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 $\mathrm{y}^{\prime}-\mathrm{y}^{\prime \prime}$ at $\mathrm{x}=-1$ is equal to

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Differentiation

Let

y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})

.
Then

y^{'} - y^{''}

x = - 1

is equal to

Answer: (C)

Multiplying the product by

(1 - x)

telescopes it:

(1 - x) y = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16}) = 1 - x^{32}

. Differentiating

y - x y = 1 - x^{32}

gives

(1 - x) y^{'} - y = - 32 x^{31}

, and differentiating once more gives

(1 - x) y^{''} - 2 y^{'} = - 992 x^{30}

. At

x = - 1

the first relation becomes

2 y^{'} - y = 32

, and the second becomes

2 y^{''} - 2 y^{'} = - 992

, that is

y^{''} - y^{'} = - 496

. Therefore

y^{'} - y^{''} = 496

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option