Let y (x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y prime-y prime prime at x=-1 is equal to :

Question

Let y (x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y prime-y prime prime at x=-1 is equal to : (A) 976 (B) 944 (C) 496 (D) 464

Tardigrade Admin · Accepted Answer

y=(1-x32/1-x) ⇒ y-x y=1-x32 y prime-x y prime-y=-32 x31 y prime prime-x y prime prime-y prime-y prime=-(32)(31) x30 at x=-1 ⇒ y prime-y prime prime=496

Q. Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y^{'} - y^{''}$ at $x = - 1$ is equal to :

976

944

496

464

$y = \frac{1 - x ^{32}}{1 - x} \Rightarrow y - x y = 1 - x^{32}$
$y^{'} - x y^{'} - y = - 32 x^{31}$
$y^{''} - x y^{''} - y^{'} - y^{'} = - (32) (31) x^{30}$
at $x = - 1 \Rightarrow y^{'} - y^{''} = 496$

Q. Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′ at x=−1 is equal to :

976

944

496

464

y=1−x1−x32​⇒y−xy=1−x32 y′−xy′−y=−32x31 y′′−xy′′−y′−y′=−(32)(31)x30 at x=−1⇒y′−y′′=496

Q. Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y^{'} - y^{''}$ at $x = - 1$ is equal to :

$y = \frac{1 - x ^{32}}{1 - x} \Rightarrow y - x y = 1 - x^{32}$
$y^{'} - x y^{'} - y = - 32 x^{31}$
$y^{''} - x y^{''} - y^{'} - y^{'} = - (32) (31) x^{30}$
at $x = - 1 \Rightarrow y^{'} - y^{''} = 496$