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: Understanding the Question:
For a function to be continuous at \( x=0 \), the limit as \( x \rightarrow 0 \) must equal the function value at \( x=0 \).
Step 3: Detailed Explanation:
Calculate the limit:
\[ \lim_{x \to 0} \frac{9^x - 2 \cdot 3^x + 1}{\log(1+3x) \cdot \tan 2x} = \lim_{x \to 0} \frac{(3^x - 1)^2}{\log(1+3x) \cdot \tan 2x} \]
Divide numerator and denominator by \( x^2 \):
\[ = \frac{\lim_{x \to 0} \left( \frac{3^x - 1}{x} \right)^2}{\lim_{x \to 0} \left( \frac{\log(1+3x)}{x} \cdot \frac{\tan 2x}{x} \right)} \]
Standard limits: \( \frac{3^x-1}{x} \to \log 3 \), \( \frac{\log(1+3x)}{3x} \to 1 \implies \frac{\log(1+3x)}{x} \to 3 \), and \( \frac{\tan 2x}{2x} \to 1 \implies \frac{\tan 2x}{x} \to 2 \).
\[ Limit = \frac{(\log 3)^2}{3 \times 2} = \frac{1}{6} (\log 3)^2 \]
Compare with \( a(\log b)^c \):
\( a = \frac{1}{6}, b = 3, c = 2 \).
Calculate sum: \( a+b+c = \frac{1}{6} + 3 + 2 = \frac{1}{6} + 5 = \frac{31}{6} \).
Step 4: Final Answer:
The sum \( a+b+c = \frac{31}{6} \).