A) minimum, when n = 1, 3, 5, ...
B) minimum, when n = 0, 2, 4, ...
C) maximum, when n = 1, 3, 5,...
D) None of the above
Correct Answer: A
Solution :
Given, \[f(x)=\int_{1}^{x}{\frac{\cos t}{t}}dt,x<1\] \[\Rightarrow \] \[f'(x)=\frac{\cos x}{x}(x<1)\] Put \[f'(x)=0,\] \[\frac{\cos x}{x}=0\] \[\Rightarrow \] \[\cos x=0\] \[\Rightarrow \] \[x=n\pi +\frac{\pi }{2}\] Now, \[f'\,'(x)=\frac{(-x\sin x-\cos x)}{{{x}^{2}}}\] \[=-\frac{x\sin x}{{{x}^{2}}}-\frac{\cos x}{{{x}^{2}}}\] \[=-\frac{\sin x}{x}\] [where\[\cos x=0\]] At \[x=n\pi +\frac{\pi }{2},\] \[f'\,'\left( n\pi +\frac{\pi }{2} \right)=-\frac{1}{\left( n\pi +\frac{\pi }{2} \right)}\sin \left( n\pi +\frac{\pi }{2} \right)\] \[=\frac{-1}{n\pi +\frac{\pi }{2}}{{(-1)}^{n}}\]\[\left[ \begin{align} & \because for\,n=0,1,2..... \\ & \sin \left( n\pi +\frac{\pi }{2} \right)=1,-1,1,-1,...... \\ \end{align} \right]\] \[=\frac{{{(-1)}^{(n+1)}}}{\left( n\pi +\frac{\pi }{2} \right)}\] If it is an even number, then \[f'\,'\left( n\pi +\frac{\pi }{2} \right)<0\] and if n is an odd number, then \[f'\,'\left( n\pi +\frac{\pi }{2} \right)>0\] Hence, the value of\[f(x)\]will be minimum, when \[n\]is odd is \[n=1,3,5......\]
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