If the domain of the function $f(x)=\frac{[x]}{1+x^2}$, where $[x]$ is greatest integer $\leq x$, is $[2,6)$, then its range is
If the domain of the function $f(x)=\frac{[x]}{1+x^2}$, where $[x]$ is greatest integer $\leq x$, is $[2,6)$, then its range is
Show Hint
- $\left(\frac{5}{37}, \frac{2}{5}\right]$
- $\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$
- $\left(\frac{5}{26}, \frac{2}{5}\right]$
- $\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$
The Correct Option is A
Solution and Explanation
![If the domain of the function f(x)=([x]/1+x2), where [x] is greatest integer ≤ x, is [2,6), then its range is](https://images.collegedunia.com/public/qa/images/content/2023_08_16/Screenshot_69b046f61692163421184.png)
So, the correct answer is (A): \(\left(\frac{5}{37}, \frac{2}{5}\right]\)
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
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