Q59 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If $f(x) = \dfrac{3^x + 3^{-x} - 2}{\tan x \cdot \log(1+x)}$ for $x \neq 0$, is continuous at $x = 0$, then the value of $f(0)$ is equal to $\ldots$
- A$2\log 3$
- Correct answerB$(\log 3)^2$
- C$\dfrac{1}{2}\log 3$
- D$\log \dfrac{1}{3}$
Solution
Since $f(x)$ is continuous at $x = 0$, $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \dfrac{3^x + 3^{-x} - 2}{\tan x \cdot \log(1+x)}$
$f(0) = \lim_{x \to 0} \dfrac{3^{2x} + 1 - 2 \cdot 3^x}{3^x \tan x \cdot \log(1+x)}$
$f(0) = \lim_{x \to 0} \dfrac{(3^x - 1)^2}{3^x \tan x \cdot \log(1+x)}$
Dividing the numerator and the denominator by $x^2$:
$f(0) = \lim_{x \to 0} \dfrac{\left(\dfrac{3^x - 1}{x}\right)^2}{3^x \left(\dfrac{\tan x}{x}\right) \left(\dfrac{\log(1+x)}{x}\right)}$
Using the standard limits $\lim_{x \to 0} \dfrac{a^x - 1}{x} = \log a$, $\lim_{x \to 0} \dfrac{\tan x}{x} = 1$, and $\lim_{x \to 0} \dfrac{\log(1+x)}{x} = 1$:
$f(0) = \dfrac{(\log 3)^2}{1 \cdot 1 \cdot 1} = (\log 3)^2$
Answer: $(\log 3)^2$
$f(0) = \lim_{x \to 0} \dfrac{3^x + 3^{-x} - 2}{\tan x \cdot \log(1+x)}$
$f(0) = \lim_{x \to 0} \dfrac{3^{2x} + 1 - 2 \cdot 3^x}{3^x \tan x \cdot \log(1+x)}$
$f(0) = \lim_{x \to 0} \dfrac{(3^x - 1)^2}{3^x \tan x \cdot \log(1+x)}$
Dividing the numerator and the denominator by $x^2$:
$f(0) = \lim_{x \to 0} \dfrac{\left(\dfrac{3^x - 1}{x}\right)^2}{3^x \left(\dfrac{\tan x}{x}\right) \left(\dfrac{\log(1+x)}{x}\right)}$
Using the standard limits $\lim_{x \to 0} \dfrac{a^x - 1}{x} = \log a$, $\lim_{x \to 0} \dfrac{\tan x}{x} = 1$, and $\lim_{x \to 0} \dfrac{\log(1+x)}{x} = 1$:
$f(0) = \dfrac{(\log 3)^2}{1 \cdot 1 \cdot 1} = (\log 3)^2$
Answer: $(\log 3)^2$
Hence, the correct answer is option: 'B'.