State with reason whether following functions have inverse h:{2,3,4,5} →{7,9,11,13} with h={(2,7),

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1.Relation and Function

easy

State with reason whether following functions have inverse $h:\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$

Option A

Option B

Option C

Option D

Solution

$h :\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ defined as

$h =\{(2,7)\,,(3,9),\,(4,11),\,(5,13)\}$

It is seen that all distinct elements of the set $\{2,3,4,5\}$ have distinct images under $h$.

$\therefore$ Function $h$ is one – one.

Also, $h$ is onto since for every element $y$ of the set $\{7,9,11,13\},$ there exists an element $x$ in the set $\{2,3,4,5\},$ such that $h ( x )= y$.

Thus, $h$ is a one-one and onto function.

Hence, $h$ has an inverse.

Standard 12

Mathematics

Let $f:\left[ {4,\infty } \right) \to \left[ {1,\infty } \right)$ be a function defined by $f\left( x \right) = {5^{x\left( {x – 4} \right)}}$ then $f^{-1}(x)$ is

normal

Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, \,f(2)=b$ and $f(3)=c .$ Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.

hard

Which of the following function is invertible

normal

If $f:IR \to IR$ is defined by $f(x) = 3x – 4$, then ${f^{ – 1}}:IR \to IR$ is

easy

It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,2),(2,1),(3,1)\}=f$

easy

State with reason whether following functions have inverse $h:\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$

Solution

Similar Questions

Let $f:\left[ {4,\infty } \right) \to \left[ {1,\infty } \right)$ be a function defined by $f\left( x \right) = {5^{x\left( {x – 4} \right)}}$ then $f^{-1}(x)$ is

Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, \,f(2)=b$ and $f(3)=c .$ Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.

Which of the following function is invertible

If $f:IR \to IR$ is defined by $f(x) = 3x – 4$, then ${f^{ – 1}}:IR \to IR$ is

It is easy to see that $f$ is one-one and onto, so that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}=\{(1,2),(2,1),(3,1)\}=f$

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