| Statement-1: \[1=0\] 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, and 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\[f(-x)=\frac{1}{1+\sin x}\] \[\therefore \]\[\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{1+\sin x}}\] Now, \[f(x)+f(-x)=2I\int\limits_{-\pi /4}^{\pi /4}{\frac{2dx}{1-{{\sin }^{2}}x}}\] \[\Rightarrow \]\[II\int\limits_{-\pi /4}^{\pi /4}{\frac{dx}{{{\cos }^{2}}x}}.\]This is an even function \[\therefore \]\[I=2\int\limits_{0}^{\pi /4}{{{\sec }^{2}}xdx\ne 0}\Rightarrow \]Statement-1 is false.
You need to login to perform this action.
You will be redirected in
3 sec
