Verified Solution • 2025–26 Session moderate QJEEM24Mabad3: JEE Main 2024 Mathematics - Class 12 JEEM24Mabad3 Solution: If , then is equal to : If y=(x+1)(x2−x)xx+x+x+115(3cos2x−5)cos3xy=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 xy=xx+x+x(x+1)(x2−x)+151(3cos2x−5)cos3x , then 96y′(π6)96 y^{\prime}\left(\frac{\pi}{6}\right)96y′(6π) is equal to : Step-by-Step Verified Solution amp;y=(x+1)(x2−x)xx+x+x+115(3cos2x−5)cos3xamp;y=(x+1)(x)((x)3−1)(x)((x)2+(x)+1)+15cos5x−13cos3xamp;y=(x+1)(x−1)+15cos5x−13cos3x\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x\end{aligned}amp;y=xx+x+x(x+1)(x2−x)+151(3cos2x−5)cos3xamp;y=(x)((x)2+(x)+1)(x+1)(x)((x)3−1)+51cos5x−31cos3xamp;y=(x+1)(x−1)+51cos5x−31cos3x amp;y′=1−cos4x⋅(sinx)+cos2x(sinxamp;y′(π6)=1−916×12+34×12amp;=32−9+1232=3532amp;∴96y′(π6)=105\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}amp;y′=1−cos4x⋅(sinx)+cos2x(sinxamp;y′(6π)=1−169×21+43×21amp;=3232−9+12=3235amp;∴96y′(6π)=105 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