MCQ
${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $
- A${\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt x $
- ✓${\sin ^{ - 1}}x - {\sin ^{ - 1}}\sqrt x $
- C${\sin ^{ - 1}}\sqrt x - {\sin ^{ - 1}}x$
- DNone of these
Hence ${\sin ^{ - 1}}(x\sqrt {1 - x} - \sqrt x \,\sqrt {1 - {x^2}} )$
$ = {\sin ^{ - 1}}(\sin \theta \sqrt {1 - {{\sin }^2}\phi } - \sin \phi \sqrt {1 - {{\sin }^2}\theta } )$
$ = {\sin ^{ - 1}}(\sin \theta \cos \phi - \sin \phi \cos \theta ) = {\sin ^{ - 1}}\sin \,(\theta - \phi )$
$ = \theta - \phi = {\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(\sqrt x )$.
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$P: x=0$ is a point of local minima of $f$
$Q: x=\sqrt{2}$ is a point of inflection of $f$
$R: f^{\prime}$ is increasing for $x>\sqrt{2}$
($A$) $g^{\prime}(2)=\frac{1}{15}$ ($B$) $h^{\prime}(1)=666$ ($C$) $h(0)=16$ ($D$) $h(g(3))=36$