Question
यदि $2{\cos ^{ - 1}}\sqrt {\frac{{1 + x}}{2}} = \frac{\pi }{2},$ तो $x = $
==> ${\cos ^{ - 1}}\sqrt {\left( {\frac{{1 + x}}{2}} \right)} = \frac{\pi }{4} $
$\Rightarrow \cos \frac{\pi }{4} = \frac{{\sqrt {1 + x} }}{{\sqrt 2 }}$
==> $\frac{1}{{\sqrt 2 }} = \frac{{\sqrt {1 + x} }}{{\sqrt 2 }} $
$\Rightarrow 1 = \sqrt {1 + x} \Rightarrow x = 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
($A$) $f\left(\frac{1}{2}\right) \geq f(1)$
($B$) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$
($C$) $f^{\prime}(2) \leq 0$
($D$) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$