Question

If \[f\left( x \right) = \cos \left( {\dfrac{\pi }{x}} \right)\] , then find the intervals in which the function decreases.
A) \[\pi < \theta < 2\pi \]
B) \[\dfrac{\pi }{2} < \theta < \pi \]
C) \[\dfrac{\pi }{4} < \theta < \dfrac{\pi }{2}\]
D) \[0 < \theta < 2\pi \]

Answer 1

Hint:
To find the interval in which the given function is decreasing, we will first differentiate both sides with respect to \[x\]. Then, to find the decreasing interval, we will rewrite our derivative function as \[f'\left( x \right) < 0\]. We will solve and interchange the inequalities to find the required interval.

Complete step by step solution:
In the given question,
\[f\left( x \right) = \cos \left( {\dfrac{\pi }{x}} \right)\]
Let \[y = f\left( x \right)\]
Now, differentiating both sides with respect to \[x\], we get,
\[\dfrac{{dy}}{{dx}} = f'\left( x \right)\]
As we know, derivative of \[\cos \theta \] is \[ - \sin \theta \] and derivative of \[\dfrac{1}{x} = \dfrac{{ - 1}}{{{x^2}}}\] , using the formula of derivative, \[\dfrac{{dy}}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \left( { - \sin \left( {\dfrac{\pi }{x}} \right)} \right).\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)\pi \]
Here, \[\pi \] being a constant, remains the same.
Rewriting the above equation, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{\pi }{{{x^2}}}\sin \left( {\dfrac{\pi }{x}} \right)\]
Since, we have to find the interval, in which the given function \[f\left( x \right)\] is decreasing. Therefore, we will write
\[\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}} < 0\\ \Rightarrow f'\left( x \right) < 0\end{array}\]
\[ \Rightarrow \sin \left( {\dfrac{\pi }{x}} \right) < 0\]
We will ignore \[\dfrac{\pi }{{{x^2}}}\] because it is greater than 0 for all values of \[x\].
Also, in this question, \[x\] is not defined when it is equal to \[0\].
Hence, \[x \ne 0\].
Now, as we know, \[\sin \theta < 0\] in the interval \[\pi < \theta < 2\pi \].
Hence, \[\left( {\dfrac{\pi }{x}} \right) < 0\] in the interval, \[\pi \left( {2k + 1} \right) < \dfrac{\pi }{x} < \pi \left( {2k + 2} \right)\].
Here, \[k \in Z\], so if we substitute \[k = 0\] in the above interval, we get
\[\pi < \dfrac{\pi }{x} < 2\pi \]
The interval obtained is the same as the interval \[\pi < \theta < 2\pi \].
Now, dividing both sides of \[\pi \left( {2k + 1} \right) < \dfrac{\pi }{x} < \pi \left( {2k + 2} \right)\] by \[\pi \] , we get,
\[\left( {2k + 1} \right) < \dfrac{1}{x} < \left( {2k + 2} \right)\]
By doing reciprocal and interchanging the inequality signs, we get
\[\dfrac{1}{{2k + 2}} < x < \dfrac{1}{{2k + 1}}\]
Hence, \[x \in \left( {\dfrac{1}{{2k + 2}},\dfrac{1}{{2k + 1}}} \right)\] , where it is decreasing.
Thus, it is proved that, \[\sin \theta \] is decreasing in the interval \[\pi < \theta < 2\pi \]

Hence, option A is the correct option.

Note:
The derivative of \[\cos \theta \] is \[ - \sin \theta \] plays a vital role because if this minus sign is neglected then it could completely change our answer. Also, to satisfy the fact that \[\sin \theta \] is decreasing in the interval \[\pi < \theta < 2\pi \], we assumed the below mentioned two sides of the inequality by our basic observations.
\[\pi \left( {2k + 1} \right) < \dfrac{\pi }{x} < \pi \left( {2k + 2} \right)\]
Such observations help us to solve the question easily and correctly.
In addition to this, we should keep in mind that when we multiply or divide both sides of an inequality then the signs must interchange.