MCQ
$\int_{ - \pi /2}^{\pi /2} {\log \left( {\frac{{2 - \sin \theta }}{{2 + \sin \theta }}} \right)\,d\theta = } $
- ✓$0$
- B$1$
- C$2$
- DNone of these
$\therefore $ $f(x)$ is an odd function of $x$.
Therefore, $2\int_0^{\pi /2} {\log \left( {\frac{{2 - \sin \theta }}{{2 + \sin \theta }}} \right){\rm{ }}d\theta = 0} $.
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Statement $2$ : A function $f : R \to R$ is discontinuous at $x_0$ if and only if, $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right)$ exists and $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right) \ne f\left( {{x_0}} \right)$