Question 62

Mathematics Continuity Easy

The point(s) at which function \( f \) is given by \( f(x)=\left\{\begin{array}{ll}\frac{x}{|x|} ; & x<0 \\ -1 ; & x \geq 0\end{array}\right\} \) is continuous is/are

(A) \( x \in R \)
(B) \( x=0 \)
(C) \( x \in R\{0\} \)
(D) -1 and 1
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

To determine the points at which the function \( f \) is continuous, we analyze the function piecewise: \[ f(x) = \begin{cases} \frac{x}{|x|}, & x < 0 \\ -1, & x \geq 0 \end{cases} \] --- Step 1: Analyze Continuity for \( x < 0 \) For \( x < 0 \), \( f(x) = \frac{x}{|x|} \). Since \( |x| = -x \) for \( x < 0 \), we have: \[ f(x) = \frac{x}{-x} = -1 \] Thus, for \( x < 0 \), \( f(x) = -1 \), which is a constant function and hence continuous for all \( x < 0 \). --- Step 2: Analyze Continuity for \( x \geq 0 \) For \( x \geq 0 \), \( f(x) = -1 \), which is also a constant function and hence continuous for all \( x \geq 0 \). --- Step 3: Check Continuity at \( x = 0 \) To check continuity at \( x = 0 \), we evaluate the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at \( x = 0 \): - Left-hand limit (LHL) as \( x \to 0^- \): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = -1 \] - Right-hand limit (RHL) as \( x \to 0^+ \): \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1 \] - Value of the function at \( x = 0 \): \[ f(0) = -1 \] Since LHL = RHL = \( f(0) \), the function is continuous at \( x = 0 \). --- Step 4: Conclusion The function \( f \) is continuous for all real numbers \( x \), including \( x = 0 \). --- Final Answer: The function \( f \) is continuous for all \( x \in \mathbb{R} \). \[ \boxed{A} \]