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 \( f(\pi) \) must be equal.

The left-hand limit is given by:

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

The right-hand limit is:

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

The function's value at \( x = \pi \) is:

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

Continuity at \( x = \pi \) requires:

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

Setting the limits equal yields:

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

Solving for \( k \):

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

Therefore, \( k \) equals:

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