Given that for each a in (0,1),the limit g(a) | vedclass

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(C) $1.$ To find $g(a)$,we evaluate the integral $g(a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$.For $a = \frac{1}{2}$,$g\left(\

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$(A, D)$

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(C) $1.$ To find $g(a)$,we evaluate the integral $g(a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$.For $a = \frac{1}{2}$,$g\left(\frac{1}{2}ight) = \int_0^1 t^{-1/2}(1-t)^{-1/2} dt = \int_0^1 \frac{dt}{\sqrt{t(1-t)}}$.Using the substitution $t = \sin^2 	heta$,$dt = 2 \sin 	heta \cos 	heta d	heta$,the integral becomes $\int_0^{\pi/2} \frac{2 \sin 	heta \cos 	heta}{\sin 	heta \cos 	heta} d	heta = \int_0^{\pi/2} 2 d	heta = \pi$.Thus,$g\left(\frac{1}{2}ight) = \pi$.$2.$ We observe that $g(1-a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{-(1-a)}(1-t)^{(1-a)-1} dt = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{a-1}(1-t)^{-a} dt$.Using the property $\int_n^{1-n} f(t) dt = \int_n^{1-n} f(1-t) dt$,we get $g(1-a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} (1-t)^{a-1} t^{-a} dt = g(a)$.Since $g(1-a) = g(a)$,differentiating both sides with respect to $a$ using the chain rule gives $-g'(1-a) = g'(a)$.At $a = \frac{1}{2}$,$-g'\left(\frac{1}{2}ight) = g'\left(\frac{1}{2}ight)$,which implies $2g'\left(\frac{1}{2}ight) = 0$,so $g'\left(\frac{1}{2}ight) = 0$.Therefore,the correct pair is $(A, D)$.

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