Question

If $f\,:\,Z\, \to \,Z$ is such that $f(x)\, = \,6x\, - \,11$ then $f$ is
$\,A)$ Injective but not Surjective
$\,B)$ Surjective but not injective
$\,C)$ Bijective
$\,D)$ Neither injective nor surjective

Answer 1

If the function is said to be injective, then it satisfies the condition that the differential function of $f$ should be in increasing order of $Z$. If the function is said to be surjective, then $f(x)$ takes all the elements in $Z$.

Complete step by step solution:
Given that, the function $f\,:\,Z\, \to \,Z$ such that $f(x)\, = \,6x\, - \,11$.
Now, we want to find what kind of function $f$ is given:
Take the function $f(x)\, = \,6x\, - \,11$
Differentiate the above equation with respect to $'x'$ as follows:
${F^{'}}(x) = \,6(1) - 0$
By differentiating the given function, we can get the function as
${F^{'}}(x) = 6$
From the answer, $f(x)\, > \,0$at each point, a function is said to be increasing on $Z$.
If the function is said to be increasing, then it is a one-one function or injective function.
Thus, $f$ is an injective function.
The given relation is $f\,:\,Z\, \to \,Z$.
Modify the given relation as $f\,:\,Z\, \to \,{Z^*}$
We want to check whether all the elements in $Z$ is present in the set in the ${Z^*}$ or not.
To check the above condition, substitute $x\, = \,0,1,2,3.......$ in the equation $f(x)\, = \,6x\, - \,11$
By substituting the value of $x$ in the equation, we will get the range of the relation:
$f(x)\, = \,6x\, - \,11$
Substitute $x = 0$in the above equation:
$f(0) = 6(0) - 11$
Thus, the value of $f(0)$is $- 11$
Substitute $x = 1$in the above equation:
$f(1) = 6(1) - 11$
Thus, the value of $f(1)$is $- 5$
Substitute $x = 2$in the above equation:
$f(2) = 6(2) - 11$
Thus, the value of $f(1)$is $1$
Substitute $x = 3$in the above equation:
$f(3) = 6(3) - 11$
Thus, the value of $f(1)$is $7$
Thus, the range values of $Z$ as follows:
Range $= \,\\{ \,..........., - 11, - 5,1,7,......\\}$
The co-domain is the values of $x$:
Co-domain $= \,\\{ \,...............,0,1,2,3,..........\\}$
From the range and co-domain not all the values are unique in it.
Thus, the function $f$ is not onto function or surjective function.
Thus, the function $f$ is injective but not surjective.

The option $A)$ injective but not surjective is the correct answer.

Note:
Differentiate the function to find whether the given function is injective or not. Substitute the value for $x$ to get the range and codomain in the given function. If all the elements in the range have unique elements in the co-domain, then it is surjective or else it is not a surjective function.