Question

Let k be a non-zero real number. If

\( f(x) = \begin{cases} \dfrac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right)\log\left(1+\frac{x}{4}\right)}, & x \ne 0 \\[2ex] 12, & x = 0 \end{cases} \)

is a continuous function, then the value of \( k \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

To determine the value of \(k\) for which the given function \(f(x)\) is continuous at \(x = 0\), we need to ensure that the left-hand limit, the right-hand limit, and the value of the function at this point are all equal. Let's analyze the problem step-by-step:

1. Given the piecewise function:

\(f(x) = \begin{cases} \frac{\left(e^x-1\right)^2}{\sin\left(\frac{x}{k}\right)\log\left(1+\frac{x}{4}\right)}, & x \neq 0 \\[2ex] 12, & x = 0 \end{cases}\)

1. For the function to be continuous at \(x = 0\), the limit of \(f(x)\) as \(x\) approaches 0 must be equal to \(12\). That is, \(\lim_{{x \to 0}} f(x) = f(0) = 12\).
2. Calculate the limit:

\(\lim_{{x \to 0}} \frac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right)\log\left(1 + \frac{x}{4}\right)}\)

1. Using the Taylor series expansions around \(x = 0\):
   • \(e^x - 1 \approx x \, \text{as} \, x \to 0\)
   • \(\sin\left(\frac{x}{k}\right) \approx \frac{x}{k} \, \text{as} \, x \to 0\)
   • \(\log\left(1 + \frac{x}{4}\right) \approx \frac{x}{4} \, \text{as} \, x \to 0\)
2. Substitute these approximations into the limit:

\(\lim_{{x \to 0}} \frac{x^2}{\frac{x}{k} \cdot \frac{x}{4}} = \lim_{{x \to 0}} \frac{x^2}{\frac{x^2}{4k}}\)

\(= \lim_{{x \to 0}} \frac{4k x^2}{x^2} = 4k\)

1. Set this limit equal to the function value at \(x = 0\):

\(4k = 12\)

1. Solve for \(k\):

\(k = \frac{12}{4} = 3\)

1. Therefore, the value of \(k\) is \(3\), which makes the function continuous at \(x = 0\).

Hence, the correct answer is 3.