Verified Solution • 2025–26 Session moderate QJEEM12Ma71d3: JEE Main 2012 Mathematics - Class 12 JEEM12Ma71d3 Solution: If then equals If g(x)=∫0xcos4t dt,g\left( x \right) = \int\limits_0^x {\cos 4t\,dt,}g(x)=0∫xcos4tdt, then g(x+π)g\left( {x + \pi } \right)g(x+π) equals Multiple Choice Options A g(x)8(π){{g\left( x \right)} \over {8\left( \pi \right)}}8(π)g(x) B g(x)+g(π)g\left( x \right) + g\left( \pi \right)g(x)+g(π) C g(x)−g(π)g\left( x \right) - g\left( \pi \right)g(x)−g(π) D g(x).g(π)g\left( x \right) . g\left( \pi \right)g(x).g(π) Step-by-Step Verified Solution (b,c)g(x+π)=∫0x+πcos 4t dt\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt}(b,c)g(x+π)=0∫x+πcos4tdt =∫0πcos4tdt+∫ππ+xcos4t dt= \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} }=0∫πcos4tdt+π∫π+xcos4tdt =g(π)+∫0xcos 4t dt= g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt}=g(π)+0∫xcos4tdt Putting t=π+yt = \pi + yt=π+y in second integral, we get ∫xπ+xcos 4t dt=∫0πcos 4t dt\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} }x∫π+xcos4tdt=0∫πcos4tdt =g(π)+g(x)=g(x)−g(π)= g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)=g(π)+g(x)=g(x)−g(π) ∴\therefore∴ g(π)=0)\left. {g\left( \pi \right) = 0} \right)g(π)=0) 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