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QJEEM23Ma1ef4: JEE Main 2023 Mathematics - Class 12 JEEM23Ma1ef4 Solution: Let . Then is equal to

Let f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡x,x∈[0,π]−{π4}f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\} . Then f(7π12)f′′(7π12)f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) is equal to

Multiple Choice Options

  • A 233\frac{2}{3 \sqrt{3}}
  • B 29\frac{2}{9}
  • C −133\frac{-1}{3 \sqrt{3}}
  • D −23\frac{-2}{3}

Step-by-Step Verified Solution

f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡xf(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}

amp;=12sin⁡x+12cos⁡x−112sin⁡x−12cos⁡xamp;=cos⁡(x−π4)−1sin⁡(x−π4)amp;=−2sin⁡2(x2−π8)2sin⁡(x2−π8)cos⁡(x2−π8)\begin{aligned}& =\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{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\& =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)}\end{aligned}

amp;⇒f(x)=−tan⁡(x2−π8)amp;⇒f′(x)=−12sec⁡2(x2−π8)\begin{aligned}& \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\& \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)\end{aligned}

amp;⇒f′′(x)=−12⋅2sec⁡(x2−π8)⋅sec⁡(x2−π8)tan⁡(x2−π8)×12amp;=−12sec⁡2(x2−π8)⋅tan⁡(x2−π8)\begin{aligned}& \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\& =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right)\end{aligned}

amp; Now, f(7π12)f′′(7π12)amp;=−tan⁡(7π24−π8)×−12sec⁡2(7π24−π8)×tan⁡(7π24−π8)amp;=12tan⁡2(π6)×sec⁡2π6amp;=12×13×43=29\begin{aligned}& \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\& =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\& =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\& =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9}\end{aligned}

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