Q28 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

Let $f (x) = \frac{x (3^{x} - 1)}{1 - \cos x}$ for $x \neq 0 .$ Then value of $f (0),$ which make $f (x)$ continuous at $x = 0,$ is

A
$\log 3$
B
$\frac{1}{2} \log 3$
C
$\frac{1}{2 \log 3}$
Correct answer
D
$2 \log 3$

Solution

\lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{x (3^{x} - 1)}{1 - \cos x}

= \lim_{x \to 0} \frac{3^{x} - 1}{x} \cdot \frac{x^{2}}{1 - \cos x}

= \log 3 \cdot (\frac{1}{\frac{1}{2}}) = 2 \log 3

Hence, the correct answer is option: 'D'.

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