If a=17−16, b=15−14 then

a=17−16,b=15−14 1a=117−16×17+1617+16=17+161 1b=115−14×15+1415+14=15+141 1a>1b ⇒a<b

If a=17−16, b=15−14 then

a=17−16,b=15−14 1a=117−16×17+1617+16=17+161 1b=115−14×15+1415+14=15+141 1a>1b ⇒a<b

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If $a = \sqrt{17} - \sqrt{16}, b = \sqrt{15} - \sqrt{14}$ then

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a > b

a < b

a = b

a + b = 0

answer is B.

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Detailed Solution

a = \sqrt{17} - \sqrt{16}, b = \sqrt{15} - \sqrt{14}

$\frac{1}{a} = \frac{1}{\sqrt{17} - \sqrt{16}} \times \frac{\sqrt{17} + \sqrt{16}}{\sqrt{17} + \sqrt{16}} = \frac{\sqrt{17} + \sqrt{16}}{1}$

$\frac{1}{b} = \frac{1}{\sqrt{15} - \sqrt{14}} \times \frac{\sqrt{15} + \sqrt{14}}{\sqrt{15} + \sqrt{14}} = \frac{\sqrt{15} + \sqrt{14}}{1}$

$\frac{1}{a} > \frac{1}{b}$

$\Rightarrow a < b$

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If a=17−16, b=15−14 then

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If $a = \sqrt{17} - \sqrt{16}, b = \sqrt{15} - \sqrt{14}$ then