MCQ
If $2{\cos ^{ - 1}}\sqrt {\frac{{1 + x}}{2}} = \frac{\pi }{2},$ then $x = $
- A$1$
- ✓$0$
- C$-1/2$
- D$1/2$
==> ${\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 }} $
$\Rightarrow 1 = \sqrt {1 + x} \Rightarrow x = 0$.
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$D^*f(x) =\mathop {Limit}\limits_{h \to 0} \frac{{{f^2}(x + h) - {f^2}(x)}}{h}$ where $f^2(x)$ means $[f(x)]^2.$ If $f(x) = x lnx$ then
${\left. {D^*f(x)} \right|_{x = e}}$ has the value