Let f(x) = _0^x t t dt, , ,x > 0 then f(x) has - Vidyadip

MCQ

Let $f(x) = \int\limits_0^x {{{\cos t} \over t}dt,\,\,x > 0} $ then $f(x)$ has

A
Maxima when $n = - 2,\, - 4,\, - 6,\,.....$
✓
Maxima when $n = - 1,\, - 3,\, - 5,\,....$
C
Minima when $n = 0,\,2,\,4,....$
D
None of these

✓

Answer

Correct option: B.

Maxima when $n = - 1,\, - 3,\, - 5,\,....$

b
(b) $f(x) = \int\limits_0^x {\frac{{\cos t}}{t}} \,dt$, $x > 0$

==> $f'(x) = \frac{{\cos x}}{x}$, $x > 0$

==>$f'(x) = 0 \Rightarrow \frac{{\cos x}}{x} = 0$

==> $x = (2n + 1){\rm{ }}\frac{\pi }{{\rm{2}}}$, 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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