Let $f(x)=[x^{2}-x]+|-x+[x]|$, where $x \in \mathbb{R}$ and [t] denotes the greatest integer less than or equal to t. Then, f is

(D) continuous at $x=1$, but not continuous at $x=0$

(A) continuous at $x=0$, but not continuous at $x=1$

(B) continuous at $x=0$ and $x=1$

(C) not continuous at $x=0$ and $x=1$

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Continuity and Differentiability

Let

f (x) = [x^{2} - x] + ∣ - x + [x] ∣

, where

x \in R

and [t] denotes the greatest integer less than or equal to t. Then, f is

Answer: (D)

Rewrite

f (x) = [x (x - 1)] + {x}

where

{x}

is the fractional part. At

x = 0

f (0^{+}) = - 1 + 0 = - 1

f (0) = 0

, so discontinuous. At

x = 1

f (1^{-}) = - 1 + 1 = 0

f (1) = 0

f (1^{+}) = 0

, so continuous. Therefore f is continuous at

x = 1

but not at

x = 0

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option