Solve $|x\,-\,2| + |x\,-\,1| = x\,-\,3$

Let $N$ be the set of positive integers. For all $n \in N$, let $f_n=(n+1)^{1 / 3}-n^{1 / 3} \text { and }$ $A=\left\{n \in N: f_{n+1}<\frac{1}{3(n+1)^{2 / 3}} < f_n\right\}$ Then,

For a suitably chosen real constant $a$, let a function, $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x} .$ Further suppose that for any real number $x \neq- a$ and $f( x ) \neq- a ,( fof )( x )= x .$ Then $f\left(-\frac{1}{2}\right)$ is equal to

Let the sets $A$ and $B$ denote the domain and range respectively of the function $f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$ where $\lceil x \rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements

$( S 1): A \cap B =(1, \infty)-N$ and

$( S 2): A \cup B=(1, \infty)$

The total number of functions,$f:\{1,2,3,4\} \cdot\{1,2,3,4,5,6\}$   such that $f (1)+ f (2)= f (3)$, is equal to .

Consider a function $f:\left[0, \frac{\pi}{2}\right]$ $ \rightarrow$ $R$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] $ $\rightarrow$ $R$ given by $g(x)=\cos x .$ Show that $f$ and $g$ are one-one, but $f\,+\,g$ is not one-one.