MCQ
$\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \,dx = $
- A$2$
- B$ - 2$
- ✓$0$
- DNone of these
Now $I = \int_0^{\pi /2} {\frac{{\cos \left( {\frac{\pi }{2} - x} \right) - \sin \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}\,dx} $
$= \int_0^{\pi /2} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}\,\,dx} $.....$(ii)$
On adding, $2I = 0 \Rightarrow I = 0$.
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If a curve passes through the point $(1, -2)$ and has slope of the tangent at any point $(x,y)$ on it as $\frac{{{x^2} - 2y}}{x}$ then the curve also passes through the point
| X: | -4 | -3 | -2 | -1 | 0 |
| P(X): | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
The value of E(X) is: