Let a function be defined by . Then is

continuous for all except for Explanation: To find where the function is continuous, let us first simplify the given function by breaking down the absolute value term . The definition of the absolute value function is: Step 1: Simplify for different intervals of Case 1: When Here, . Substituting this into the function: Case 2: When Here, . Substituting this into the function: Case 3: When The problem explicitly states that . Now, we can rewrite the piecewise function clearly as: Step 2: Check continuity at For a function to be continuous at a point , the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at that point must all be equal: Let's test this at : Left-Hand Limit (LHL) at : Right-Hand Limit (RHL) at : Value of the function at : Since the Left-Hand Limit ( ) is not equal to the Right-Hand Limit (

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MAH-CET 2026 — Mathematics PYQ

MAH-CET | Mathematics | 2026

Let a function $f$ be defined by $f (x) = \frac{x - ∣ x ∣}{x}, x \neq = 0, f (0) = 2$ . Then $f$ is

Choose the correct answer:

Explanation

To find where the function is continuous, let us first simplify the given function $f (x)$ by breaking down the absolute value term $∣ x ∣$ .

The definition of the absolute value function is:

$∣ x ∣ = {x - x am p; if x \geq 0 am p; if x l t; 0$

Step 1: Simplify $f (x)$ for different intervals of $x$

Case 1: When $x > 0$
Here, $∣ x ∣ = x$ . Substituting this into the function:
$f (x) = \frac{x - x}{x} = \frac{0}{x} = 0$
Case 2: When $x < 0$
Here, $∣ x ∣ = - x$ . Substituting this into the function:
$f (x) = \frac{x - ( - x )}{x} = \frac{x + x}{x} = \frac{2 x}{x} = 2$
Case 3: When $x = 0$
The problem explicitly states that $f (0) = 2$ .

Now, we can rewrite the piecewise function clearly as:

$f (x) = ⎩ ⎨ ⎧ 022 am p; if x am p; if x = 0 am p; if x g t; 0 l t; 0$

Step 2: Check continuity at $x = 0$

For a function to be continuous at a point $x = a$ , the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at that point must all be equal:

$x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = f (a)$

Let's test this at $x = 0$ :

Left-Hand Limit (LHL) at $x = 0$ :
$LHL = x \to 0^{-} lim f (x) = x \to 0^{-} lim (2) = 2$
Right-Hand Limit (RHL) at $x = 0$ :
$RHL = x \to 0^{+} lim f (x) = x \to 0^{+} lim (0) = 0$
Value of the function at $x = 0$ :
$f (0) = 2$

Since the Left-Hand Limit ( $LHL = 2$ ) is not equal to the Right-Hand Limit ( $RHL = 0$ ), the limit does not exist at $x = 0$ :

$x \to 0^{-} lim f (x) \neq = x \to 0^{+} lim f (x)$

Therefore, the function is not continuous at $x = 0$ .

Step 3: Check continuity for other values of $x$

For any $x > 0$ , $f (x) = 0$ , which is a constant function and therefore continuous.
For any $x < 0$ , $f (x) = 2$ , which is also a constant function and therefore continuous.

Conclusion

The function is perfectly continuous everywhere except at the single point where the split happens, which is $x = 0$ .

Hence, the correct option is (d) continuous for all $x$ except for $x = 0$ .