Question

\(\text{ If } f(x), \text{ defined by } f(x) = \begin{cases}  kx + 1 & \text{if } x \leq \pi \\  \cos x & \text{if } x > \pi  \end{cases} \text{ is continuous at } x = \pi, \text{ then the value of } k \text{ is:}\)

• A. 0
• B. \( \pi \)
• C. \( \frac{2}{\pi} \)
• D. \( -\frac{2}{\pi} \)

Hint:

When solving for constants in a function to ensure continuity, it’s essential to set the left-hand limit, right-hand limit, and the value of the function at the point equal to each other. Make sure to equate the limits and solve for the unknown constant. For trigonometric functions like \( \cos \), remember the known values at standard angles (e.g., \( \cos \pi = -1 \)). This approach is critical in piecewise functions and ensuring they are continuous at specific points.

Answer 1

The Correct Option is D

For \( f(x) \) to be continuous at \( x = \pi \), the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at \( x = \pi \) must all be equal.

The left-hand limit is:

\[LHL = \lim_{{x \to \pi^-}} f(x) = k\pi + 1.\]

The right-hand limit is:

\[RHL = \lim_{{x \to \pi^+}} f(x) = \cos \pi = -1.\]

The value of the function at \( x = \pi \) is:

\[f(\pi) = k\pi + 1.\]

Since \( f(x) \) is continuous at \( x = \pi \), we must have:

\[LHL = RHL = f(\pi).\]

Equating the limits:

\[k\pi + 1 = -1.\]

Simplify to solve for \( k \):

\[k\pi = -2 \quad \Rightarrow \quad k = \frac{-2}{\pi}.\]

Thus, the value of \( k \) is:

\[\frac{-2}{\pi}.\]

Answer 2

For \( f(x) \) to be continuous at \( x = \pi \), the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at \( x = \pi \) must all be equal. Continuity of a function at a point means that the function is both defined at that point and that the limits from both sides agree with the function's value at that point. Mathematically, this condition can be written as:

\[ \lim_{{x \to \pi^-}} f(x) = f(\pi) = \lim_{{x \to \pi^+}} f(x). \]

The left-hand limit (LHL) is:

The left-hand limit represents the value that the function approaches as \( x \) approaches \( \pi \) from the left. For the given function, this is expressed as:

\[ LHL = \lim_{{x \to \pi^-}} f(x) = k\pi + 1. \]

The right-hand limit (RHL) is:

The right-hand limit represents the value that the function approaches as \( x \) approaches \( \pi \) from the right. For this problem, we are given that \( \cos \pi = -1 \), so the right-hand limit is:

\[ RHL = \lim_{{x \to \pi^+}} f(x) = \cos \pi = -1. \]

The value of the function at \( x = \pi \) is:

The function value at \( x = \pi \) is simply the function evaluated at that point. Here, it is the same as the left-hand limit expression, i.e.,:

\[ f(\pi) = k\pi + 1. \]

Since \( f(x) \) is continuous at \( x = \pi \), we must have:

To ensure continuity, the left-hand limit, right-hand limit, and function value must be equal. Thus, we equate the limits and function value:

\[ LHL = RHL = f(\pi). \]

Equating the limits:

We now equate the expressions for the left-hand limit and the right-hand limit to solve for \( k \):

\[ k\pi + 1 = -1. \]

Simplify to solve for \( k \):

Now, we can solve the equation for \( k \):

\[ k\pi = -2 \quad \Rightarrow \quad k = \frac{-2}{\pi}. \]

Thus, the value of \( k \) is: 

\[ \frac{-2}{\pi}. \]