JEE Main 2023DifferentiationDifferentiationHardQuestionLet f(x)=sinx+cosx−2sinx−cosx,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\}f(x)=sinx−cosxsinx+cosx−2,x∈[0,π]−{4π}. Then f(7π12)f′′(7π12)f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)f(127π)f′′(127π) is equal toOptionsA233\frac{2}{3 \sqrt{3}}332B29\frac{2}{9}92C−133\frac{-1}{3 \sqrt{3}}33−1D−23\frac{-2}{3}3−2Check AnswerHide SolutionSolutionf(x)=sinx+cosx−2sinx−cosxf(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}f(x)=sinx−cosxsinx+cosx−2 =12sinx+12cosx−112sinx−12cosx=cos(x−π4)−1sin(x−π4)=−2sin2(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}=21sinx−21cosx21sinx+21cosx−1=sin(x−4π)cos(x−4π)−1=2sin(2x−8π)cos(2x−8π)−2sin2(2x−8π) ⇒f(x)=−tan(x2−π8)⇒f′(x)=−12sec2(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}⇒f(x)=−tan(2x−8π)⇒f′(x)=−21sec2(2x−8π) ⇒f′′(x)=−12⋅2sec(x2−π8)⋅sec(x2−π8)tan(x2−π8)×12=−12sec2(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}⇒f′′(x)=−21⋅2sec(2x−8π)⋅sec(2x−8π)tan(2x−8π)×21=−21sec2(2x−8π)⋅tan(2x−8π) Now, f(7π12)f′′(7π12)=−tan(7π24−π8)×−12sec2(7π24−π8)×tan(7π24−π8)=12tan2(π6)×sec2π6=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} Now, f(127π)f′′(127π)=−tan(247π−8π)×2−1sec2(247π−8π)×tan(247π−8π)=21tan2(6π)×sec26π=21×31×34=92