JEE Main 2023 (Online) 25th January Morning ShiftQuestion 1
Let $y(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left(1+x^{16}\right)$. Then $y'-y''$ at $x=-1$ is equal to
Differentiation and its Application
Solution
Concept. $y=(1+x)P$, and $1+x=0$ at $x=-1$; so only a few terms of $y'$, $y''$ survive there. Formulas used. (1) $y'=P+(1+x)P'$, $y''=2P'+(1+x)P''$(2) $\frac{P'}{P}=\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\frac{16x^{15}}{1+x^{16}}$Step 1 - $P(-1)$. $P=(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$, so $P(-1)=2^4=16$. Step 2 - $P'(-1)$. By (2), $\frac{P'}{P}=\frac{-2-4-8-16}{2}=-15$, so $P'(-1)=-240$. Step 3 - At $x=-1$. By (1), $y'=16$ and $y''=2(-240)=-480$. Step 4 - Value. $y'-y''=16+480=496$. Answer: (C)