MCQ
$\sqrt {[x + 2\sqrt {(x - 1)} ]} + \sqrt {[x - 2\sqrt {(x - 1)} ]} = $
- A$2$, if $1 \le x \le 2$
- B$2$, if $x > 2$
- C$2\sqrt {(x - 1)} $, if $1 \le x \le 2$
- ✓$2\sqrt {(x - 1)} $, if $x > 2$
Next, $x \pm 2\sqrt {x - 1} \ge 0$
==> ${x^2} \ge 4(x - 1) \Rightarrow {x^2} - 4x + 4 \ge 0 \Rightarrow {(x - 2)^2} \ge 0$,
which is true $\forall \,\,x$, $\therefore x \ge 1$.
For $1 \le x \le 2$, $\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } $
= $\sqrt {1 + (x - 1) + 2\sqrt {x - 1} } + \sqrt {1 + (x - 1) - 2\sqrt {x - 1} } $
= $(1 + \sqrt {x - 1} ) + (1 - \sqrt {x - 1} ) = 2$
For $x > 2$, $\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } $
= $(1 + \sqrt {x - 1} ) + (\sqrt {x - 1} - 1) = 2\sqrt {x - 1} $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(\sqrt 3 - 1)\,\sin \,\theta \, + \,(\sqrt 3 + 1)\,\cos \theta \, = \,2$ is