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, $a-b=2-(-1)=3$.

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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