Let k be a non-zero real number. If
$f(x) = \begin{cases} \frac{\left(e^x-1\right)^2}{sin\left(\frac{x}{k}\right)log\left(1+\frac{x}{4}\right)}, & \text{x $\ne$ 0} \\[2ex] 12, & \text{x = 0} \end{cases}$
is a continuous function, then the value of $k$ is :
- 1
- 2
- 3
- 4
The Correct Option is C
Solution and Explanation
$ \displaystyle\lim _{x \rightarrow \theta}\left\{\frac{\left(e^{x}-1\right)^{2}}{\sin \left(\frac{x}{k}\right) \cdot \ln \left(1+\frac{x}{4}\right)}\right\}=12$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0}\left[\frac{\left(\frac{e^{x}-1}{x}\right)^{2}}{\frac{\sin \left(\frac{x}{k}\right)}{k\left(\frac{x}{k}\right)} \cdot \frac{\ln \left(1+\frac{x}{4}\right)}{4 \cdot\left(\frac{x}{4}\right)}}\right]$
$\Rightarrow \left\{\frac{(1)^{2}}{\left(\frac{1}{k}\right)} \cdot \frac{1}{\frac{1}{4}(1)}\right\}=12$
$\Rightarrow 4 k=12 \Rightarrow k=3$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.