Question

: Number of roots of $\left| \sin \left| x \right| \right|=x+\left| x \right|$ in $\left[ -2\pi ,2\pi \right]$is
(a) 2
(b) 3
(c) 4
(d) 6

Answer 1

Hint: First, before proceeding for this, we must consider the following function given as y to get the graph of the given function as $y=\left| \sin \left| x \right| \right|$. Then, we know that the total number of solutions for the given equation is equal to the number of points of intersection of the curves and then by using the conditions of modulus of functions. Then, from the calculations of the two conditions, we get the result that the only values of the roots of the equation are existing only when the equation gives the value of 0w hcih gives the number of roots.

Complete step by step answer:
In this question, we are supposed to find the number of roots of $\left| \sin \left| x \right| \right|=x+\left| x \right|$ in $\left[ -2\pi ,2\pi \right]$.
So, before proceeding for this, we must consider the following function given as y to get the graph of the given function as:
$y=\left| \sin \left| x \right| \right|$
So, we should plot the graph of the function to get the shape of the curve and the points at which that curve is touching the real axis as:

Then, we know that the total number of solutions for the given equation is equal to the number of points of intersection of the curves.
So, we get the value of the function$y=x+\left| x \right|$ for which we use the conditions of the modulus of the functions as:
For $x\ge 0$, $\begin{align}
  & y=x+x \\
 & \Rightarrow y=2x \\
\end{align}$
For x<0, $\begin{align}
  & y=x-x \\
 & \Rightarrow y=0 \\
\end{align}$
Now, we can see clearly that the curve intersects at five points which are as follows in the given range $\left[ -2\pi ,2\pi \right]$ as:
$x=-2\pi ,-\pi ,0,\pi ,2\pi $
However, from the calculations of the two conditions, we get the result that the only values of the roots of the equation are existing only when the equation gives the value of 0.
So, we get the value of roots only at three points as:
$x=-2\pi ,-\pi ,0$
So, there are three solutions of the function $\left| \sin \left| x \right| \right|=x+\left| x \right|$ in the range $\left[ -2\pi ,2\pi \right]$.
So, the correct answer is “Option b”.

Note:
Now, to solve these type of the questions we must be aware of the functioning of the modulus function, so that e can easily get the conditions for the given question. So, the conditions for the modulus of the function is given by:
$\left| x \right|=x,\text{ }x\ge 0$ and $\left| x \right|=-x,\text{ }x<0$