JEE Main 2023 — Mathematics Calculus

Given,

Function $y(x)=(1+x)(1+{x}^{2})(1+{x}^{4})(1+{x}^{8})(1+{x}^{16})$y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)

Now multiplying and dividing by $(1-x)$(1−x) both side we get,

$$(1-x)y(x)=(1-x)(1+x)(1+{x}^{2})(1+{x}^{4})(1+{x}^{8})(1+{x}^{16})$$(1−x)y(x)=(1−x)(1+x)(1+x2)(1+x4)(1+x8)(1+x16)

$$\Rightarrow y(1-x)=1-{x}^{32}$$⇒y(1−x)=1−x32

$$\Rightarrow y-xy=1-{x}^{32}......(1)$$⇒y−xy=1−x32......(1)

Now differentiating equation $(1)$(1) on both sides with respect to $x$x we get,

$${y}^{'}-x{y}^{'}-y=-32{x}^{31}.......(2)$$y′−xy′−y=−32x31.......(2)

Now at $x=-1$x=−1

$$\Rightarrow {y}^{'}(-1)=16$$⇒y′(−1)=16

Now differentiating equation $(2)$(2) with respect to $x$x on both sides we get,

$$\Rightarrow {y}^{"}-x{y}^{"}-{y}^{'}-{y}^{'}=-(32)(31){x}^{30}$$⇒y"−xy"−y′−y′=−(32)(31)x30

Now at $x=-1$x=−1

$$\Rightarrow {y}^{''}(-1)=-480$$⇒y′′(−1)=−480

So, at $x=-1$x=−1, ${y}^{'}-{y}^{"}=16-(-480)=496$y′−y"=16−(−480)=496