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# In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.(i)  $f:R\to R$  defined by $f(x)=3-4x$(ii) $f:R\to R$  defined by $f(x)=1+{{x}^{2}}$

Last updated date: 13th Jul 2024
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Hint: To  find the function is one-one let  ${{x}_{1}},{{x}_{2}}\in R$ such that $f({{x}_{1}})=f({{x}_{2}})$  after that check whether the function is invertible. If the function is invertible then it is one-one onto function and if any function is one -one and onto then it is also known as bijective function.

A function is a relation which describes that there should be only one output for each input. OR We can say that, a special kind of relation (a set of ordered pairs) which follows a rule i.e every X-value should be associated to only one y-value is called a Function.

To recall, a function is something, which relates elements/values of one set to the elements/values of another set, in such a way that elements of the second set are identically determined by the elements of the first set. A function has many types which define the relationship between two sets in a different pattern. There are various types of functions like one to one function, onto function, many to one function, etc.

A function has many types and one of the most common functions used is the one-to-one function or injective function. Also, we will be learning here the inverse of this function.
One to one function basically denotes the mapping of two sets. A function g is one-to-one if every element of the range of g corresponds to exactly one element of the domain of g. One-to-one is also written as 1-1. A function f() is a method, which relates elements/values of one variable to the elements/values of another variable, in such a way that elements of the second variable are identically determined by the elements of the first variable.
Onto function could be explained by considering two sets, Set A and Set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto function or surjective function.
A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. If the function satisfies this condition, then it is known as one-to-one correspondence.
It is given that $f:R\to R$defined by $f(x)=3-4x$ .
Let  ${{x}_{1}},{{x}_{2}}\in R$ such that $f({{x}_{1}})=f({{x}_{2}})$ .
$3-4{{x}_{1}}=3-4{{x}_{2}}$
$-4{{x}_{1}}=-4{{x}_{2}}$
$f$ is one-one.
We know one thing: if a function $f(x)$ is invertible then $f(x)$  is definitely a bijective function. means, $f(x)$ will be one - one and onto.
Let's try to do the inverse of  $f(x)=3-4x$.
$y=3-4x$
$x=\dfrac{y-3}{4}$
Hence,  $f(x)$ is invertible.
So $f(x)$ is one - one and onto function.
Hence, $f(x)$ is a bijective function (if any function is one -one and onto then it is also known as a bijective function.)

It is given that $f:R\to R$defined by $f(x)=1+{{x}^{2}}$ .
Let  ${{x}_{1}},{{x}_{2}}\in R$ such that $f({{x}_{1}})=f({{x}_{2}})$ .
$1+{{x}_{1}}^{2}=1+{{x}_{2}}^{2}$
${{x}_{1}}=\pm {{x}_{2}}$
Now, $f(1)=f(-1)=2$ .
$f$ is not one-one.
Also for all real values of  $x$ , $f(x)$ is always greater than $1$ . so, the range of  $f(x)\in (1,\infty )$ but co-domain belongs to  $R$ .
e.g., Co - domain $\ne$ range
so, $f$  is not onto function.
Also f is not a bijective function.

Note: Read the question and see what is asked.  Your concept regarding functions should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.