Question

If \[f\left( x \right) = \cos \left( {\pi /x} \right)\] , then find the intervals in which the function decreases.
$\left( a \right){\text{ }}\pi {\text{ < }}\theta {\text{ < 2}}\pi $
$\left( b \right){\text{ }}\pi /2{\text{ < }}\theta {\text{ < }}\pi $
\[\left( c \right){\text{ }}\pi /4{\text{ < }}\theta {\text{ < }}\pi /2\]
\[\left( d \right){\text{ 0 < }}\theta {\text{ < }}2\pi \]

Answer 1

Hint: For solving the interval question, we have to morph the inequality and for this firstly we will differentiate the \[f\left( x \right) = \cos \left( {\pi /x} \right)\] this with respect to $x$ . And then by using the inequality of $\sin \theta $ which is ${\text{0 < }}\theta {\text{ < }}\pi $ and with this we will get to the solution.

Complete step-by-step answer:
First of all we have the equation \[f\left( x \right) = \cos \left( {\pi /x} \right)\] .
On differentiating the above equation with respect to $x$ , we get
$ \Rightarrow \frac{{dy}}{{dx}} = \left( { - \sin \frac{\pi }{x}} \right)\left( { - \frac{\pi }{{{x^2}}}} \right)$
Therefore, on solving the above equation we get
$ \Rightarrow \frac{{dy}}{{dx}} = \frac{\pi }{{{x^2}}}\sin \frac{\pi }{x}$
As we know that the inequality of $\sin \theta $ is ${\text{0 < }}\theta {\text{ < }}\pi $ and will always be positive
So from this we can say that the $\frac{{dy}}{{dx}}$ will also be positive when the $\sin \frac{\pi }{x}$ will be positive.
So the inequality can be written as
$ \Rightarrow 0 < \frac{\pi }{x} < \pi $ , or it can also be written as
$ \Rightarrow 2n\pi < \pi /x < \left( {2n\pi + \pi } \right)$ , or in terms of $n$ we can write it as
$ \Rightarrow 2n < 1/x < (2n + 1)$
Now on taking the reciprocal of the above inequality, we will get
$ \Rightarrow \frac{1}{{2n + 1}} < x < \frac{1}{{2n}}$
Here $n$will be the subset of an integer, so mathematically it can be written as
$ \Rightarrow \left[ {\frac{1}{{2n + 1}},\frac{1}{{2n}}} \right]$
Similarly by using the above equations we can also get the inequality where the value of $\sin \theta $ will be negative.
Therefore, it will be decreasing in the inequality
$ \Rightarrow \left[ {\frac{1}{{2n + 2}},\frac{1}{{2n + 1}}} \right]$
Therefore, the sine will be negative when $\pi {\text{ < }}\theta {\text{ < 2}}\pi $
Therefore, the option $\left( a \right)$ will be correct.


Note: For solving this type of question, the inequality is changed over into a mathematical condition by utilizing the equality sign in the spot of the imbalance sign. The geometrical condition is explained and arrangements are gotten as point esteems in the stretch $\left( {0,2\pi } \right)$ . The positive point more noteworthy than $\pi $ is changed over to the identical negative, an incentive for the way that the rehashed essential stretch may lie on the negative side of the source. The base stretch is created between two qualities. If the capacity asymptotes inside the stretch are grown, at that point the point and incentive at which the capacity asymptotes, restricts the estimation of the essential span. Finally, the arrangement is summed up.