Question

Let $f(x) = \begin{cases} \cos x & \text{if } x \geq 0 \\ -\cos x & \text{if } x<0 \end{cases}$. Which one of the following statements is not true?

• A. $f(x)$ is continuous at $x = 1$
• B. $f(x)$ is continuous at $x = -1$
• C. $f(x)$ is continuous at $x = 2$
• D. $f(x)$ is continuous at $x = -2$
• E. $f(x)$ is continuous at $x = 0$

Hint:

When looking for discontinuities in piecewise functions, focus your attention on the points where the function definition changes. If the limits from both sides don't "meet," the function is not continuous there.

Answer 1

Answer: (E)

Concept: A function is continuous at a point if the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function's value at that point are all equal. Trigonometric functions like $\cos x$ are continuous everywhere in their domain; the only potential point of discontinuity for a piecewise function is at the "break" point ($x = 0$ in this case).

Step 1: Check continuity at $x = 0$.
Calculate the Left-Hand Limit ($x \to 0^-$): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} (-\cos x) = -(\cos 0) = -1 \]
Calculate the Right-Hand Limit ($x \to 0^+$): \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} (\cos x) = \cos 0 = 1 \]

Step 2: Compare the limits.
Since $\text{LHL} (-1) \neq \text{RHL} (1)$, the limit does not exist at $x = 0$. Therefore, the function is discontinuous at $x = 0$.

Step 3: Evaluate the other options.
For any $x \neq 0$, the function follows either $\cos x$ or $-\cos x$, both of which are continuous for all real numbers. Thus, statements (A), (B), (C), and (D) are all true. Statement (E) is the false one.