A figure is bounded by the curves y = $| \sqrt{2} \sin \frac{π x}{4} |, y = 0, x = 2, x = 4$ . Two straight lines passing through (4, 0) divide the above bounded figure in to 3-equal parts then inclination of the lines are

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$\tan^{- 1} (\frac{4 \sqrt{2}}{3 π})$

$π - \tan^{- 1} (\frac{4 \sqrt{2}}{3 π})$

$π - \tan^{- 1} (\frac{2 \sqrt{2}}{3 π})$

answer is C, D.

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Detailed Solution

$y = | \sqrt{2} \sin \frac{π x}{4} | x = 2 \Rightarrow y = \sqrt{2} x = 3 \Rightarrow y = 1 x = 4 \Rightarrow y = 0$
Area of triangle ABC $= \int_{2}^{4} \sqrt{2} \sin \frac{π x}{4} d x$
Question Image
$= \sqrt{2} {(\frac{- \cos \frac{π x}{4}}{\frac{π}{4}})}_{2}^{4}$
$= \frac{4 \sqrt{2}}{11}$
Slope of AM is say m
Equation of AM is $y - 0 = m (x - 4)$
$∴ M = (2, - 2 m)$
Area of triangle AMB $= | \frac{1}{2} \times 2 \times (- 2 m) | = \frac{1}{3} \times \frac{4 \sqrt{2}}{11}$
$m = \frac{2 \sqrt{2}}{3 π}$
Slope of AM $= π - \tan^{- 1} (\frac{2 \sqrt{2}}{3 π})$
Similarly Slope of AN $= π - \tan^{- 1} (\frac{4 \sqrt{2}}{3 π})$