JEE Main 2020DifferentiationDifferentiationEasyQuestionIf f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈Rf(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R, thenOptionsA2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)2f(0)−f(1)+f(3)=f(2)Bf(1)+f(2)+f(3)=f(0)f(1) + f(2) + f(3) = f(0)f(1)+f(2)+f(3)=f(0)Cf(3)−f(2)=f(1)f(3) - f(2) = f(1)f(3)−f(2)=f(1)D3f(1)+f(2)=f(3)3f(1) + f(2) = f(3)3f(1)+f(2)=f(3)Check AnswerHide SolutionSolutionf(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 Rf(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R Let f′(1)=a,f′′(2)=b,f′′′(3)=c\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}f′(1)=a,f′′(2)=b,f′′′(3)=c f(x)=x3−ax2+bx−cf′(x)=3x2−2ax+bf′′(x)=6x−2af′′′(x)=6c=6,a=3,b=6f(x)=x3−3x2+6x−6f(1)=−2,f(2)=2,f(3)=12,f(0)=−62f(0)−f(1)+f(3)=2=f(2)\begin{aligned} & f(x)=x^3-a x^2+b x-c \\\\ & f^{\prime}(x)=3 x^2-2 a x+b \\\\ & f^{\prime \prime}(x)=6 x-2 a \\\\ & f^{\prime \prime \prime}(x)=6 \\\\ & c=6, a=3, b=6 \\\\ & f(x)=x^3-3 x^2+6 x-6 \\\\ & f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\ & 2 f(0)-f(1)+f(3)=2=f(2) \end{aligned}f(x)=x3−ax2+bx−cf′(x)=3x2−2ax+bf′′(x)=6x−2af′′′(x)=6c=6,a=3,b=6f(x)=x3−3x2+6x−6f(1)=−2,f(2)=2,f(3)=12,f(0)=−62f(0)−f(1)+f(3)=2=f(2)