Let f(x)=[x2−x]+|−x+[x]| where x∈R and [t] denotes the greatest integer less than or equal to t. Then, f is

f(x)=[x2−x]+|−x+[x]| Ltx→0− f(x)=Ltx→0−− [x2−x]+|[x]−x| Ltx→0−− 0+|−1−x| = 1 Ltx→0+− f(x)=Ltx→0+− [x2−x]+|[x]−x|=0 ⇒ f discontinuous at x=0 Ltx→1−f(x)=Ltx→1−(−1+|−x|) = −1+1=0 Ltx→1+f(x)=Ltx→1+(0+|1−x|) = 0 f(1)=0 ∴ fis continuous at x=1

Let f(x)=[x2−x]+|−x+[x]| where x∈R and [t] denotes the greatest integer less than or equal to t. Then, f is

f(x)=[x2−x]+|−x+[x]| Ltx→0− f(x)=Ltx→0−− [x2−x]+|[x]−x| Ltx→0−− 0+|−1−x| = 1 Ltx→0+− f(x)=Ltx→0+− [x2−x]+|[x]−x|=0 ⇒ f discontinuous at x=0 Ltx→1−f(x)=Ltx→1−(−1+|−x|) = −1+1=0 Ltx→1+f(x)=Ltx→1+(0+|1−x|) = 0 f(1)=0 ∴ fis continuous at x=1

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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

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Continuous at x=0, but bot continuous at x = 1

Continuous at x = 0 and x = 1

Continuous at x = 1, but not continuous at x = 0

Continuous at x = 0 , but not continuous at x = 1

answer is C.

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Detailed Solution

$f (x) = [x^{2} - x] + | - x + [x] | \underset{x \to 0^{-}}{Lt} f (x) = \underset{x \to {0^{-}}^{-}}{Lt} [x^{2} - x] + | [x] - x | \underset{x \to {0^{-}}^{-}}{Lt} 0 + | - 1 - x | = 1 \underset{x \to {0^{+}}^{-}}{Lt} f (x) = \underset{x \to {0^{+}}^{-}}{Lt} [x^{2} - x] + | [x] - x | = 0 \Rightarrow f discontinuous at x = 0 \underset{x \to 1^{-}}{Lt} f (x) = \underset{x \to 1^{-}}{Lt} (- 1 + | - x |) = - 1 + 1 = 0 \underset{x \to 1^{+}}{Lt} f (x) = \underset{x \to 1^{+}}{Lt} (0 + | 1 - x |) = 0 f (1) = 0 ∴ fis continuous at x = 1$