Question

If

$f(x) = \begin{cases} (x^2/a)-a, \text{ when } x < a \\ 0, \text{ when } x = a, \text{ then }\\ a-(x^2/a), \text{ when } x > a \end{cases}$

• A. $\lim_{x \to a} f(x) = a$
• B. $f(x)$ is continuous at $x=a$
• C. $f(x)$ is discontinuous at $x=a$
• D. None of these

Answer 1

Answer: (B)

(b)

Answer 2

To determine the correct answer, we need to analyze the continuity of the function $f(x)$ at $x=a$.

First, we find the left-hand limit (LHL) as $x$ approaches $a$:

$\lim_{x \to a^-} f(x) = \lim_{x \to a^-} (\frac{x^2}{a} - a) = \frac{a^2}{a} - a = a - a = 0$

Next, we find the right-hand limit (RHL) as $x$ approaches $a$:

$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} (a - \frac{x^2}{a}) = a - \frac{a^2}{a} = a - a = 0$

Since LHL = RHL = 0, the limit exists and $\lim_{x \to a} f(x) = 0$.

Now, we check the value of the function at $x=a$:

$f(a) = 0$

Since $\lim_{x \to a} f(x) = f(a) = 0$, the function is continuous at $x=a$. Therefore, option (b) is correct.