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 → 0 must equal f(0). We factorize the numerator and use standard limits.

Step 2: Key Formula or Approach:

1. lim (x → 0) (ax - 1)/x = ln a
2. lim (x → 0) (1 - cos kx)/x2 = k2/2

Step 3: Detailed Explanation:

Numerator:
6x - 3x - 2x + 1
= 3x(2x - 1) - (2x - 1)
= (3x - 1)(2x - 1)

Denominator:
1 - cos(x/a) ≈ (x/a)2 / 2 as x → 0

Evaluate limit:

lim (x → 0) [(3x - 1)(2x - 1)] / (x2 / (2a2))

= 2a2 × lim (x → 0) [(3x - 1)/x] × [(2x - 1)/x]

= 2a2 (ln 3)(ln 2)

Given:
f(0) = log 3 × log 4 = ln 3 × 2ln 2 = 2 ln 3 ln 2

Equating limit to f(0):

2a2 ln 3 ln 2 = 2 ln 3 ln 2

a2 = 1

Since a > 0:
a = 1

Step 4: Final Answer:

The value of a is 1.