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The area bounded by the X-axis, the curve $y = ⨍(x)$ and the lines x = 1 and x = b is equal to $\sqrt{b^{2} + 1} - \sqrt{2}$ for all b > 1 then $⨍(x)$ is

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$\frac{x^{2}}{\sqrt{x^{2} + 1}}$

$\frac{x}{\sqrt{x^{2} + 1}}$

$\sqrt{x^{2} + 1}$

$\sqrt{x + 1}$

answer is C.

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Detailed Solution

$Area = \int_{1}^{b} f (x) d x = \sqrt{b^{2} + 1} - \sqrt{2}$

$= \sqrt{b^{2} + 1} - \sqrt{1 + 1}$

$= {|\sqrt{x^{2} + 1}|}_{1}^{b}$

$∴ f (x) = \frac{d}{d x} (\sqrt{x^{2} + 1}) = \frac{1}{2} \frac{2 x}{\sqrt{x^{2} + 1}} = \frac{x}{\sqrt{x^{2} + 1}}$

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