Question

Let 'a' be a positive real number. If a real valued function \( f(x) = \begin{cases}\frac{6^x - 3^x - 2^x + 1}{1 - \cos\left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x = 0 \end{cases} \) is continuous at \( x=0 \), then \( a = \)

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Factorization of terms like \( (ab)^x - a^x - b^x + 1 \) into \( (a^x-1)(b^x-1) \) is a common pattern in limit problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

For continuity at \( x=0 \), the limit of the function as \( x \to 0 \) must equal \( f(0) \). We factorize the numerator and use standard limits.
Step 2: Key Formula or Approach:

1. \( \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \). 2. \( \lim_{x \to 0} \frac{1-\cos kx}{x^2} = \frac{k^2}{2} \).
Step 3: Detailed Explanation:

Numerator: \( 6^x - 3^x - 2^x + 1 = 3^x(2^x - 1) - 1(2^x - 1) = (3^x - 1)(2^x - 1) \). Denominator: \( 1 - \cos(x/a) \approx \frac{(x/a)^2}{2} \) as \( x \to 0 \). Evaluate Limit: \[ \lim_{x \to 0} \frac{(3^x - 1)(2^x - 1)}{\frac{x^2}{2a^2}} = 2a^2 \lim_{x \to 0} \left( \frac{3^x - 1}{x} \right) \left( \frac{2^x - 1}{x} \right) \] \[ = 2a^2 (\ln 3)(\ln 2) \] Given \( f(0) = \log 3 \log 4 = \ln 3 \cdot 2\ln 2 = 2 \ln 3 \ln 2 \). Equating limit to \( f(0) \): \[ 2a^2 \ln 3 \ln 2 = 2 \ln 3 \ln 2 \] \[ a^2 = 1 \] Since \( a \textgreater 0 \), \( a = 1 \).
Step 4: Final Answer:

The value of \( a \) is 1.