[x + 2(x - 1) ] + [x - 2(x - 1) ] = - Vidyadip

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 $x > 2$

✓

Answer

Correct option: D.

$2\sqrt {(x - 1)} $, if $x > 2$

d
(d) $x - 1 \ge 0 \Rightarrow x \ge 1$

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} $.

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