Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \([x]\) denotes the greatest integer less than or equal to \(x\). If \(f(x)=[x \sin p x]\), then \(f(x)\) is (A) continuous at \(x=0\) (B) continuous in \((-1,0)\) (C) differentiable at \(x=1\) (D) differentiable in \((-1,1)\)

Step-by-Step Solution

Answer

(A) f(x) is continuous at x=0.

Step 1: Understanding the Function

The function is given as \( f(x) = [x \sin(px)] \), where \([ \cdot ]\) denotes the greatest integer function (also known as the floor function). The floor function outputs the greatest integer less than or equal to a given number. \( f(x) \) involves taking the floor of the product of \( x \) and \( \sin(px) \).

Step 2: Analyze Continuity at Specific Points

To determine continuity at \( x = 0 \), evaluate the left-hand limit, right-hand limit, and \( f(0) \). Since \( \sin(0) = 0 \), \( f(0) = [0 \cdot \sin(0)] = 0 \). The left and right limits as \( x \to 0 \) are also 0 due to the product with \( x \sin(px) \) approaching zero, thus \( f(x) \) is continuous at \( x = 0 \).

Step 3: Examine Continuity on Intervals

For \( x \sin(px) \) to change values as \( x \) moves through \((-1,0)\), it can take on any real value, leading \( [x \sin(px)] \) to possibly change values at every small interval. Hence, \( f(x) \) is not continuous over the interval \((-1,0)\).

Step 4: Evaluate Differentiability at Points

Differentiability at \( x = 1 \) requires continuity at that point and smooth behavior (i.e., no jumps or corners). However, since the floor function creates discrete jumps, \( f(x) \) cannot be differentiable at \( x = 1 \) or anywhere else where \( x \sin(px) \) crosses an integer value.

Step 5: Final Assessment of Differentiability on Intervals

Similarly, because \( f(x) \) is not continuous across small intervals due to the floor function, \( f(x) \) is not differentiable in \((-1,1)\) since differentiability implies continuity.

Key Concepts

Continuity

Continuity refers to a function that does not make abrupt jumps or gaps as its input changes smoothly. For a function to be continuous at a point, the left-hand limit, right-hand limit, and function value at that point must all be equal. Let's consider the function given in the exercise,

• Function: \[f(x) = [x \sin(px)]\],

where \([\cdot ]\) denotes the floor function, which rounds down to the nearest integer.
To check if \(f(x)\) is continuous at \(x = 0\), we compute the limits from both sides and the function value at \(x = 0\):

• Left-hand limit as \(x\) approaches 0 from the negative side: \(x \sin(px)\) gets very close to zero, leading to \([0]\).
• Right-hand limit as \(x\) approaches 0 from the positive side: \(x \sin(px)\) also approaches zero, resulting in \([0]\).
• The function value at \(x = 0\) is simply \(f(0) = [0 \sin(0)] = 0.\)

Since all these values match, \(f(x)\) is continuous at \(x = 0\). However, in the interval \((-1, 0)\), \(x \sin(px)\) varies freely through real values, causing different outputs of the floor function. This makes \(f(x)\) discontinuous in any small span within this interval.

Differentiability

Differentiability of a function at a point implies that the function has a well-defined tangent (slope) at that point. It indicates that the function is smooth and continuous, with no sharp turns or corners. For a function to be differentiable everywhere within an interval, it must also be continuous throughout the entire interval.
Looking at the given function \(f(x) = [x \sin(px)]\), the floor function is inherently non-differentiable because it produces sudden jumps.
To determine if \(f(x)\) is differentiable at \(x = 1\), let's examine the continuity and smooth behavior:

• Since \([x \sin(px)]\) forms steps due to the integers involved, the function cannot be smooth.
• These step changes indicate broken continuity at transition points like integer crossings, confirming it's not differentiable.

Conclusively, since \(f(x)\) is not smooth at integer points like \(x = 1\) and has jumps across any interval, it is non-differentiable both at specific points like \(x = 1\) and over intervals such as \((-1, 1)\).

Floor Function

The floor function, denoted as \([x]\), is a mathematical function that takes a real number and returns the greatest integer less than or equal to that number. It effectively rounds down a number to the nearest whole integer.

• If \( x = 3.6 \), then \([x] = 3\).
• If \( x = -2.4 \), then \([x] = -3\).

In the context of the function \(f(x) = [x \sin(px)]\), the floor function adds complexity by rounding the product of \(x\) and \(\sin(px)\) to the nearest lower integer.
This behavior introduces discontinuities:

• For any small change in \(x\) that causes \(x \sin(px)\) to cross an integer, \([x \sin(px)]\) can abruptly shift values.

Such properties make the floor function especially interesting yet challenging when analyzing the continuity and differentiability of more complex expressions where it is a component.