Let f(x)=sinx+cosx−2sinx−cosx, x∈[0, π]−π4. Then f7π12f''7π12 is equal to

f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡xf(x)=2sin⁡x+π4−22sin⁡x−π4f(x)=sin⁡x+π4−1sin⁡x−π4fx+π4=cos⁡x−1sin⁡x=−tan⁡x2⇒f(x)=−tan⁡x2−π8f′(x)=−12sec2⁡x2−π8f′′(x)=12sec2⁡x2−π8tan⁡x2−π8f7π12=−tan⁡7π24−π8=−tan⁡4π24=−tan⁡π6=−13f′′7π12=−12×232×13=−233f7π12f′′7π12=29

Let f(x)=sinx+cosx−2sinx−cosx, x∈[0, π]−π4. Then f7π12f''7π12 is equal to

f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡xf(x)=2sin⁡x+π4−22sin⁡x−π4f(x)=sin⁡x+π4−1sin⁡x−π4fx+π4=cos⁡x−1sin⁡x=−tan⁡x2⇒f(x)=−tan⁡x2−π8f′(x)=−12sec2⁡x2−π8f′′(x)=12sec2⁡x2−π8tan⁡x2−π8f7π12=−tan⁡7π24−π8=−tan⁡4π24=−tan⁡π6=−13f′′7π12=−12×232×13=−233f7π12f′′7π12=29

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

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

$\frac{- 2}{3}$

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

$\frac{2}{9}$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x} \\ f (x) = \frac{\sqrt{2} \sin (x + \frac{π}{4}) - \sqrt{2}}{\sqrt{2} (\sin (x - \frac{π}{4}))} \\ f (x) = \frac{\sin (x + \frac{π}{4}) - 1}{\sin (x - \frac{π}{4})} \\ f (x + \frac{π}{4}) = \frac{\cos x - 1}{\sin x} \\ = - \tan \frac{x}{2} \\ \Rightarrow f (x) = - \tan (\frac{x}{2} - \frac{π}{8}) \\ \begin{array}{l} f^{'} (x) = \frac{- 1}{2} \sec^{2} (\frac{x}{2} - \frac{π}{8}) \\ f^{''} (x) = \frac{1}{2} (\sec^{2} (\frac{x}{2} - \frac{π}{8}) \tan (\frac{x}{2} - \frac{π}{8})) \\ f (\frac{7 π}{12}) = - \tan (\frac{7 π}{24} - \frac{π}{8}) \\ = - \tan (\frac{4 π}{24}) \\ = - \tan (\frac{π}{6}) = - \frac{1}{\sqrt{3}} \end{array} \end{array}$
$\begin{array}{l} f^{''} (\frac{7 π}{12}) = - \frac{1}{2} \times {(\frac{2}{\sqrt{3}})}^{2} \times \frac{1}{\sqrt{3}} \\ = \frac{- 2}{3 \sqrt{3}} \\ f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12}) = \frac{2}{9} \end{array}$