A) Both statements are True, Statement-2 explains Statement-1.
B) Both statements are True, Statement-2 does not explain Statement-1.
C) Statement-1 is True, Statement-2 is False.
D) Statement-1 is False, Statement-2 is true.
Correct Answer: A
Solution :
\[{{A}_{n}}={{U}_{n}}-{{U}_{n-1}}\] \[=\int\limits_{0}^{\pi /2}{\frac{{{\sin }^{2}}nx-{{\sin }^{2}}(n-1)}{{{\sin }^{2}}x}dx}\] \[=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x\sin x}{{{\sin }^{2}}x}dx=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)dx}{\sin x}}}\] \[{{A}_{n}}-{{A}_{n-1}}=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x-\sin (2n-3)x}{\sin x}dx}\] \[=2\,\int\limits_{0}^{\pi /2}{\cos 2(n-1)xdx=0}\] \[\Rightarrow \] \[{{A}_{n}}={{A}_{n-1}}={{A}_{n-2}}=.....={{A}_{1}}\] \[\Rightarrow \] \[{{A}_{1}}=\frac{\pi }{2}\] \[\Rightarrow \] \[{{U}_{n}}-{{U}_{n-1}}=\frac{\pi }{2}\] \[{{U}_{0}}=0,\,{{U}_{1}}=0\,+\frac{\pi }{2}=\frac{\pi }{2},\,{{U}_{2}}=\frac{\pi }{2}+\frac{\pi }{2}=\pi \]\[\Rightarrow \] \[{{U}_{n}}=\frac{n\pi }{2}\]
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