Question ID: #353
Let $f:\mathbb{R}-\{0\}\rightarrow\mathbb{R}$ be a function such that $f(x)-6f\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}$. If $\lim_{x\rightarrow0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta$, where $\alpha, \beta \in \mathbb{R}$, then $\alpha+2\beta$ is equal to
- (1) 3
- (2) 5
- (3) 4
- (4) 6
Solution:
Given the functional equation:
$$ f(x) – 6f\left(\frac{1}{x}\right) = \frac{35}{3x} – \frac{5}{2} \quad \dots(1) $$
Replace $x$ with $\frac{1}{x}$ in equation (1):
$$ f\left(\frac{1}{x}\right) – 6f(x) = \frac{35x}{3} – \frac{5}{2} \quad \dots(2) $$
To find $f(x)$, eliminate $f\left(\frac{1}{x}\right)$. Multiply equation (2) by 6:
$$ 6f\left(\frac{1}{x}\right) – 36f(x) = 70x – 15 \quad \dots(3) $$
Add equation (1) and equation (3):
$$ (f(x) – 36f(x)) + (-6f(1/x) + 6f(1/x)) = \frac{35}{3x} – \frac{5}{2} + 70x – 15 $$
$$ -35f(x) = \frac{35}{3x} + 70x – \frac{35}{2} $$
Divide by $-35$:
$$ f(x) = -\frac{1}{3x} – 2x + \frac{1}{2} $$
Now, substitute $f(x)$ into the given limit:
$$ \lim_{x\rightarrow0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta $$
$$ \lim_{x\rightarrow0} \left( \frac{1}{\alpha x} – \frac{1}{3x} – 2x + \frac{1}{2} \right) = \beta $$
$$ \lim_{x\rightarrow0} \left( \frac{1}{x}\left(\frac{1}{\alpha} – \frac{1}{3}\right) – 2x + \frac{1}{2} \right) = \beta $$
For the limit $\beta$ to exist (be finite), the coefficient of the $\frac{1}{x}$ term must be zero:
$$ \frac{1}{\alpha} – \frac{1}{3} = 0 \Rightarrow \alpha = 3 $$
With this value, the singular term vanishes, and we evaluate the limit:
$$ \beta = \lim_{x\rightarrow0} \left( -2x + \frac{1}{2} \right) = \frac{1}{2} $$
Finally, find the value of $\alpha + 2\beta$:
$$ \alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4 $$
Ans. (3)
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