The area bounded by the curve $y = s e c x,$ the x-axis and the lines $x = 0$ and $x = \frac{π}{4}$ is

\sqrt{2}

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\frac{1}{2}

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(\sqrt{2} + 1)

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(\sqrt{2} - 1)

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Solution

The correct option is C ln $(\sqrt{2} + 1)$
$p o int o f int e r sec t i o n o f c u r v e s y = sec x a n d x = \frac{π}{4}$
$y = sec (\frac{π}{4})$
$= \sqrt{2}$
$H e n c e p o int i s (\frac{π}{4}, \sqrt{2})$
$Re q u i r e d a r e a = \frac{π}{4} \int 0 sec x d x$
$= {[ln (tan x + sec x)]}_{0}^{\frac{π}{4}}$
$= ln (1 + \sqrt{2}) - ln (0 + 1)$
$= ln (1 + \sqrt{2}) [ln (1) = 0]$

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