Question

The number of real solutions of the equation ${{x}^{2}}=1-\left| x-5 \right|$ is
A. 1
B. 2
C. 4
D. No solution.

Answer 1

Hint: To solve this question, we should have the knowledge of modulus, that is, \left| x \right|=\left\\{ \begin{aligned} & -x\text{ , }x\lt 0 \\\ & x\text{ , }x\ge 0 \\\ \end{aligned} \right.. Also, we should know that when we have to solve a quadratic equation of type $a{{x}^{2}}+bx+c=0$, then $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. And real solutions are those that have no imaginary terms, that is, ${{b}^{2}}-4ac\ge 0$.

Complete step-by-step solution -
In this question, we have to find the number of real solutions of ${{x}^{2}}=1-\left| x-5 \right|$. To solve this, we should know that
\left| \alpha \right|=\left\\{ \begin{aligned} & -\alpha\text{ , }\alpha \lt 0 \\\ & \alpha \text{ , }\alpha \ge 0 \\\ \end{aligned} \right..
So, we can write,
| x-5 |=\Bigg\\{ \begin{aligned} & -( x-5), x-5 < \text{0} \\\ & x-5, \text{ }x-5\ge 0 \\\ \end{aligned} \\\
\Rightarrow \left| x-5 \right|=\Bigg\\{ \begin{aligned} & -( x-5 ), x < \text{5} \\\ & ( x-5 ), x\ge 5 \\\ \end{aligned} \\\
So, we will find the solution for both cases separately, that is for $x<5$ and $x\ge 5$.
Let us consider case 1 for $x<5$. So, we can write the given equation as,
${{x}^{2}}=1-\left( -\left( x-5 \right) \right)$
We will simplify it further to form a quadratic equation. So, we will get,
$\begin{aligned} & {{x}^{2}}=1+x-5 \\\ & \Rightarrow {{x}^{2}}-x+4=0 \\\ \end{aligned}$
Now, we know that for real solutions in quadratic equation ${{b}^{2}}-4ac\ge 0$. So, we will calculate the value of ${{b}^{2}}-4ac$. So, we get,
$\begin{aligned} & {{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( 4 \right) \\\ & \Rightarrow 1-8\Rightarrow -7<0 \\\ \end{aligned}$
As we can see that, ${{b}^{2}}-4ac<0$ for $x<5$, that means for $x<5$, the equation has no real roots.
Let us consider case 2 for $x\ge 5$. So, we can write the given equation as,
$\begin{aligned} & {{x}^{2}}=1-\left( x-5 \right) \\\ & \Rightarrow {{x}^{2}}=1-x+5 \\\ \end{aligned}$
We will simplify it further to form a quadratic equation. So, we will get,
${{x}^{2}}+x-6=0$
Now, we will calculate the value of ${{b}^{2}}-4ac$ as we know that for quadratic equations, if ${{b}^{2}}-4ac\ge 0$, then both roots are real. So, we get,
$\begin{aligned} & {{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( -6 \right) \\\ & \Rightarrow 1+24\Rightarrow 25>0 \\\ \end{aligned}$
As we can see that ${{b}^{2}}-4ac>0$, we can say that for $x\ge 5$, the equation has 2 real roots.
So, we can conclude from both the cases, that only 2 real solutions exist for the given equation.
Therefore, the correct answer is option B.

Note: The possible mistake that one can make in this question is by not writing \left| x-5 \right|=\left\\{ \begin{aligned} & -\left( x-5 \right)\text{ , }x\lt 5 \\\ & \left( x-5 \right)\text{ , }x\ge 5 \\\ \end{aligned} \right., because if we directly write $\left| x-5 \right|$ as $\left( x-5 \right)$, we might get the correct answer in this question, but we may get wrong answers in further questions.