Question

Assertion (A) : $f(x) = 1 + {x^2}$is a one to one faction from ${R^ + } \to R,$
Reason (R) : Every strictly monotonic function is a one to one function.
(A) Both A and R are true and R is the correct explanation of A,
(B) Both A and R are true but R is not the correct explanation of A,
(C) A is true but R is false,
(D) A is false but R is true.

Answer 1

Hint:-One to one function or an injective function is a function that maps distinct elements of its domain to a distinct element of its codomain or range. In other words, every element of the function's codomain is the image of at most one element of its domain. One to one function can also be written as 1-1. Formally, it is stated as, if $f\left( x \right) = f\left( y \right)$ implies $x = y$, then $f$ is one to one mapped or $f$is 1-1.

Complete step by step solution:
A monotonic function is a function between ordered sets that preserves or reverses the given order.
Given, $f\left( x \right) = 1 + {x^2},$domain is ${R^ + }$and range is R
Now, for one to one function, we know, if $f\left( x \right) = f\left( y \right)$ then $x = y$
$\therefore 1 + x_1^2 = 1 + x_2^2$
$ \Rightarrow x_1^2 = x_2^2$ [1 is cancelled in both sides]
Thus, no negative value is considered.
A strictly monotonic function is either entirely non-increasing or non-decreasing, i.e., every element in the range maps to only one element in the domain.
$\therefore f\left( x \right)$ is strictly monotonic function and one-one function.
The correct option is (A)

Note: Let R be a relation from a set A to B. Then the domain of the relation R is the set of all first element of the ordered pairs which belong to R. Thus,
Dom.(R) [i.e., domain of R] $ = \left\{ {x:\left( {x,y} \right) \in R} \right\}.$
The range of the relation R is the set of all second elements of the ordered pairs which belong to R. thus,
Range (R) [i.e., range of R] $ = \left\{ {y:\left( {x,y} \right) \in R} \right\}.$