MAH-CET 2026 — Mathematics PYQ
MAH-CET | Mathematics | 2026Let a function be defined by . Then is

Let a function f be defined by f(x)=xx−∣x∣,x=0,f(0)=2. Then f is
not continuous anywhere
continuous everywhere
continuous for all x except for x=1
continuous for all x except for x=0
(Correct Answer)continuous for all x except for x=0
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−xamp;if x≥0amp;if xlt;0
Case 1: When x > 0
Here, ∣x∣=x. Substituting this into the function:
f(x)=xx−x=x0=0
Case 2: When x < 0
Here, ∣x∣=−x. Substituting this into the function:
f(x)=xx−(−x)=xx+x=x2x=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)=⎩⎨⎧022amp;if xamp;if x=0amp;if xgt;0lt;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→a−limf(x)=x→a+limf(x)=f(a)
Let's test this at x=0:
Left-Hand Limit (LHL) at x=0:
LHL=x→0−limf(x)=x→0−lim(2)=2
Right-Hand Limit (RHL) at x=0:
RHL=x→0+limf(x)=x→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→0−limf(x)=x→0+limf(x)
Therefore, the function is not continuous at x=0.
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.
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.
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−xamp;if x≥0amp;if xlt;0
Case 1: When x > 0
Here, ∣x∣=x. Substituting this into the function:
f(x)=xx−x=x0=0
Case 2: When x < 0
Here, ∣x∣=−x. Substituting this into the function:
f(x)=xx−(−x)=xx+x=x2x=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)=⎩⎨⎧022amp;if xamp;if x=0amp;if xgt;0lt;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→a−limf(x)=x→a+limf(x)=f(a)
Let's test this at x=0:
Left-Hand Limit (LHL) at x=0:
LHL=x→0−limf(x)=x→0−lim(2)=2
Right-Hand Limit (RHL) at x=0:
RHL=x→0+limf(x)=x→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→0−limf(x)=x→0+limf(x)
Therefore, the function is not continuous at x=0.
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.
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.