Show that function f: N → N , given by f(1)=f(2)=1 and f(x)=x-1 for every x>2, is onto but not

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1.Relation and Function

medium

Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.

Option A

Option B

Option C

Option D

Solution

$f$ is not one-one, as $f(1)=f(2)=1 .$ But $f$ is onto, as given any $y \in N ,\, y \neq 1$ we can choose $x$ as $y+1$ such that $f(y+1)$ $=y+1-1=y .$ Also for $1 \in N$, we have $f(1)=1$.

Standard 12

Mathematics

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has

hard

(KVPY-2019)

If $f(x)$ is a quadratic expression such that $f(1) + f (2)\, = 0$ , and $-1$ is a root of $f(x)\, = 0$, then the other root of $f(x)\, = 0$ is

hard

(JEE MAIN-2018)

Domain of $f (x)$ = $\sqrt {{{\log }_2}\left( {\frac{{10x – 4}}{{4 – {x^2}}}} \right) – 1} $ , is

normal

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

$f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1 .$ Then, the value of $\sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+\mathrm{f}(\mathrm{k}))}$ is equal to:

hard

(JEE MAIN-2021)

Let $f(x) = sin\,x,\,\,g(x) = x.$

Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$

Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$

hard

(AIEEE-2012)

Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.

Solution

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