Question:

If $D$ is the set of all real $x$ such that $1-e^{\left(1/ x\right)-1}$ is positive, then $D$ is equal to

Updated On: Aug 13, 2024
  • $(-\infty, 1]$
  • $\left(-\infty , 0\right)$
  • $\left(1, \infty\right)$
  • $\left(-\infty , 0\right)\cup\left(1, \infty\right)$
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The Correct Option is D

Solution and Explanation

$1-e^{\frac{1}{x}-1} >0$ $\Rightarrow e^{\frac{1}{x}-1} < 1$ $ \Rightarrow \frac{1}{x}-1 < log\,1$ $\Rightarrow \frac{1}{x}-1 < 0 $ $\Rightarrow \frac{1}{x} < 1$ $\Rightarrow x\,\in\left(-\infty, 0\right)\cup\left(1, \infty\right)$
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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.

Read More: Relations and Functions