Question

Let $f\left(x\right)={\int }_0^x\frac{\cos t}{t}dt(x>0);$ then for $x=(2n+1)\frac{\pi }{2},f\left(x\right)$ has

• A. minima when $n=0,2,4,...$
• B. maxima when $n=0,2,4,6,...$
• C. neither maxima nor minima when $n=-1,-3,-5,...$
• D. None of these

Answer 1

The correct option is B maxima when $n=0,2,4,6,...$

We have, $f^{\prime }\left(x\right)=\frac{\cos x}{x}.$

$\therefore f^{\prime }\left(x\right)=0\Rightarrow \cos x=0\Rightarrow x=(2n+1)\frac{\pi }{2},n\in I.$

Also, $f^{\prime \prime }\left(x\right)=\frac{-x\sin x-\cos x}{x^2}$

$\therefore {\left[f^{\prime \prime }\left(x\right)\right]}_{x=(2n+1)\frac{\pi }{2}}=\frac{-(2n+1)\frac{\pi }{2}\sin (2n+1)\frac{\pi }{2}-0}{{\left[(2n+1)\frac{\pi }{2}\right]}^2}=\frac{-2{(-1)}^n}{(2n+1)\pi }<0$

for $n=0,2,4,6,...$

$\therefore f\left(x\right)$ has maxima when $n=0,2,4,6,...$