For what value of $k$ , the function defined by $ f(x) = \begin{cases} \frac{log(1+2x)sin\,x^\circ}{x^2} & \text{for } x \ge \text {0}\\ k & \text{for } x = \text{ 0} \end{cases}$ is continuous at $x = 0$ ?

Given, $p(x) = \begin{cases} \frac{log (1+2x) sin \,x^{\circ}}{x^{2}} & \text{for $x \neq 0$ } \\[2ex] k, & \text{for $x=0$ } \end{cases}$

For what value of $k$, the function defined by $

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Concepts Used:

Continuity

A function is said to be continuous at a point x = a, if

lim_x→a

f(x) Exists, and

lim_x→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.

For what value of $k$, the function defined by $ f(x) = \begin{cases} \frac{log(1+2x)sin\,x^\circ}{x^2} & \text{for } x \ge \text {0}\\ k & \text{for } x = \text{ 0} \end{cases}$ is continuous at $x = 0$ ?

The Correct Option is C

Solution and Explanation

Top Questions on Differentiability

Questions Asked in MHT CET exam

Concepts Used:

Continuity