Question

Let the function
\(f(x)\)=\(\begin{cases}     \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}, & \text{if } x ∈ 0 \\    \space        10 & \text;{if } x = 0 \\ \end{cases}\)
be continuous at x = 0. Then α is equal to

• A. 10
• B. -10
• C. 5
• D. -5

Answer 1

The Correct Option is D

To determine the value of \(\alpha\) that makes the function \(f(x)\) continuous at \(x = 0\), we need to ensure that the left-hand limit (as \(x \to 0\)), the right-hand limit, and the value of the function at \(x = 0\) are equal.

The function is defined as:

• If \(x \neq 0\): \(f(x) = \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}\)
• If \(x = 0\): \(f(x) = 10\)

For continuity, we require:

• \(\lim_{{x \to 0}} f(x) = f(0)\)

Let's calculate the limit \(\lim_{{x \to 0}} f(x)\):

Using the given formula for \(|x \neq 0\):

\(f(x) = \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}\)

This simplifies to:

\(= \frac{1}{x} \left(\log_e\left(\frac{1+5x}{1+\alpha x}\right)\right)\)

As \(x \to 0\), we can use the property \(\log_e(1+y) \approx y\) for small \(y\) and expand the logarithm:

\(= \frac{1}{x} \times \frac{5x - \alpha x}{1} = 5 - \alpha\)

Therefore, \(\lim_{{x \to 0}} f(x) = 5 - \alpha\)

For the function to be continuous at \(x = 0\), this must equal the value of \(f(0)\), which is 10:

\(5 - \alpha = 10\)

Solving for \(\alpha\):

\(\alpha = 5 - 10\)

\(\alpha = -5\)

Thus, the correct value of \(\alpha\) that ensures continuity is \(-5\).

Hence, the correct answer is -5.