| Consider\[I=\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{1-\sin x}}\] |
| Statement-1:\[I=0\] because |
| Statement-2: \[\int\limits_{-a}^{a}{f(x)}\,dx=0\], wherever \[f(x)\] is an odd function. |
A) Statement-1 is false, Statement-2 is true.
B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D) Statement-1 is true, Statement-2 is false.
Correct Answer: A
Solution :
\[f(x)=\frac{1}{1-\sin x}\]and\[f(-x)=\frac{1}{1+\sin x}\] \[\therefore \] \[I=\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{1+\sin x}}\] Now, \[f(x)+f(-x)=2I=\int\limits_{-\pi /4}^{\pi /4}{\frac{2dx}{{{\sin }^{2}}x}}\] \[\Rightarrow \]\[I=\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{{{\cos }^{2}}x}}\]. This is an even function \[\therefore \]\[I=2\int\limits_{0}^{\pi /4}{{{\sec }^{2}}x}\,\,dx\ne 0\Rightarrow \]Statement-1 is false.
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