Let \[f\left( x \right) = \int_0^x {\frac{{\cos t}}{t}dt} \], \[x > 0\]. Which of the following statement is correct for \[f\left( x \right)\]?
A. Maxima when \[n = - 2, - 4, - 6, \cdots \]
B. Maxima when \[n = - 1, - 3, - 5, \cdots \]
C. Minima when \[n = 0,2,4, \cdots \]
D. Minima when \[n = 1,3,5, \cdots \]
E. B and D

Hint: First we will calculate \[f'\left( x \right)\] from the integration \[f\left( x \right) = \int_0^x {\frac{{\cos t}}{t}dt} \]. Then we will calculate the derivative \[f'\left( x \right)\] and apply the maxima and minima condition on \[f''\left( x \right)\].

Formula used:
The Fundamental Theorem of Calculus:
If \[f\] is continuous on \[\left[ {a,b} \right]\] , and \[F\left( x \right) = \int_a^x {f\left( t \right)dt} \], then \[F'\left( x \right) = f\left( x \right)\] for all \[x\] in \[\left( {a,b} \right)\].
The general solution of \[\cos \theta = 0\] is \[\theta = \left( {2n + 1} \right)\frac{\pi }{2}\], for \[\forall n \in I\].

Complete step by step solution:
Given integration is \[f\left( x \right) = \int_0^x {\frac{{\cos t}}{t}dt} \].
Now we will apply the fundamental theorem:
\[f'\left( x \right) = \frac{{\cos x}}{x}\]
To find the maxima and minima we will equate \[f'\left( x \right)\] with zero
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{{\cos x}}{x} = 0\]
Since \[x \ne 0\]
\[ \Rightarrow \cos x = 0\]
Now find the general solution of the equation
\[ \Rightarrow x = \left( {2n + 1} \right)\frac{\pi }{2}\]
Now we will find the derivative of \[f'\left( x \right)\]
\[f''\left( x \right) = \frac{{x\frac{d}{{dx}}\left( {\cos x} \right) - \cos x\frac{d}{{dx}}\left( x \right)}}{{{x^2}}}\]
\[ \Rightarrow f''\left( x \right) = \frac{{ - x\sin x - \cos x}}{{{x^2}}}\]
Now putting \[x = \left( {2n + 1} \right)\frac{\pi }{2}\] in \[f''\left( x \right)\]
\[f''\left( x \right) = \frac{{ - \left( {2n + 1} \right)\frac{\pi }{2}\sin \left( {2n + 1} \right)\frac{\pi }{2} - \cos \left( {2n + 1} \right)\frac{\pi }{2}}}{{{{\left[ {\left( {2n + 1} \right)\frac{\pi }{2}} \right]}^2}}}\]
Since \[\cos \left( {2n + 1} \right)\frac{\pi }{2} = 0\] for \[\forall n \in I\] and \[\sin \left( {2n + 1} \right)\frac{\pi }{2} = {\left( { - 1} \right)^n}\] for \[\forall n \in I\].
\[ \Rightarrow f''\left( x \right) = \frac{{0 - \left( {2n + 1} \right)\frac{\pi }{2}{{\left( { - 1} \right)}^n}}}{{{{\left[ {\left( {2n + 1} \right)\frac{\pi }{2}} \right]}^2}}}\]
\[ \Rightarrow f''\left( x \right) = \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{\left( {2n + 1} \right)\frac{\pi }{2}}}\]
So, when \[n = - 2, - 4, - 6, \cdots \]
\[f''\left( x \right) > 0\]
When \[n = - 1, - 3, - 5, \cdots \]
\[f''\left( x \right) < 0\]
When \[n = 0,2,4, \cdots \]
\[f''\left( x \right) < 0\]
When \[n = 1,3,5, \cdots \]
\[f''\left( x \right) > 0\]
We know that when \[f''\left( x \right) < 0\] at \[x = a\] then \[f\left( x \right)\] has maxima \[x = a\].
When \[n = - 1, - 3, - 5, \cdots \], \[f\left( x \right)\] has maxima.
We know that when \[f''\left( x \right) > 0\] at \[x = a\] then \[f\left( x \right)\] has minima \[x = a\].
When \[n = 1, 3, 5, \cdots \], \[f\left( x \right)\] has minima \[n = 1,3,5, \cdots \].
Hence option E is correct.

Note: Students often confused with maxima and minima. When \[f''\left( x \right) < 0\] at \[x = a\] then \[f\left( x \right)\] has maxima \[x = a\]. When \[f''\left( x \right) > 0\] at \[x = a\] then \[f\left( x \right)\] has minima \[x = a\].