If the area enclosed between x2+y2≤9(pi)2 and sin(x-y)≥0 is m(pi)3n sq units,then find [m]-[n]+[m+nm-n]?
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\hspace{-16}$Now from Graph We have seen that the area of shaded region is \\\\\\ Half of Total Area of the Circle. So Shaded area is.........\\\\\\ $\bf{=\frac{1}{2}.\pi.9\pi^2=\frac{9\pi^3}{2}}$\\\\\\ So after Camparing , We Get $\bf{m=9}$ and $\bf{n=2}$\\\\\\ So $\bf{[m]-[n]+\left\[\frac{m+n}{m-n}\right\]=9-2+\left\[\frac{9+2}{9-2}\right\]=7+1=8}$
