Let f ( x ) = sin x + cos x - 2 sin x - cos x , x [ 0 ] - { 4 } . Then f ( 7 12 ) f ( 7 12 ) is equal to [2023]

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Solution

Q.
$Let f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}, x \in [0, π] - {\frac{π}{4}} .$ Then $f (\frac{7 π}{12}) f'' (\frac{7 π}{12})$ is equal to [2023]

1 $\frac{- 1}{3 \sqrt{3}}$

2 $\frac{2}{9}$

3 $\frac{2}{3 \sqrt{3}}$

4 $\frac{- 2}{3}$

Ans.
(2)

$Given f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$

$\Rightarrow$ $f (x) = \frac{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x - 1}{\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x}$

   $= \frac{\sin (\frac{π}{4}) \cdot \sin x + \cos (\frac{π}{4}) \cdot \cos x - 1}{\cos \frac{π}{4} \cdot \sin x - \sin \frac{π}{4} \cdot \cos x}$

$\Rightarrow f (x) = \frac{\cos (x - \frac{π}{4}) - 1}{\sin (x - \frac{π}{4})} = \frac{- 2 \sin^{2} (\frac{x}{2} - \frac{π}{8})}{2 \sin (\frac{x}{2} - \frac{π}{8}) \cdot \cos (\frac{x}{2} - \frac{π}{8})}$

$\Rightarrow f (x) = - \tan (\frac{x}{2} - \frac{π}{8}) \Rightarrow f' (x) = - \frac{1}{2} \sec^{2} (\frac{x}{2} - \frac{π}{8})$

$and f'' (x) = - \frac{1}{2} \times 2 \sec (\frac{x}{2} - \frac{π}{8}) \cdot \sec (\frac{x}{2} - \frac{π}{8}) \cdot \tan (\frac{x}{2} - \frac{π}{8}) \times \frac{1}{2}$

$\Rightarrow f'' (x) = - \frac{1}{2} \sec^{2} (\frac{x}{2} - \frac{π}{8}) \cdot \tan (\frac{x}{2} - \frac{π}{8})$

$∴$    $f'' (\frac{7 π}{12}) = - \frac{1}{2} \sec^{2} (\frac{7 π}{24} - \frac{π}{8}) \cdot \tan (\frac{7 π}{24} - \frac{π}{8})$

   $= - \frac{1}{2} \sec^{2} (\frac{π}{6}) \cdot \tan (\frac{π}{6}) = - \frac{1}{2} \times \frac{4}{3} \cdot \frac{1}{\sqrt{3}} = - \frac{2}{3 \sqrt{3}}$

$Also, f (\frac{7 π}{12}) = - \tan (\frac{π}{6}) = - \frac{1}{\sqrt{3}}$

$∴$ $f (\frac{7 π}{12}) \cdot f'' (\frac{7 π}{12}) = \frac{- 2}{3 \sqrt{3}} \times \frac{- 1}{\sqrt{3}} = \frac{2}{9}$

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