Question

If $x \in R:\dfrac{{2x - 1}}{{{x^3} + 4{x^2} + 3x}}$ is equal to:
$\left( A \right)R - \left\{ 0 \right\}$
$\left( B \right)R - \left\{ {0,1,3} \right\}$
$\left( C \right)R - \left\{ {0, - 1, - 3} \right\}$
$\left( D \right)R - \left\{ {0, - 1, - 3, - \dfrac{1}{2}} \right\}$

Answer 1

Hint: Here, $x \in R$ , means set of real values. Here, the domain of $x$ is given and a function is given.
Based on this information, we have to find the domain of the given function. For the function to take all the real values, the denominator must not be equal to zero. Here the denominator is ${x^3} + 4{x^2} + 3x$
Therefore, it must be not equal to zero, i.e. ${x^3} + 4{x^2} + 3x \ne 0$ .

Complete step by step answer:
Let us define $f\left( x \right)$ in terms of numerator and denominator;
$ \Rightarrow f\left( x \right) = \dfrac{{Numerator}}{{Deno\min ator}}$
$ \Rightarrow f\left( x \right) = \dfrac{{2x - 1}}{{{x^3} + 4{x^2} + 3x}}$
To satisfy the given condition, the denominator i.e.
$ \Rightarrow {x^3} + 4{x^2} + 3x \ne 0$
Taking $x$ common outside the equation;
$ \Rightarrow x\left( {{x^2} + 4x + 3} \right) \ne 0$
Factorizing the above equation, we get;
$ \Rightarrow x\left( {x + 3} \right)\left( {x + 1} \right) \ne 0$
From the above equation , we can conclude that ;
$ \Rightarrow x \ne 0$ , $x \ne - 1$ , $x \ne - 3$
Therefore to satisfy the condition given in the question, i.e. $\dfrac{{2x - 1}}{{{x^3} + 4{x^2} + 3x}} \in R$, we have to exclude these three values $x \ne 0$ , $x \ne - 1$ , $x \ne - 3$ from the set of real numbers.
Therefore, the correct answer for this question is option $\left( 3 \right)R - \left\{ {0, - 1, - 3} \right\}$.

Note:
In the given question they have asked about the domain of the given function. Therefore, we must have an idea about the concept. The domain of a function is the set of all values for which it is defined . For example: The given function $f\left( x \right) = \dfrac{{2x - 1}}{{{x^3} + 4{x^2} + 3x}}$ is defined for all the set of real numbers except for $0$, $ - 1$, $ - 3$ which is represented by $R - \left\{ {0, - 1, - 3} \right\}$. For these 3 values, the function becomes undefined. There is also the concept of range of a function which is defined by the set of values that the function can take. Domain and range are like input and output for a particular function respectively.