If $f(x)=\sqrt{x}$ and $g(x)=x$ be two functions defined in the domain $R^{+} \cup\{0\}$, then find the value of
i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) $(f g)(x)$
(iv) $\left(\frac{f}{g}\right)(x)$
We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+} \cup\{0\}$.
(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$
(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$
(ii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$
(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$
Find the domain and range of the function $f(x)=\frac{1}{\sqrt{x-5}}$.
We have, $\quad f(x)=\frac{1}{\sqrt{x-5}}$
$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$
$\therefore \quad$ Domain of $f=(5, \infty)$
Let $$f(x)=y$$
$$\therefore \quad y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}$$
$$\begin{array}{ll} \Rightarrow & x-5=\frac{1}{y^2} \\ \therefore & x=\frac{1}{y^2}+5 \\ \because & x \in(5, \infty) \Rightarrow y \in R^{+} \end{array}$$
Hence, range of $f=R^{+}$
If $f(x)=y=\frac{a x-b}{c x-a}$, then prove that $f(y)=x$.
We have, $\quad f(x)=y=\frac{a x-b}{c x-a}$
$\begin{aligned} \therefore \quad f(y) & =\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\ & =\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)}=\frac{a^2 x-a b-b c x+a b}{a c x-b c-a c x+a^2} \\ & =\frac{a^2 x-b c x}{a^2-b c}=\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x \\ \therefore \quad f(y) & =x\quad \text{Hence proved.}\end{aligned}$
Let $n(A)=m$ and $n(B)=n$. Then, the total number of non-empty relations that can be defined from $A$ to $B$ is
If $[x]^2-5[x]+6=0$, where $[\cdot]$ denote the greatest integer function, then