Consider identity function I _{ N }: N → N defined as I

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1.Relation and Function

easy

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.

Option A

Option B

Option C

Option D

Solution

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 $\left( I _{ N }+ I _{ N }\right)(x)=2 x=3$

Standard 12

Mathematics

The domain of the function $f(x) = \frac{{{{\sin }^{ – 1}}(x – 3)}}{{\sqrt {9 – {x^2}} }}$ is

easy

(AIEEE-2004)

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

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

hard

(JEE MAIN-2023)

Let $A=\{1,3,7,9,11\}$ and $B=\{2,4,5,7,8,10,12\}$. Then the total number of one-one maps $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$, such that $\mathrm{f}(1)+\mathrm{f}(3)=14$, is :

hard

(JEE MAIN-2024)

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $ f(30) = 20,$ then the value of $f(40)$ is-

normal

If $f({x_1}) – f({x_2}) = f\left( {\frac{{{x_1} – {x_2}}}{{1 – {x_1}{x_2}}}} \right)$ for ${x_1},{x_2} \in [ – 1,\,1]$, then $f(x)$ is

easy

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.

Solution

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