Verified Solution • 2025–26 Session moderate QJEEM24Ma9c9b: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9c9b Solution: Let . Then is equal to ____________. Let f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈Rf(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R . Then f′(10)f^{\prime}(10)f′(10) is equal to ____________. Step-by-Step Verified Solution amp;f(x)=x3+x2⋅f′(1)+x⋅f′′(2)+f′′′(3)amp;f′(x)=3x2+2xf′(1)+f′′(2)amp;f′′(x)=6x+2f′(1)amp;f′′′(x)=6amp;f′(1)=−5,f′′(2)=2,f′′′(3)=6amp;f(x)=x3+x2⋅(−5)+x⋅(2)+6amp;f′(x)=3x2−10x+2amp;f′(10)=300−100+2=202\begin{aligned}& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\& f^{\prime \prime \prime}(x)=6 \\& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\& f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\& f^{\prime}(x)=3 x^2-10 x+2 \\& f^{\prime}(10)=300-100+2=202\end{aligned}amp;f(x)=x3+x2⋅f′(1)+x⋅f′′(2)+f′′′(3)amp;f′(x)=3x2+2xf′(1)+f′′(2)amp;f′′(x)=6x+2f′(1)amp;f′′′(x)=6amp;f′(1)=−5,f′′(2)=2,f′′′(3)=6amp;f(x)=x3+x2⋅(−5)+x⋅(2)+6amp;f′(x)=3x2−10x+2amp;f′(10)=300−100+2=202 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