NIMCET 2018 — Mathematics PYQ

Let $x=\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\cdots \infty }}}$

$\Rightarrow x=\frac{1}{\sqrt{2}}\sqrt{1-x}$

Squaring both sides, we get:

$x^2=\frac{1}{2}(1-x)$

$\Rightarrow 2x^2+x-1=0$

$\Rightarrow (x+1)(2x-1)=0$

$\Rightarrow x=-1\text{ or }x=\frac{1}{2}$

Discarding $x=-1$, the value of $x$ is $\frac{1}{2}$.

$\therefore 6+{\log }_{\frac{1}{4}}\left(\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{\sqrt{2}}\cdots \infty }}}\right)$

$=6+{\log }_{\frac{1}{4}}\left(\frac{1}{2}\right)$

Let ${\log }_{\frac{1}{4}}\left(\frac{1}{2}\right)=m$

$\Rightarrow {\left(\frac{1}{4}\right)}^m=\frac{1}{2}$

$\Rightarrow {\left(\frac{1}{2}\right)}^{2m}={\left(\frac{1}{2}\right)}^1$

$\Rightarrow 2m=1$

$\Rightarrow m=\frac{1}{2}$

$\therefore$ The required value is $6+m$

$=6+\frac{1}{2}$

$=\frac{13}{2}$