Aptitude Question ID : 93352

If 1 < x < 2, then the value of \sqrt{\left ( x-1 \right )^{2}}+\sqrt{\left ( x-3 \right )^{2}} is:
[A]2
[B]2x – 4
[C]1
[D]3

2
Since 1 < x < 2, we have x – 1 > 0 and x – 3 > 0 or 3 – x > 0
\therefore \sqrt{\left ( x - 1 \right )^{2}} + \sqrt{\left ( x - 3 \right )^{2}}
= \sqrt{\left ( x - 1 \right )^{2}} + \sqrt{\left ( 3 - x \right )^{2}}
\left [ \because \left ( x-3 \right )^{2} = \left ( 3-x \right )^{2} \right ]
= x -1 +3-x = 2
Hence option [A] is the right answer.

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