JEE Main 2023DifferentiationDifferentiationMediumQuestionLet y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′y' - y''y′−y′′ at x=−1x = - 1x=−1 is equal toOptionsA496B976C464D944Check AnswerHide SolutionSolutiony=1−x321−x=1+x+x2+x3+…+x31y′=1+2x+3x2+…+31x30y′(−1)=1−2+3−4+…+31=16y′′(x)=2+6x+12x2+…+31.30x29y′′(−1)=2−6+12…31.30=480y′′(−1)−y′(−1)=−496\begin{aligned} & y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\ & y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\ & y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\ & y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\ & y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\ & y^{\prime \prime}(-1)-y^{\prime}(-1)=-496 \end{aligned}y=1−x1−x32=1+x+x2+x3+…+x31y′=1+2x+3x2+…+31x30y′(−1)=1−2+3−4+…+31=16y′′(x)=2+6x+12x2+…+31.30x29y′′(−1)=2−6+12…31.30=480y′′(−1)−y′(−1)=−496