MCQ
Let $f(x) = \int\limits_0^x {{{\cos t} \over t}dt,\,\,x > 0} $ then $f(x)$ has
- AMaxima when $n = - 2,\, - 4,\, - 6,\,.....$
- ✓Maxima when $n = - 1,\, - 3,\, - 5,\,....$
- CMinima when $n = 0,\,2,\,4,....$
- DNone of these
==> $f'(x) = \frac{{\cos x}}{x}$, $x > 0$
==>$f'(x) = 0 \Rightarrow \frac{{\cos x}}{x} = 0$
==> $x = (2n + 1){\rm{ }}\frac{\pi }{{\rm{2}}}$, for $n \in z$.
Now $f''(x) = \frac{{ - x\sin x - \cos x}}{{{x^2}}}$
$\therefore$ $f''[(2n + 1)\pi /2] = \frac{{ - 2}}{{(2n + 1)\pi }}\,{( - 1)^n}$ = $\frac{{2{{( - 1)}^{n + 1}}}}{{(2n + 1)\pi }}$.
Thus $f''(x) > 0$$n = - 2,\, - 4,\, - 6,........$
$f''(x) < 0$ $n = 0,\,2,\,4,\,........$
$f''(x) > 0$ $n = 1,\,3,\,5........$
$f''(x) < 0$ $n = - 1,\, - 3,\, - 5........$
Thus $f(x)$ attain maximum for $n = - 1,\, - 3,\, - 5$,…. and minimum for $n = 1,\,3,\,5$,…..
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where $[x]$ is the greatest integerr function of $x$,