MCQ
If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $
- A$5$
- B$\sqrt 5 $
- C$1/5$
- ✓$1/\sqrt 5 $
$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$ $3x - 2 = 9 - 3\sqrt 5 - 2 = 7 - 3\sqrt 5 = {{14 - 6\sqrt 5 } \over 2}$
= ${{{{(3 - \sqrt 5 )}^2}} \over 2}$;
$ \Rightarrow $ $\sqrt 2 + \sqrt {3x - 2} = \sqrt 5 \,.\,\sqrt x $;
$\therefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$