Question

Consider a function \(f(x)\) which has exactly two roots at \(x=a\). If \[ \lim_{x\rightarrow a}\left(\frac{\lambda f^{\prime}(x)}{f(x)}-\frac{1}{x-a}\right)=m \ (\ne0), \] then the value of \(\lambda\) is:

• A. 2
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{1}{4}$

Hint:

In general, if $f(x)$ has a root of multiplicity $k$ at $x=a$, the expression $\frac{f'(x)}{f(x)}$ will behave like $\frac{k}{x-a}$ near that root. To balance out an external $-\frac{1}{x-a}$ term, the scaling factor $\lambda$ must always satisfy $\lambda \cdot k = 1 \implies \lambda = \frac{1}{k}$.

Answer 1

The Correct Option is B

Concept: If a differentiable function $f(x)$ possesses a root at $x = a$ with a multiplicity of $k$, it can be analytically factored as $f(x) = (x-a)^k \cdot g(x)$, where $g(a) \neq 0$. When working with limits containing logarithmic derivatives $\frac{f'(x)}{f(x)}$, substituting this factored representation helps isolate the indeterminate pole of order $1$ from the smooth analytical component. Step 1: Factor the function and compute its derivative.
Since $x = a$ is a root of multiplicity $2$, we can write: $$f(x) = (x-a)^2 \cdot g(x)$$ where $g(x)$ is differentiable and $g(a) \neq 0$. Applying the product rule to find $f'(x)$: $$f'(x) = 2(x-a)g(x) + (x-a)^2 g'(x)$$

Step 2: Deconstruct the fractional ratio component.
Form the ratio of the derivative to the original function: $$\frac{f'(x)}{f(x)} = \frac{2(x-a)g(x) + (x-a)^2 g'(x)}{(x-a)^2 g(x)}$$ Divide out the common factor $(x-a)g(x)$ from each individual term in the numerator: $$\frac{f'(x)}{f(x)} = \frac{2}{x-a} + \frac{g'(x)}{g(x)}$$

Step 3: Substitute back into the limit expression and evaluate $\lambda$.
Now, substitute this expanded fractional definition back into the given limit sequence condition: $$\lim_{x\rightarrow a} \left( \lambda \left[ \frac{2}{x-a} + \frac{g'(x)}{g(x)} \right] - \frac{1}{x-a} \right) = m$$ Group the terms sharing the common denominator element $(x-a)$: $$\lim_{x\rightarrow a} \left( \frac{2\lambda - 1}{x-a} + \lambda \frac{g'(x)}{g(x)} \right) = m$$ As $x \rightarrow a$, the second term converges to a finite stable value $\lambda \frac{g'(a)}{g(a)}$. For the total limit to converge to a finite non-zero value $m$ instead of diverging to infinity, the singular term's coefficient must vanish completely: $$2\lambda - 1 = 0 \quad \Rightarrow \quad \lambda = \frac{1}{2}$$