Let k be a non-zero real number If f(x) = * 20 c ( e^x - 1 )^2 1m… - Vidyadip

MCQ

Let $k$ be a non-zero real number

If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$

is a continuous function then the value of $k$ is

A
$4$
B
$1$
✓
$3$
D
$2$

✓

Answer

Correct option: C.

$3$

c
Since $f(x)$ is a continuous function therefore

lime it of $x \to 0 = $ value of $f(x)$ at $0$.

$\therefore \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{e^x} - 1} \right)}^2}}}{{\sin \left( {\frac{x}{k}} \right)\log \left( {1 + \frac{x}{4}} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}{{\left( {\frac{{{e^x} - 1}}{x}} \right)}^2}}}{{\frac{x}{R}\left[ {\frac{{\sin \left( {\frac{x}{R}} \right)}}{{\frac{x}{R}}}} \right]\log \frac{{\left( {1 + \frac{x}{4}} \right)}}{{\left( {\frac{x}{4}} \right)}}}} \times \left( {\frac{x}{4}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}{{\left( {\frac{{{e^x} - 1}}{x}} \right)}^2}4k}}{{\frac{{\sin \frac{x}{k}}}{{\frac{x}{k}}}.\frac{{\log \left( {1 + \frac{x}{4}} \right)}}{{\frac{x}{4}}}}}$

on applying limit we get

$4k = 12 \Rightarrow k = 3$

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