MCQ
${\tan ^{ - 1}}\,\,\left( {\sin \,\left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{2}{3}} } \right)} \right)} \right)\,\,=$
- A$\frac {\pi }{4}$
- B$\frac {\pi }{2}$
- C$\frac {\pi }{3}$
- ✓$\frac {\pi }{6}$
Let ${\cos ^{ - 1}}\sqrt {\frac{2}{3}} = \theta \Rightarrow \cos \theta = \sqrt {\frac{2}{3}} $
$ \Rightarrow \sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - \frac{2}{3}} = \sqrt {\frac{1}{3}} $
$\therefore {\tan ^{ - 1}}\left[ {\sin \left( {{{\cos }^{ - 1}}\sqrt {\frac{2}{3}} } \right)} \right] = {\tan ^{ - 1}}\left[ {\sin \theta } \right]$
$ = {\tan ^{ - 1}}\left[ {\sqrt {\frac{1}{3}} } \right] = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$
$ = \frac{\pi }{6}$
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$f(x)=\left\{\begin{array}{ll} \min \left\{(x+6), x^{2}\right\}, & -3 \leq x \leq 0 \\ \max \left\{\sqrt{x}, x^{2}\right\}, & 0 \leq x \leq 1 \end{array}\right.$ આપેલ છે.
જો $y = f ( x )$ અને $x$ -અક્ષ દ્વારા આવૃત પ્રદેશનું ક્ષેત્રફળ $A$ હોય તો $6 A$ ની કિમંત મેળવો.