lll} x & -x \mid, \mathrm{f}(x) \end{array}=\left\lfloor\begin{array}{ll} x & -x \end{array}\right.\right.\)
\(f(0) =0-0=0 \)
\(f(6.5) =6-6.5=|-0.5|=0.5 \)
\(f(-7.2) =8-7.2=0.8 \)
\(\therefore \text { Range is }[0,1)\)

\(-1 \leq \sin x \leq 1 \)
\(-2 \leq 2 \sin x \leq 2 \)
\(2 \geq-2 \sin x \geq-2 \)
\(-2 \leq-2 \sin x \leq 2 \)
Adding 1,
\(1-2 \leq 1-2 \sin x \leq 1+2 \)
\(-1 \leq 1-2 \sin x \leq 3 \)
\(-1 \geq \frac{1}{1-2 \sin x} \geq \frac{1}{3} \)
\(\frac{1}{3} \leq \frac{1}{1-2 \sin x} \leq-1 \)
\(\therefore \text { Range is }(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)\)

\(\text { If } 4 \in X \text { then }(4,4) \notin \mathrm{R}\)
R is not reflexive
Symmetric can be easily checked

Let X = (a,b,c)
Then R = Universal relation
= {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.
It is transitive

It is clear that the curve can be obtained from that of y = sin x using translation and dilation. So first we draw y = sin x. From that it is easy to draw the curve y = sin (x - 1), then draw y = 2 sin (x - 1) and finally y = 2 sin (x - 1) + 3.

The graphs figures (i) and (ii) look identical until we compare the scales on the y-axis. The scale in figure(ii) is four times as large, reflecting the multiplication of the original function by 4 (i).
The effect looks different when the functions are plotted on the same scale as in figure(iii).
The graph of y = (4x)² - 1 is shown in figure (iv). Can you spot the difference between figure (i) and figure (iv)? In this case, x-scale has now changed, by the same factor of 4 as in the function (figure (iv)).
To see this, note that substituting \(x={1\over 4}\) into (4x)² - 1 produces 12 - 1, 1, exactly the same as substituting x = 1 into the original function (figure (i)), When plotted on the same set of axes (as in figure (v» the parabola y = (4x)² - 1 looks thinner.
Here, the x-intercepts are different, but y-intercepts are the same.

Since \(P(\phi)\) contains 1 element, \(P(P(\phi))\) contains 2¹ elements and hence \(P(P(P(\phi)))\) contains 2² elements. That is, 4 elements.

\((A\cap B)'\cap A=(A'\cup B')\cap A=(A'\cap A)\cup(B'\cap A)\)
\(=\oslash\cup(B'\cap A)=(B'\cap A)\)
= A - B
So \(n((A\cap B)'\cap A)=n(A-B))=n(A)-n(A\cap B)=7\)

We have \(n(A \cup B)=6,n(A\cap B)=2\) and \(n(A \triangle B)=4.\)
So, \(n((A\cup B)\times(A\cap B)\times(A \triangle B))=n(A \cup B)\times n(A\cap B)\times n(A\triangle B)= 6 \times 2 \times 4 = 48.\)

Method 1:
One-to-one: Let f(x) = f(y). Then 2x - 3 = 2y - 3; this implies that x = y. That is, f(x) = f(y) implies that x = y. Thus f is one-to-one.
Onto: Let y \(\in\) R. Let x \(={y+3 \over 2}.\)  Then \(f(x)=2\left( {y+3\over2} \right)-3=y.\)  Thus f is onto. This also can be proved by saying the following statement. The range of f is R (how?) which is equal to the co-domain and hence f is onto.
Inverse: Let y = 2x - 3. Then y + 3 = 2x and hence \(x={y+3\over 2}.\)  Thus \({f}^{-1}(y)={y+3\over2}\). By replacing y as x, we get \({f}^{-1}(x)={x+3\over 2}.\)
Method 2:
Let y = 2x − 3. Then \(x=\frac{y+3}{2}\). Let \(g(y)=\frac{y+3}{2}\)
Now
\((g \circ f)(x)=g(f(x))=g(2 x-3)=\frac{(2 x-3)+3}{2}=x .\)
\((f \circ g)(y)=f(g(y))=f\left(\frac{y+3}{2}\right)=2\left(\frac{y+3}{2}\right)-3=y\)
Thus, \(g \circ f=I_{X} \text { and } f \circ g=I_{Y}\)
This implies that f and g are bijections and inverses to each other. Hence f is a bijection and \(f^{-1}(y)=\frac{y+3}{2}\). Replacing y by x we get \(f^{-1}(x)=\frac{x+3}{2}\)

We know \(|x|=\begin{cases} -x\ \ if\ x\le0 \\ x\quad if\ x>0 \end{cases}\)
So, \(f(x)=\begin{cases} 2x-(-x)\quad if x \le 0\\ 2x-x\quad if\ x>0 \end{cases}\)
Thus, \(f(x)=\begin{cases}3x\quad ifx\le0\\x\quad if\ x>0 \end{cases}\)
Also, \(g(x)=\begin{cases} 2x+(-x)\quad if \ x\le 0\\2x+x\quad if\ x>0 \end{cases}\)
Thus, \(g(x)=\begin{cases} x\quad if x \le 0\\3x\quad if x >0 \end{cases}\)
Let \(x\le0.\) Then 
(f o g) (x) = f(g{x)) = f(x) = 3x.,
The last equality is taken because \(3x\le0\) whenever \(x\le0.\)
Let x > 0. Then
(f o g)(x) = f(g(x)) = f(3x) = 3x.
Thus (f o g)(x) = 3x for all x.