Question

For ${x^2} - \left( {a + 3} \right)\left| x \right| + 4 = 0$ to have real solutions, the range of $a$ is
A) $\left( { - \infty , - 7} \right] \cup \left[ {1,\infty } \right)$
B) $\left( { - 3,\infty } \right)$
C) $\left( { - \infty , - 7} \right]$
D) $\left[ {1,\infty } \right)$

Answer 1

Here we will first simplify the equation using basic mathematical operation. We will then rewrite the equation in such a way that it takes the form of some algebraic identity. We will then apply a suitable identity and solve that equation to get the range of $a$.

Complete step by step solution:
The given equation is ${x^2} - \left( {a + 3} \right)\left| x \right| + 4 = 0$.
Rewriting the equation, we get
$\Rightarrow \left( {a + 3} \right)\left| x \right| = {x^2} + 4$
Dividing both sides by $\left| x \right|$, we get
$\Rightarrow a + 3 = \dfrac{{{x^2} + 4}}{{\left| x \right|}}$
Subtracting 3 from both the sides, we get
$\Rightarrow a = \dfrac{{{x^2} + 4}}{{\left| x \right|}} - 3$
Now we will simplify the above equation of $a$.
$\Rightarrow a = \left| x \right| + \dfrac{4}{{\left| x \right|}} - 3$
Rewriting the above equation, we get
$\Rightarrow a = \left| x \right| + \dfrac{4}{{\left| x \right|}} - 4 + 1$
$\Rightarrow a = {\left( {\sqrt {\left| x \right|} } \right)^2} + {\left( {\dfrac{2}{{\sqrt {\left| x \right|} }}} \right)^2} - \left( {2 \times \sqrt {\left| x \right|} \times \dfrac{2}{{\sqrt {\left| x \right|} }}} \right) + 1$
Now, by using the simple algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get
$\Rightarrow a = {\left( {\sqrt {\left| x \right|} - \dfrac{2}{{\sqrt {\left| x \right|} }}} \right)^2} + 1$
We can see that the minimum value of $a$ will be 1. Therefore, we get
$\Rightarrow a \ge 1$
So the upper limit of $a$ will be positive infinity i.e. $+ \infty$.
Now we will write the range of the value of $a$ in the brackets. Therefore, we get
$\Rightarrow \left[ {1,\infty } \right)$
Hence the range of $a$ is equal to $\left[ {1,\infty } \right)$.

So, option D is the correct option.

Note:
We should know that the natural domain of a function is the range of the function where its value can lie. Also we have to keep in mind the type of bracket which should be used to show the natural domain of the function. The brackets may be either open bracket or closed bracket.
Open bracket is used to show the natural domain of the function when the end points are not included in it. For example: $(2,15)$ which means$2 < x < 15$.
Closed bracket is used to show the natural domain of the function when the end points are included in it. For example: $[2,15]$ which means$2 \le x \le 15$.
Mixed brackets are used to show the natural domain of the function when one end point are included and one endpoint is not included in it. For example: $\left[ {1,5} \right)$ which means $1 \le x < 5$.