Verified Solution • 2025–26 Session moderate QJEEM22Mac754: JEE Main 2022 Mathematics - Class 12 JEEM22Mac754 Solution: If the function f ( x ) = { log e ( 1 − x + x 2 ) + log e ( 1 + … If the function f(x)={loge(1−x+x2)+loge(1+x+x2)secx−cosx,x∈(−π2,π2)−{0}k,x=0 is continuous at x = 0, then k is equal to: Multiple Choice Options A 1 B −-− 1 C e D 0 Step-by-Step Verified Solution f(x)={loge(1−x+x2)+loge(1+x+x2)secx−cosx,x∈(−π2,π2)−{0}k,x=0 for continuity at x=0x = 0x=0 limx→0f(x)=k\mathop {\lim }\limits_{x \to 0} f(x) = kx→0limf(x)=k ∴\therefore∴ k=limx→0loge(x4+x2+1)secx−cosx(00 form)k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)k=x→0limsecx−cosxloge(x4+x2+1)(00form) =limx→0cosxloge(x4+x2+1)sin2x= \mathop {\lim }\limits_{x \to 0} {{\cos x{{\log }_e}({x^4} + {x^2} + 1)} \over {{{\sin }^2}x}}=x→0limsin2xcosxloge(x4+x2+1) =limx→0loge(x4+x2+1)x2= \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}=x→0limx2loge(x4+x2+1) =limx→0ln(1+x2+x4)x2+x4 . x2+x4x2= \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}=x→0limx2+x4ln(1+x2+x4).x2x2+x4 =1= 1=1 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