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If $f (x) = x + \sin x, g (x) = e^{- x}, u = \sqrt{c + 1} - \sqrt{c} v = \sqrt{c} - \sqrt{c - 1} (c > 1)$ , then

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$fog (u) < fog (v)$

$gof (u) < gof (v)$

$g o f (u) < g o f (v)$

answer is C.

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Detailed Solution

$u = \sqrt{c + 1} - \sqrt{c}$

$\Rightarrow u = \frac{1}{\sqrt{c + 1} + \sqrt{c}} and v = \frac{1}{\sqrt{c - 1} + \sqrt{c}}$

Clearly, $u < v$ . also, f is increasing where as g is decreasing ,

Thus $, u < v ⇒ f (u) < f (v) \Rightarrow g o f (u) < g o f (v)$

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