MCQMediumJEE 2023Continuity

JEE Mathematics 2023 Question with Solution

Let f(x)=x2x+xf(x) = \lfloor x^2 - x \rfloor + \lfloor x \rfloor, where xRx \in \mathbb{R} and t| t | denotes the greatest integer less than or equal to tt. Then, ff is:

  • A

    Not continuous at x=0x = 0 and at x=1x = 1

  • B

    Continuous at x=0x = 0 and at x=1x = 1

  • C

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

  • D

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

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: f(x)=x2x+xf(x) = \lfloor x^2 - x \rfloor + \lfloor x \rfloor.

Find: Whether ff is continuous at x=0x = 0 and x=1x = 1.

At x=0x = 0,

f(0)=020+0=0+0=0f(0) = \lfloor 0^2 - 0 \rfloor + \lfloor 0 \rfloor = \lfloor 0 \rfloor + \lfloor 0 \rfloor = 0

As x0x \to 0^- and x0+x \to 0^+, the function approaches the same value. Hence, ff is continuous at x=0x = 0.

At x=1x = 1,

f(1)=121+1=0+1=0+1=1f(1) = \lfloor 1^2 - 1 \rfloor + \lfloor 1 \rfloor = \lfloor 0 \rfloor + \lfloor 1 \rfloor = 0 + 1 = 1

Now check the limits from both sides:

limx1f(x)=0\lim_{x \to 1^-} f(x) = 0

and

limx1+f(x)=0\lim_{x \to 1^+} f(x) = 0

The left-hand limit and right-hand limit are equal, but this common limit is not equal to f(1)=1f(1) = 1. Therefore, ff is not continuous at x=1x = 1.

So, ff is continuous at x=0x = 0, but not continuous at x=1x = 1. The correct option is D.

Common mistakes

  • Confusing the floor function with the modulus function. Here t| t | is explicitly defined as the greatest integer less than or equal to tt, so it means t\lfloor t \rfloor, not absolute value. Always use the given definition.

  • Checking only the function value and not the one-sided limits. Continuity at a point requires left-hand limit, right-hand limit, and function value to be equal. At x=1x = 1, the limits are equal to 00 but f(1)=1f(1) = 1.

  • Assuming floor functions are discontinuous at every integer input without examining the full expression. The combination x2x+x\lfloor x^2 - x \rfloor + \lfloor x \rfloor can still be continuous at some points such as x=0x = 0, so evaluate the actual behavior near the point.

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