If I = int_{0}^{frac{pi}{6}} frac{cos x}{x} d | vedclass

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(B) For $0  x$ (since $\cos x$ is decreasing and $x$ is increasing,and at $x=0$,$\cos 0 = 1 > 0$).Thus,$\fr

Vedclass · Accepted Answer

$I > \frac{\pi}{6}, J < \frac{\pi}{6}$

Vedclass · Answer

(B) For $0  x$ (since $\cos x$ is decreasing and $x$ is increasing,and at $x=0$,$\cos 0 = 1 > 0$).Thus,$\frac{\cos x}{x} > 1$.Integrating both sides from $0$ to $\frac{\pi}{6}$:$I = \int_{0}^{\pi/6} \frac{\cos x}{x} dx > \int_{0}^{\pi/6} 1 dx = \frac{\pi}{6}$.So,$I > \frac{\pi}{6}$.For $\frac{\pi}{3} Thus,$\frac{\cos x}{x} Integrating both sides from $\frac{\pi}{3}$ to $\frac{\pi}{2}$:$J = \int_{\pi/3}^{\pi/2} \frac{\cos x}{x} dx So,$J Therefore,the correct option is $I > \frac{\pi}{6}$ and $J < \frac{\pi}{6}$.

If $I = \int_{0}^{\frac{\pi}{6}} \frac{\cos x}{x} dx$ and $J = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos x}{x} dx$,which of the following is $CORRECT$?

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