Question:

The domain of the function $f\left(x\right)=\frac{1+2\left(x+4\right)^{-0.5}}{2-\left(x+4\right)^{0.5}}+5\left(x+4\right)^{0.5}$ is

Updated On: Jun 23, 2023
  • $R$
  • $(-4, 4)$
  • $\left(0, \infty\right)$
  • $\left(-4, 0\right) \cup\left(0, \infty\right)$
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The Correct Option is D

Solution and Explanation

$f\left(x\right)=\frac{1+2\left(x+4\right)^{-0.5}}{2-\left(x+4\right)^{0.5}}+5\left(x+4\right)^{0.5}$ $=\frac{1+2\cdot \frac{1}{\sqrt{x+4}}}{2-\sqrt{x+4}}+5 \sqrt{x+4}=\frac{\sqrt{x+4}+2}{\sqrt{x+4}\left(2-\sqrt{x+4}\right)}+5\sqrt{x+4}$ $f\left(x\right)$ is defined if $x + 4 > 0$ and $2-\sqrt{x+4}\ne 0$. $\therefore f\left(x\right)$ is defined if $x >- 4$ and $x \ne 0$ $\therefore D\left(f\right)=\left(-4,0\right) \cup\left(0, \infty\right)$.
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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