Let Tn be the area bounded by y=tannx,x=0,y=0 and x=π4 where n is a integer greater than 2, then T100 is
A
1200<T100<1196
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B
1206<T100<1204
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C
1204<T100<1202
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D
1202<T100<1198
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Solution
The correct option is D1202<T100<1198 We know that, Tn=π/4∫0tannxdxTn−2=π/4∫0tann−2xdx Therefore, Tn+Tn−2=π/4∫0tann−2x(tan2x+1)dx⇒Tn+Tn−2=π/4∫0tann−2x(sec2x)dx⇒Tn+Tn−2=[tann−1xn−1]π/40⇒Tn+Tn−2=1n−1 We know that for 0≤x≤π/40≤tanx≤1tann+2x<tannx<tann−2x⇒π/4∫0tann+2xdx<π/4∫0tannxdx<π/4∫0tann−2xdx⇒Tn+2<Tn<Tn−2⇒Tn+Tn+2<2Tn<Tn+Tn−2⇒12(n+1)<Tn<12(n−1)⇒1202<T100<1198