Verified Solution • 2025–26 Session moderate QJEEM25Ma45de: JEE Main 2025 Mathematics - Class 12 JEEM25Ma45de Solution: Let for some function and . Then is equal to Let for some function \mathrm{y}=f(x), \int_0^x t f(t) d t=x^2 f(x), x>0 and f(2)=3f(2)=3f(2)=3 . Then f(6)f(6)f(6) is equal to Multiple Choice Options A 1 B 6 C 2 D 3 Step-by-Step Verified Solution amp;∫0xtf(t)dt=x2+(x),xgt;0amp; Diff. both side w.r. to xamp;amp;xf(x)=x2f′(x)+2xf(x)amp;−xf(x)=x2f′(x)amp;∫f′(x)f(x)dx=∫−1xdxamp;logdf(x)=−logx+logcamp;f(x)=cxamp;f(2)=3⇒3=c2⇒c=6amp;f(x)=6xamp;f(6)=1∴(1)\begin{aligned}&\int_0^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^2+(\mathrm{x}), \mathrm{x}>0\\&\text { Diff. both side w.r. to } x\\&\begin{aligned}& x f(x)=x^2 f^{\prime}(x)+2 x f(x) \\& -x f(x)=x^2 f^{\prime}(x) \\& \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x \\& \operatorname{logdf}(x)=-\log x+\log c \\& f(x)=\frac{c}{x}\end{aligned}\\&\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6\\&f(x)=\frac{6}{x}\\&f(6)=1\quad\therefore (1)\end{aligned}amp;∫0xtf(t)dt=x2+(x),xamp; Diff. both side w.r. to xamp;amp;xf(x)=x2f′(x)+2xf(x)amp;−xf(x)=x2f′(x)amp;∫f(x)f′(x)dx=∫x−1dxamp;logdf(x)=−logx+logcamp;f(x)=xcamp;f(2)=3⇒3=2c⇒c=6amp;f(x)=x6amp;f(6)=1∴(1)gt;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