Question

For \(f(x) = [x^2 + 1]^{[x + 1]}\), where \([x]\) denotes the greatest integer less than or equal to \(x\), which of the following is not true: 

• A. f is a discontinuous function
• B. f(1) = 4
• C. \(\lim_{x \to 1} f(x)\) exists
• D. f is discontinuous at \(x = 2\)

Hint:

Exam Tip:
For functions involving floor functions:

• Check the behavior at integer points.
• Evaluate left and right limits separately.
• The function is often discontinuous at integer points.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We have a function with greatest integer functions (floor functions) in both the base and the exponent. We need to evaluate the function at specific points and check for limits and continuity.

Step 2: Analyzing the Function:
\(f(x) = [x^2 + 1]^{[x + 1]}\).
Let's denote \([ \cdot ]\) as the floor function.
• (B) f(1) = 4:
At \(x = 1\): \([1^2 + 1] = [2] = 2\), \([1 + 1] = [2] = 2\).
So, \(f(1) = 2^2 = 4\). This is true.
• (D) f is discontinuous at \(x = 2\):
At \(x = 2\): \([2^2 + 1] = [5] = 5\), \([2 + 1] = [3] = 3\).
So, \(f(2) = 5^3 = 125\).
We need to check the left-hand limit as \(x \to 2^-\):
For \(x \in [1, 2)\), \([x + 1] = 2\) (since \(x + 1 \in [2, 3)\)).
\([x^2 + 1]\) for \(x \to 2^-\): \(x^2 \to 4^-\), so \(x^2 + 1 \to 5^-\), so \([x^2 + 1] = 4\).
Thus, \(\lim_{x \to 2^-} f(x) = 4^2 = 16\).
But \(f(2) = 125\), so the left-hand limit is not equal to \(f(2)\).
So, f is discontinuous at \(x = 2\). This is true.
• (C) \(\lim_{x \to 1} f(x)\) exists:
We need to check the left-hand and right-hand limits at \(x = 1\).
• Right-hand limit (\(x \to 1^+\)):
For \(x \in [1, 2)\), \([x + 1] = 2\).
\([x^2 + 1]\) for \(x \to 1^+\): \(x^2 \to 1^+\), so \(x^2 + 1 \to 2^+\), so \([x^2 + 1] = 2\).
Thus, \(\lim_{x \to 1^+} f(x) = 2^2 = 4\).
• Left-hand limit (\(x \to 1^-\)):
For \(x \in (0, 1)\), \([x + 1] = 1\) (since \(x + 1 \in (1, 2)\)).
\([x^2 + 1]\) for \(x \to 1^-\): \(x^2 \to 1^-\), so \(x^2 + 1 \to 2^-\), so \([x^2 + 1] = 1\).
Thus, \(\lim_{x \to 1^-} f(x) = 1^1 = 1\). The left-hand limit is 1 and the right-hand limit is 4. They are not equal.
So, \(\lim_{x \to 1} f(x)\) does not exist.
Therefore, option (C) is not true.
• (A) f is a discontinuous function:
Since f is discontinuous at \(x = 1\) (and \(x = 2\)), it is a discontinuous function. This is true.

Step 3: Final Answer:
Option (C) is not true. Therefore, option (C) is correct.