Question

Period of $\left| \sin x \right|+\left| \cos x \right|$ is:
(a) $\dfrac{\pi }{2}$
(b) $\pi $
(c) $\dfrac{3\pi }{2}$
(d) $2\pi $

Answer 1

Hint: We solve this problem by using the graph of given expression.
We draw the graph of the given expression using the maximum and minimum values of the given expression then we check for the value on the $x-$axis where the graph is repeating which is the required period of the given expression. We use the condition that the maximum value of the expression of the form $a\left| \sin x \right|+b\left| \cos x \right|+c$ is given as,
$M=c+\sqrt{{{a}^{2}}+{{b}^{2}}}$

Complete step by step solution:
We are asked to find the period of $\left| \sin x \right|+\left| \cos x \right|$
Let us assume that the given expression as,
$\Rightarrow f\left( x \right)=\left| \sin x \right|+\left| \cos x \right|$
We know that sine and cosine functions will have phase differences of $\dfrac{\pi }{2}$ such that if one function obtains a value of 0 then the other function obtains 1.
Here, we can see that the given expression has modulus for all terms so that their sum will not be less than 1.
Here, we can see that the given function obtains value of 1 when any one of the two terms is 0 so that other term will be 1 which is possible when,
$x=0,\dfrac{\pi }{2},\pi ,\dfrac{3\pi }{2},.....$
Therefore, we can say that the minimum value of the given function $f\left( x \right)$ is 1
Now, let us find the maximum value of the above expression.
We know that the maximum value of $a\left| \sin x \right|+b\left| \cos x \right|+c$ is given as,
$M=c+\sqrt{{{a}^{2}}+{{b}^{2}}}$
By using this we get the maximum value of the given function as,
$\begin{align}
  & \Rightarrow M=0+\sqrt{{{1}^{2}}+{{1}^{2}}} \\
 & \Rightarrow M=\sqrt{2} \\
\end{align}$
Now, we know that minimum value, maximum value and the points where we get the minimum value.
So, let us draw the graph of given function by using these three values then we get the graph as,
Here, we can see that the green line represents the graph of given function.
From the graph we can see that there are two parts repeating in the interval $\left[ 0,\pi \right]$
So, we can say that the part of the function in half of the interval is the repeating.
So, the interval in which the graph is repeating is given as $\left[ 0,\dfrac{\pi }{2} \right]$
Therefore, we can conclude that the period of the given function is,
$\therefore p=\dfrac{\pi }{2}$

So, the correct answer is “Option a”.

Note: We have a shortcut for finding the period of the given function.
Let us assume that the given expression as,
$\Rightarrow f\left( x \right)=\left| \sin x \right|+\left| \cos x \right|$
We know that the period of $\left| \sin x \right|,\left| \cos x \right|$ separately are $\pi ,\pi $ respectively.
We have a condition that if there are two periodic functions having period $a,b$ then the period of function which is sum of those two periodic functions is given as LCM of $a,b$
Here, we can see that the given function $f\left( x \right)$ is the sum of two periodic functions whose periods are given as $\pi ,\pi $
By using the above condition of periods we get the period of $f\left( x \right)$ as LCM of $\pi ,\pi $ that is,
$\begin{align}
  & \Rightarrow p=LCM\left( \pi ,\pi \right) \\
 & \Rightarrow p=\pi \\
\end{align}$
Here, we can see that the two functions that are added to form $f\left( x \right)$ are sine and cosine functions.
We know that in trigonometry the sine and cosine functions are complementary functions.
We have the exception for the condition that if the two periodic functions are complementary then the final period is obtained by dividing the LCM of $a,b$ by 2.
By using the above exception we get the period of $f\left( x \right)$ as,
$\Rightarrow P=\dfrac{p}{2}=\dfrac{\pi }{2}$
Therefore, we can conclude that the period of the given function is,
$\therefore P=\dfrac{\pi }{2}$
So, option (a) is the correct answer.