Let f: mathbbR - \0\ to mathbbR$f: \mathbb{R} - \{0\} \to \mathbb{R}$ be a function such that f(x) - 6fleft(frac1xright) = frac353x - frac52$$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$$ If lim_x to 0 left( frac1alpha x + f(x) right) = beta$\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$ for some alpha, beta in mathbbR$\alpha, \beta \in \mathbb{R}$, then alpha + 2beta$\alpha + 2\beta$ is equal to :

### Related Formula For functional equations with inversion, substituting x to frac1x$x \to \frac{1}{x}$ establishes a solvable system of algebraic equations to isolate f(x)$f(x)$ directly. ### Core Logic The given equation is: f(x) - 6fleft(frac1xright) = frac353x - frac52 quad dots (1)$$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \quad \dots (1)$$ Substitute x to frac1x$x \to \frac{1}{x}$ in equation (1): fleft(frac1xright) - 6f(x) = frac35x3 - frac52 quad dots (2)$$f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \quad \dots (2)$$ ### Step 1: Eliminate f(1/x)$f(1/x)$ Multiply equation (2) by 6$6$ and add it to equation (1): left[ f(x) - 6fleft(frac1xright) right] + 6 left[ fleft(frac1xright) - 6f(x) right] = left( frac353x - frac52 right) + 6 left( frac35x3 - frac52 right)$$\left[ f(x) - 6f\left(\frac{1}{x}\right) \right] + 6 \left[ f\left(\frac{1}{x}\right) - 6f(x) \right] = \left( \frac{35}{3x} - \frac{5}{2} \right) + 6 \left( \frac{35x}{3} - \frac{5}{2} \right)$$ f(x) - 36f(x) = frac353x - frac52 + 70x - 15$$f(x) - 36f(x) = \frac{35}{3x} - \frac{5}{2} + 70x - 15$$ -35f(x) = 70x + frac353x - frac352$$-35f(x) = 70x + \frac{35}{3x} - \frac{35}{2}$$ Divide across by -35$-35$: f(x) = -2x - frac13x + frac12$$f(x) = -2x - \frac{1}{3x} + \frac{1}{2}$$ ### Step 2: Evaluate the Limit We are given that the following limit evaluates to a finite constant beta$\beta$: lim_x to 0 left( frac1alpha x + f(x) right) = beta$$\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$$ lim_x to 0 left( frac1alpha x - 2x - frac13x + frac12 right) = beta$$\lim_{x \to 0} \left( \frac{1}{\alpha x} - 2x - \frac{1}{3x} + \frac{1}{2} \right) = \beta$$ lim_x to 0 left( left[ frac1alpha - frac13 right] frac1x - 2x + frac12 right) = beta$$\lim_{x \to 0} \left( \left[ \frac{1}{\alpha} - \frac{1}{3} \right] \frac{1}{x} - 2x + \frac{1}{2} \right) = \beta$$ For the limit to be a finite value, the coefficient of frac1x$\frac{1}{x}$ must vanish completely: frac1alpha - frac13 = 0 implies alpha = 3$$\frac{1}{\alpha} - \frac{1}{3} = 0 \implies \alpha = 3$$ When alpha = 3$\alpha = 3$, the limit simplifies directly to the constant term: beta = lim_x to 0 left( -2x + frac12 right) = frac12$$\beta = \lim_{x \to 0} \left( -2x + \frac{1}{2} \right) = \frac{1}{2}$$ ### Step 3: Calculate Final Value Substitute the determined parameters alpha$\alpha$ and beta$\beta$: alpha + 2beta = 3 + 2left(frac12right) = 3 + 1 = 4$$\alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4$$ ### Pattern Recognition In limit problems involving fractional components where x to 0$x \to 0$, any term like frac1x$\frac{1}{x}$ or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives