If a+alpha=1, b+beta=2 and af(x)+alpha fleft( | vedclass

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(A) Given the equation: $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x} \quad .....(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$af

Vedclass · Accepted Answer

$2$

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(A) Given the equation: $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x} \quad .....(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$af\left(\frac{1}{x}\right)+\alpha f(x)=b\left(\frac{1}{x}\right)+\beta x \quad .....(2)$Adding equation $(1)$ and equation $(2)$:$(a+\alpha)f(x)+(a+\alpha)f\left(\frac{1}{x}\right) = bx+\frac{\beta}{x}+\frac{b}{x}+\beta x$$(a+\alpha)\left[f(x)+f\left(\frac{1}{x}\right)\right] = (b+\beta)x + (b+\beta)\frac{1}{x}$$(a+\alpha)\left[f(x)+f\left(\frac{1}{x}\right)\right] = (b+\beta)\left(x+\frac{1}{x}\right)$Now,rearrange to find the value of the expression:$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}} = \frac{b+\beta}{a+\alpha}$Given $a+\alpha=1$ and $b+\beta=2$,substitute these values:$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}} = \frac{2}{1} = 2$

If $a+\alpha=1, b+\beta=2$ and $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x}$ for $x \neq 0$,then the value of the expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is ..... .

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