JEE Main 2020DifferentiationDifferentiationEasyQuestionLet 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 ____________.Answer: 3Hide SolutionSolutionf(x)=x3+x2⋅f′(1)+x⋅f′′(2)+f′′′(3)f′(x)=3x2+2xf′(1)+f′′(2)f′′(x)=6x+2f′(1)f′′′(x)=6f′(1)=−5,f′′(2)=2,f′′′(3)=6f(x)=x3+x2⋅(−5)+x⋅(2)+6f′(x)=3x2−10x+2f′(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}f(x)=x3+x2⋅f′(1)+x⋅f′′(2)+f′′′(3)f′(x)=3x2+2xf′(1)+f′′(2)f′′(x)=6x+2f′(1)f′′′(x)=6f′(1)=−5,f′′(2)=2,f′′′(3)=6f(x)=x3+x2⋅(−5)+x⋅(2)+6f′(x)=3x2−10x+2f′(10)=300−100+2=202