NIMCET 2018 Question Paper with solutions

Given that,
... (i)
let y = [y > 0]
Squaring both sides, we get
$2{\mathit{y}}^2$ = 1 - y ⇒ $2{\mathit{y}}^2$ + y - 1 = 0
⇒ y = $\frac{-1\pm \sqrt{1+8}}{4}$
⇒ = $\frac{-1+3}{4.}\frac{-1-3}{4}$ = $\frac{1}{2},-1$
y > 0 ⇒ y = $\frac{1}{2}$
So, putting the value of y in Eq, (i),
6 + $\mathit{l}\mathit{o}{\mathit{g}}_{\frac{1}{4}}\frac{1}{2}$ = 6 + $\mathit{l}\mathit{o}{\mathit{g}}_{\left(\frac{1}{2}\right)}^2\frac{1}{2}$
= 6 + $\frac{1}{2}$ = $\frac{12+1}{2}$ = $\frac{13}{2}$