| Statement 1: \[I = 0\] |
| 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 and Statement-2 is correct explanation for Statement-1
C) Statement-1 is true, Statement-2 is true and Statement-2 is NOT 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, \[f(x)+f(-x)=2I=\int\limits_{-\pi /4}^{\pi /4}{\frac{2dx}{1-{{\sin }^{2}}x}}\] \[\Rightarrow \]\[I=\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{{{\cos }^{2}}x}}.\]This is an even function. \[\therefore \]\[I=2\int\limits_{0}^{\pi /4}{{{\sec }^{2}}xdx\ne 0}\Rightarrow \]Statement-1 is false.
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