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KCET 2019 Math Solved Paper

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Question : 53 of 60

Marks: +1, -0

If f (x) =

\begin{cases} \frac{\sin 3x}{e^{2x}-1}, & x\neq 0 \\ k-2, & x=0 \end{cases}

is continuous at x = 0, then k =

$\frac{9}{5}$
$\frac{2}{3}$
$\frac{3}{2}$
$\frac{1}{2}$
none

Solution:

Given Options are not matching

f (x) =

\begin{cases} \frac{\sin 3x}{e^{2x}-1}, & x\neq 0 \\ k-2, & x=0 \end{cases}

Since f is continuous at x = 0

⇒

\lim\limits_{x\to 0} f(x) = f(0)

\lim\limits_{x\to 0} \frac{\sin 3x}{e^{2x}-1}

= k - 2

⇒

\lim\limits_{x\to 0} \frac{ \frac{\sin 3x}{x} }{ \frac{e^{2x}-1}{x} }

= k - 2

⇒

\frac{ \lim\limits_{x\to 0} \frac{\sin 3x}{x} }{ \lim\limits_{x\to 0} \frac{e^{2x}-1}{x} }

= k - 2

⇒

\frac{3}{2}

= k - 2 ⇒ k =

\frac{3}{2} + 2 = \frac{7}{2}

∴ k =

\frac{7}{2}

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