Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$ $\forall $ $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$ $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.
Clearly $I_{N}$ is onto. But $I_{N}+I_{N}$ is not onto, as we can find an element $3$ in the co-domain $N$ such that there does not exist any $x$ in the domain $N$ with $\left( I _{ N }+ I _{ N }\right)(x)=2 x=3$
Let $\sum\limits_{k = 1}^{10} {f\,(a\, + \,k)} \, = \,16\,({2^{10}}\, - \,1),$ where the function $f$ satisfies $f(x + y) = f(x) f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $‘ a '$ is
Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{\,{x^3} - {x^2} + 10x - 5\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \le 1\,\,\,\,\,\,\,\,\,\,\,\,}\\
{ - 2x + {{\log }_2}({b^2} - 2),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, > 1\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}} \right.$ the set of values of $b$ for which $f(x)$ has greatest value at $x = 1$ is given by
Let $f :R \to R$ be defined by $f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R.$ Then the range of $f$ is
Let $f, g: N -\{1\} \rightarrow N$ be functions defined by $f(a)=\alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$, and $g(a)=a+1$, for all $a \in N -\{1\}$. Then, the function $f+ g$ is.
Period of $f(x) = nx + n - [nx + n]$, $n \in N$
where [ ] denotes the greatest integer function is :