If a f(x)+b fleft(frac{1}{x}right)=x+1, and f | vedclass

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(B) Given $a f(x)+b f\left(\frac{1}{x}\right)=x+1$ $(i)$ Replacing $x$ with $\frac{1}{x}$, we get $a f\left(\frac{1}{x}\right)+b f(x)=\frac{1}{x}+1$ $

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$3$

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(B) Given $a f(x)+b f\left(\frac{1}{x}\right)=x+1$ $(i)$ Replacing $x$ with $\frac{1}{x}$, we get $a f\left(\frac{1}{x}\right)+b f(x)=\frac{1}{x}+1$ $(ii)$ Multiplying $(i)$ by $a$ and $(ii)$ by $b$, we get: $a^2 f(x)+a b f\left(\frac{1}{x}\right)=a x+a$ $b^2 f(x)+a b f\left(\frac{1}{x}\right)=\frac{b}{x}+b$ Subtracting the two equations: $(a^2-b^2) f(x)=a x-\frac{b}{x}+a-b$ $f(x)=\frac{a x}{a^2-b^2}-\frac{b}{x(a^2-b^2)}+\frac{1}{a+b}$ Now, $x^2 f(x)=\frac{a x^3}{a^2-b^2}-\frac{b x}{a^2-b^2}+\frac{x^2}{a+b}$ Differentiating with respect to $x$: $\frac{d}{d x}\left(x^2 f(x)\right)=\frac{3 a x^2}{a^2-b^2}-\frac{b}{a^2-b^2}+\frac{2 x}{a+b}$ Comparing with $2 x^2+2 x+\frac{1}{3}$: $\frac{3 a}{a^2-b^2}=2$ $(iii)$ $\frac{2}{a+b}=2 \Rightarrow a+b=1$ $(iv)$ $-\frac{b}{a^2-b^2}=\frac{1}{3}$ $(v)$ From $(iv)$, $a^2-b^2=(a-b)(a+b)=a-b$. Substituting into $(iii)$ and $(v)$: $\frac{3 a}{a-b}=2 \Rightarrow 3 a=2 a-2 b \Rightarrow a=-2 b$ $-\frac{b}{a-b}=\frac{1}{3} \Rightarrow -3 b=a-b \Rightarrow a=-2 b$ Since $a+b=1$ and $a=-2 b$, we have $-2 b+b=1 \Rightarrow b=-1$ and $a=2$. Therefore, $a-b=2-(-1)=3$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

If $a f(x)+b f\left(\frac{1}{x}\right)=x+1$, and $\frac{d}{d x}\left(x^2 f(x)\right)=2 x^2+2 x+\frac{1}{3}$, then $a-b=$

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