Question

If $f(x) = \begin{cases} \frac{9^x - 2 \cdot 3^x + 1}{\log(1+3x) \cdot \tan 2x} & , \text{if } x \neq 0 \\ a(\log b)^c & , \text{if } x = 0 \end{cases}$ is continuous at $x = 0$, then $a+b+c =$

• A. $\frac{31}{6}$
• B. $\frac{1}{6}$
• C. $\frac{5}{6}$
• D. $\frac{3}{20}$

Hint:

Standard Limits: $\lim_{x\to 0} \frac{a^x-1}{x} = \log a$; $\lim_{x\to 0} \frac{\log(1+x)}{x} = 1$; $\lim_{x\to 0} \frac{\tan x}{x} = 1$.

Answer 1

The Correct Option is A

Step 1: Simplify Numerator
$9^x - 2 \cdot 3^x + 1 = (3^x)^2 - 2(3^x) + 1 = (3^x - 1)^2$.

Step 2: Evaluate Limit
$\lim_{x \to 0} \frac{(3^x - 1)^2}{\log(1+3x) \cdot \tan 2x} = \lim_{x \to 0} \frac{\left(\frac{3^x - 1}{x}\right)^2 \cdot x^2}{\left(\frac{\log(1+3x)}{3x}\right) \cdot 3x \cdot \left(\frac{\tan 2x}{2x}\right) \cdot 2x}$.
$= \frac{(\log 3)^2 \cdot x^2}{1 \cdot 3x \cdot 1 \cdot 2x} = \frac{(\log 3)^2}{6} = \frac{1}{6}(\log 3)^2$.

Step 3: Compare with $f(0)$
Given $f(0) = a(\log b)^c$. So, $a = 1/6, b = 3, c = 2$.

Step 4: Calculation
$a + b + c = \frac{1}{6} + 3 + 2 = \frac{1}{6} + 5 = \frac{31}{6}$.
Final Answer: (A)