Let f(x) = intlimits_0^x frac{cos t}{t} dt, x | vedclass

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(D) Given $f(x) = \int\limits_0^x \frac{\cos t}{t} dt$ for $x > 0$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{\cos x}{x}$.For critical poin

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None of these

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(D) Given $f(x) = \int\limits_0^x \frac{\cos t}{t} dt$ for $x > 0$.By the Fundamental Theorem of Calculus,$f'(x) = \frac{\cos x}{x}$.For critical points,set $f'(x) = 0$,which implies $\cos x = 0$ (since $x > 0$).Thus,$x = (2n + 1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$.Now,$f''(x) = \frac{-x \sin x - \cos x}{x^2}$.At critical points $x_n = (2n + 1)\frac{\pi}{2}$,we have $\cos x_n = 0$ and $\sin x_n = (-1)^n$.So,$f''(x_n) = \frac{-x_n \sin x_n}{x_n^2} = \frac{-\sin x_n}{x_n} = \frac{-(-1)^n}{(2n + 1)\frac{\pi}{2}} = \frac{2(-1)^{n+1}}{(2n + 1)\pi}$.For a maximum,we need $f''(x_n) For a minimum,we need $f''(x_n) > 0$. This occurs when $(-1)^{n+1} > 0$,which means $n+1$ is even,so $n$ is odd $(n = 1, 3, 5, \dots)$.Since none of the provided options match these conditions correctly,the answer is $(d)$.

Let $f(x) = \int\limits_0^x \frac{\cos t}{t} dt, x > 0$. Then $f(x)$ has:

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