(B) For $x > 0$, $|x| = x$, so $f(x) = \frac{x - x}{x} = 0$.For $x < 0$, $|x| = -x$, so $f(x) = \frac{x - (-x)}{x} = \frac{2x}{x} = 2$.At $x = 0$, $f(

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continuous for all $x$ except at $x = 0$

(B) For $x > 0$, $|x| = x$, so $f(x) = \frac{x - x}{x} = 0$.For $x At $x = 0$, $f(0) = 2$.Now, check continuity at $x = 0$:$\lim_{x \to 0^+} f(x) = 0$.$\lim_{x \to 0^-} f(x) = 2$.Since $\lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x)$, the limit does not exist at $x = 0$, so $f$ is discontinuous at $x = 0$.For $x > 0$, $f(x) = 0$ (a constant function), which is continuous.For $x Thus, $f$ is continuous for all $x$ except at $x = 0$.

Class 12 Mathematics Continuity and Differentiation Continuity

Medium

Let the function $f$ be defined by $f(x) = \frac{x - |x|}{x}$ for $x \neq 0$ and $f(0) = 2$. Then $f$ is:

MHT CET 2026 All MHT CET

A
continuous nowhere
B
continuous for all $x$ except at $x = 0$
C
continuous everywhere
D
continuous for all $x$ except at $x = 1$

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Continuity and Differentiation

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Let the function $f$ be defined by $f(x) = \frac{x - |x|}{x}$ for $x \neq 0$ and $f(0) = 2$. Then $f$ is:

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Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$.

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