Question

Let k be a non – zero real number and the function is defined as
\[f\left( x \right)=\left\{ \begin{matrix}
\dfrac{{{\left( {{e}^{x}}-1 \right)}^{2}}}{\sin \left( \dfrac{x}{k} \right)\log \left( 1+\dfrac{x}{4} \right)}, & x\ne 0 \\
12, &x=0 \\
\end{matrix} \right.\]
If the function \[f\left( x \right)\] is continuous function, then the value of \[k\] is
(a) 2
(b) 4
(c) 3
(d) 1

Answer 1

Hint: We solve this problem by using continuous theorem.
A function \[f\left( x \right)\] is said to be continuous if at the point \[x=a\] where the function is divided satisfies the equation
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
By using the above theorem we find the value of \[k\]
We use some standard results of limits that are
(1) \[\displaystyle \lim_{x \to 0}\dfrac{{{e}^{nx}}-1}{x}=n\]
(2) \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n\]
(3) \[\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+nx \right)}{x}=n\]

Complete step by step answer:
We are given that the function as
\[f\left( x \right)=\left\{ \begin{matrix}
\dfrac{{{\left( {{e}^{x}}-1 \right)}^{2}}}{\sin \left( \dfrac{x}{k} \right)\log \left( 1+\dfrac{x}{4} \right)}, & x\ne 0 \\
12, &x=0 \\
\end{matrix} \right.\]
We are given that this function is a continuous function.
Here we can see that the function is divided at \[x=0\]
We know that a function \[f\left( x \right)\] is said to be continuous if at the point \[x=a\] where the function is divided satisfies the equation
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
By using the above theorem to the given function we get
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}f\left( x \right)=f\left( 0 \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{{{\left( {{e}^{x}}-1 \right)}^{2}}}{\sin \left( \dfrac{x}{k} \right)\log \left( 1+\dfrac{x}{4} \right)}=12 \\
\end{align}\]
Now, let us divide the numerator and denominator with \[{{x}^{2}}\] then we get
\[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\left( \dfrac{{{\left( {{e}^{x}}-1 \right)}^{2}}}{{{x}^{2}}} \right)}{\left( \dfrac{\sin \left( \dfrac{x}{k} \right)\log \left( 1+\dfrac{x}{4} \right)}{{{x}^{2}}} \right)}=12\]
Now, let us separate the terms in such a way that we get the terms in the numerator and denominator as the standard results then we get
\[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{{{\left( \dfrac{{{e}^{x}}-1}{x} \right)}^{2}}}{\left( \dfrac{\sin \left( \dfrac{x}{k} \right)}{x} \right)\left( \dfrac{\log \left( 1+\dfrac{x}{4} \right)}{x} \right)}=12\]

We know that the standard results of limits as
(1) \[\displaystyle \lim_{x \to 0}\dfrac{{{e}^{nx}}-1}{x}=n\]
(2) \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n\]
(3) \[\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+nx \right)}{x}=n\]
By using these standard results to above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( 1 \right)}^{2}}}{\left( \dfrac{1}{k} \right)\left( \dfrac{1}{4} \right)}=12 \\
 & \Rightarrow 4k=12 \\
 & \Rightarrow k=3 \\
\end{align}\]
Therefore we can conclude that the value of \[k\] is 3
So, option (c) is the correct answer.

Note:
We should note that there is a condition for standard results.
We have the standard results as
(1) \[\displaystyle \lim_{x \to 0}\dfrac{{{e}^{nx}}-1}{x}=n\]
(2) \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n\]
(3) \[\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+nx \right)}{x}=n\]
Here, the results are applicable only when the value of \[x\] tends to 0
If the value of \[x\] tends to some other value then these should not be used. In that case we use the simple formula of limits that is
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
This means that we substitute \[x=a\] in the given function.