JEE Main 2023 — Mathematics Calculus

Give that:

$$f(x)=\frac{\mathrm{sin}x+\mathrm{cos}x-\sqrt{2}}{\mathrm{sin}x-\mathrm{cos}x}$$f(x)=sinx−cosxsinx+cosx−2​​

Convert the numerator and denominator in the form of $\mathrm{sin}(A\pm B)$sin(A±B).

$$\Rightarrow f(x)=\frac{\sqrt{2}s\mathrm{in}(x+\frac{\pi }{4})-\sqrt{2}}{\sqrt{2}(\mathrm{sin}(x-\frac{\pi }{4}))}$$⇒f(x)=2​(sin(x−4π​))2​sin(x+4π​)−2​​

$$\Rightarrow f(x)=\frac{\mathrm{sin}(x+\frac{\pi }{4})-1}{\mathrm{sin}(x-\frac{\pi }{4})}$$⇒f(x)=sin(x−4π​)sin(x+4π​)−1​

$$\Rightarrow f(x+\frac{\pi }{4})=\frac{\mathrm{cos}x-1}{\mathrm{sin}x}$$⇒f(x+4π​)=sinxcosx−1​

$$\Rightarrow f(x+\frac{\pi }{4})=-\mathrm{tan}\frac{x}{2}$$⇒f(x+4π​)=−tan2x​

$$\Rightarrow f(x)=-\mathrm{tan}(\frac{x}{2}-\frac{\pi }{8})$$⇒f(x)=−tan(2x​−8π​)

$$\Rightarrow {f}^{'}(x)=\frac{-1}{2}{\mathrm{sec}}^{2}(\frac{x}{2}-\frac{\pi }{8})$$⇒f′(x)=2−1​sec2(2x​−8π​)

$$\Rightarrow {f}^{"}(x)=-\frac{1}{2}({\mathrm{sec}}^{2}(\frac{x}{2}-\frac{x}{8})\mathrm{tan}(\frac{x}{2}-\frac{\pi }{8}))$$⇒f"(x)=−21​(sec2(2x​−8x​)tan(2x​−8π​))

$\Rightarrow f(\frac{7\pi }{12})=-\mathrm{tan}(\frac{7\pi }{24}-\frac{\pi }{8})$⇒f(127π​)=−tan(247π​−8π​)$=-\mathrm{tan}(\frac{4\pi }{24})$=−tan(244π​)

$$\Rightarrow -\mathrm{tan}(\frac{\pi }{6})=-\frac{1}{\sqrt{3}}$$⇒−tan(6π​)=−3​1​

Also,

$$\Rightarrow {f}^{"}(\frac{7\pi }{12})=-\frac{1}{2}\times {(\frac{2}{\sqrt{3}})}^{2}\times \frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$$⇒f"(127π​)=−21​×(3​2​)2×3​1​=−33​2​

$$\Rightarrow f(\frac{7\pi }{12}){f}^{"}(\frac{7\pi }{12})=\frac{2}{9}$$⇒f(127π​)f"(127π​)=92​

Hence this is the correct option.