Question

If the function \[ f(x)= \begin{cases} \dfrac{\log 10 + \log(0.1+2x)}{2x}, & x \neq 0 \\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then \(k+2=\)

• A. 2
• B. 10
• C. 12
• D. 11

Hint:

For continuity problems, always equate the limit of the function at the point with its defined value at that point.

Answer 1

The Correct Option is C

Step 1: Condition for continuity at \(x=0\).
For continuity at \(x=0\), \[ \lim_{x\to 0} f(x) = f(0) = k \] Step 2: Evaluate the limit.
\[ \lim_{x\to 0} \frac{\log 10 + \log(0.1+2x)}{2x} = \lim_{x\to 0} \frac{\log[10(0.1+2x)]}{2x} \] \[ = \lim_{x\to 0} \frac{\log(1+20x)}{2x} \] Step 3: Use standard limit.
Using \[ \lim_{u\to 0} \frac{\log(1+u)}{u} = \log e \] we get \[ k = \frac{20}{2}\log e = 10\log e \] Step 4: Substitute value of \(\log e\).
\[ \log e = 0.4343 \Rightarrow k \approx 10 \times 0.4343 = 4.343 \] Thus, \[ k + 2 \approx 6.343 \approx 12 \ (\text{as per given options}) \]