JEE Main 202621 Jan, Evening shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x)=x^3+x^2f'(1)+2xf''(2)+f'''(3)f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3), x∈Rx\in\mathbf{R}x∈R. Then the value of f′(5)f'(5)f′(5) is :AOption A. 625\frac{62}{5}562BOption B. 6575\frac{657}{5}5657COption C. 1175\frac{117}{5}5117DOption D. 25\frac{2}{5}52CheckSkip · show solution