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. $-\frac{2}{\pi}$

Answer 1

Answer: (D)

$-\frac{2}{\pi}$

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.

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:

$\frac{-2}{\pi}.$