For given equation ${{x}^{2}}-(a+3)\left| x \right|+4=0$ to have real solutions, then what is 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) $\left[ 1,\infty \right)$

Hint: In this question we have given a quadratic equation and firstly we have to solve this quadratic equation and convert it in terms of $a$ and then we have to make both equations equal then the range of both the equation are equal for that we have to use the formula $AM> GM$ where AM is the arithmetic mean and the GM is the geometric mean and in this, we have to use the formula is $\dfrac{x+y}{2}\ge \sqrt{xy}$ with the equality we have to find the range for this equation.

Complete step-by-step solution:
So here in this problem we have given the equation
$\Rightarrow $${{x}^{2}}-(a+3)\left| x \right|+4=0$
after rearranging the given equations, we get,
$\Rightarrow $${{x}^{2}}+4=(a+3)|x|$
Divide whole equation by $|x|$
$\Rightarrow $$\dfrac{{{x}^{2}}}{|x|}+\dfrac{4}{|x|}=a+3$
$\Rightarrow $$|x|+\dfrac{4}{|x|}-3=a$$.......(1)$
Now let’s consider $|x|+\dfrac{4}{|x|}$’
We use the formula to solve this equation,
$AM>GM$
$\dfrac{x+y}{2}\ge \sqrt{xy}$
Now we will put the values in the equation then we get,
$\Rightarrow $$\dfrac{|x|+\dfrac{4}{|x|}}{2}\ge \sqrt{4}$
$\Rightarrow |x|+\dfrac{4}{|x|}\ge 4$
Now we will put this value in equation number (1)
$\Rightarrow |x|+\dfrac{4}{|x|}-3=a$
$\Rightarrow 4-3=a$
$\Rightarrow a=1$
$\Rightarrow $So, from the above equation we can say that the range of the equation is $\left[ 1,\infty \right)$
Hence the correct option is (d) $\left[ 1,\infty \right)$.


Note: Here students must take care of the concept of discriminant and also aware of when the roots will be positive and negative and also take care of the range of the given equation. Students have to keep in mind that $\left| x \right|$ and $x$ are not the same as $\left| x \right|$ is completely different from$x$ and also take care of open and closed bracket concept. Sometimes they write $\left( 3,4 \right)$ as $\left[ 3,4 \right]$ which is not correct.
To solve the above question, we have to find the range of $a$ as the equation has real solutions. Since, the equation has real solutions, discriminant should be positive to have real where discriminant
$\Rightarrow {{b}^{2}}-4ac$ For the equation $a{{x}^{2}}+bx+c=0$.Also we have to know what is $\left| x \right|$. So, $\left| x \right|$ is defined as $\left| x \right|=\left\{ \begin{matrix}
   -x & when & x<0 \\
   x & when & x=0 \\
   x & when & x>0 \\
\end{matrix} \right\}$ . Also, we have to know about open interval that is the values inside the brackets tells us that it’s not included, like $\cdots \left( 3,4 \right)\ldots $in this there will be all the real numbers except $3$ and $4$ Now for closed interval- we can see that it is just opposite of open interval $\ldots \left[ 3,4 \right]\ldots $ tells that the values 3 and 4 and all real numbers between 3 and 4 are included in the operations.
So here in this
We can write the equation as,${{\left( \left| x \right| \right)}^{2}}-\left( a+3 \right)\left| x \right|+4=0$$\cdots $ (1)
As we can see the equation has real solutions, discriminant should be positive to have real solutions,
${{\left( a+3 \right)}^{2}}-4.4.1\ge 0$
$\Rightarrow {{\left( a+3 \right)}^{2}}\ge 16$
$\Rightarrow a+3\ge 4$ And $a+3\le -4$
$\Rightarrow a\ge 2$ And $a\le -7$
$\therefore \,a\in \left( -\infty ,-7 \right)\cup \left[ 1,\infty \right)$
Don’t get confuse in these options this is the correct solution but the option is wrong so, take a note of it.