Consider the following statements in respect of the function y = [x], x ∈ (-1, 1) where [.] is the greatest integer function: 1. Its derivative is 0 at x = 0.5 2. It is continuous at x = 0 Which of the above statements is/are correct?
1 only
The question asks us to analyze the properties of the function \(y = [x]\) within the interval \(x \in (-1, 1)\). The notation \([x]\) represents the greatest integer less than or equal to \(x\). Let's define the function explicitly for the given interval:
We need to evaluate two statements about this function.
Statement 1 says: Its derivative is 0 at \(x = 0.5\).
Let's consider the function \(y = [x]\) in the neighborhood of \(x = 0.5\). The value \(x = 0.5\) falls within the interval \(0 \le x < 1\). In this entire interval, the function is defined as \(y = [x] = 0\).
So, for any value of \(x\) slightly less than or slightly greater than \(0.5\) (as long as it stays within \(0 \le x < 1\)), the function value is a constant, \(0\).
The derivative of a constant function is always 0. Since the function \(y = [x]\) is a constant \(0\) in an open interval around \(x = 0.5\), its derivative at \(x = 0.5\) is 0.
Mathematically, the derivative at \(x=0.5\) is given by the limit: \[ f'(0.5) = \lim_{h \to 0} \frac{f(0.5+h) - f(0.5)}{h} \] For small values of \(h\) (positive or negative) such that \(0 \le 0.5+h < 1\), we have \(f(0.5+h) = [0.5+h] = 0\), and \(f(0.5) = [0.5] = 0\). \[ f'(0.5) = \lim_{h \to 0} \frac{0 - 0}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \]
Therefore, Statement 1 is correct.
Statement 2 says: It is continuous at \(x = 0\).
For a function to be continuous at a point \(x=c\), three conditions must be met:
Let's check these conditions for \(y = [x]\) at \(x = 0\):
Because the limit does not exist, the function is not continuous at \(x=0\). The third condition (limit equals function value) also cannot be met.
Therefore, Statement 2 is incorrect.
Based on our analysis:
Only Statement 1 is correct.
| Statement | Analysis | Correctness |
|---|---|---|
| 1. Derivative at \(x=0.5\) is 0 | Function is \(y=0\) near \(x=0.5\). Derivative of a constant is 0. | Correct |
| 2. Continuous at \(x=0\) | LHL at \(x=0\) is \(-1\). RHL at \(x=0\) is \(0\). LHL \(\ne\) RHL. Limit does not exist. | Incorrect |
| Concept | Description | Relevance to Question |
|---|---|---|
| Greatest Integer Function \([x]\) | Returns the largest integer less than or equal to \(x\). Jumps at integer values. | The function being analyzed. Its definition is key to determining behavior. |
| Derivative of a Function | Measures the instantaneous rate of change. For constant functions, it's 0. | Used to check Statement 1 at a non-integer point. |
| Continuity at a Point | Function must be defined, limit must exist (LHL=RHL), and limit must equal function value. | Used to check Statement 2 at an integer point (where jumps occur). |
| Left-Hand Limit (LHL) | Limit as \(x\) approaches a point from values less than the point. | Crucial for checking continuity at integer points for step functions like \([x]\). |
| Right-Hand Limit (RHL) | Limit as \(x\) approaches a point from values greater than the point. | Crucial for checking continuity at integer points for step functions like \([x]\). |
The greatest integer function, also known as the floor function, has several important properties:
In the given problem, \(x=0.5\) is a non-integer, so the derivative exists and is 0. The point \(x=0\) is an integer, where the function is discontinuous.
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Select the correct answer using the code given below:
\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {3{{\rm{x}}^2} + 12{\rm{x}} - 1,{\rm{\;\;}} - 1 \le {\rm{x}} \le 2}\\ {37 - {\rm{x}},{\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}2 < {\rm{x}} \le 3} \end{array}} \right.\)
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2. f(x) attains greatest value at x = 2
3. f(x) is differentiable at x = 2
Select the correct answer using the code given below:
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