Verified Solution • 2025–26 Session moderate QJEEM22Ma8fcf: JEE Main 2022 Mathematics - Class 12 JEEM22Ma8fcf Solution limx→π482−(cosx+sinx)72−2sin2x\lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}x→4πlim2−2sin2x82−(cosx+sinx)7 is equal to Multiple Choice Options A 14 B 7 C 14 2\sqrt22 D 7 2\sqrt22 Step-by-Step Verified Solution limx→π482−(cosx+sinx)72−2sin2x (00 form)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)x→4πlim2−2sin2x82−(cosx+sinx)7(00form) =limx→π4−7(cosx+sinx)6(−sinx+cosx)−22cos2x= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\sqrt 2 \cos 2x}}=x→4πlim−22cos2x−7(cosx+sinx)6(−sinx+cosx) using L−H\mathrm{L-H}L−H Rule =limx→π456(cosx−sinx)22cos2x (00)= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{56(\cos x - \sin x)} \over {2\sqrt 2 \cos 2x}}\,\,\left( {{0 \over 0}} \right)=x→4πlim22cos2x56(cosx−sinx)(00) =limx→π4−56(sinx+cosx)−42sin2x= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 56(\sin x + \cos x)} \over { - 4\sqrt 2 \sin 2x}}=x→4πlim−42sin2x−56(sinx+cosx) using L−H\mathrm{L-H}L−H Rule =72 . 2=14= 7\sqrt 2 \,.\,\sqrt 2 = 14=72.2=14 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring