Problem 21 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=|3-x|+(3+x)\), where \((x)\) denotes the least integer greater than or equal to \(x\), then (A) \(f(x)\) is continuous as well as differentiable at \(x=3\) (B) \(f(x)\) is continuous but not differentiable at \(x=3\) (C) \(f(x)\) is differentiable but not continuous at \(x=3\) (D) \(f(x)\) is neither differentiable nor continuous at \(x=3\)

Step-by-Step Solution

Answer

(D) \(f(x)\) is neither differentiable nor continuous at \(x=3\).

Step 1: Analyze Function Components

The given function is \( f(x) = |3-x| + \lceil 3+x \rceil \). The function consists of two parts: an absolute value function \(|3-x|\) and the ceiling function \( \lceil 3+x \rceil \). We need to consider each part separately.

Step 2: Check Continuity of |3-x|

The absolute value function \(|3-x|\) is defined for all real numbers and is continuous everywhere. At \(x=3\), it simplifies to \(|3-3|=0\), hence it is continuous at \(x=3\).

Step 3: Check Continuity of Ceiling Function

The ceiling function \( \lceil 3+x \rceil \) can cause discontinuities because it jumps values at integer inputs. At \(x=3\), \(3+x=6\) which is an integer value, therefore the function has a jump discontinuity here.

Step 4: Conclusion on Continuity of f(x)

Since \(f(x)\) includes the discontinuous ceiling function at \(x=3\), \(f(x)\) is not continuous at \(x=3\).

Step 5: Evaluating Differentiability

Since \(f(x)\) is not continuous at \(x=3\), it cannot be differentiable at \(x=3\) because differentiability implies continuity.

Step 6: Final Conclusion

The correct option based on the dysfunction in continuity and differentiability is that \(f(x)\) is neither differentiable nor continuous at \(x=3\). Therefore, the correct answer is (D).

Key Concepts

Discontinuous Functions

Functions that make sudden jumps or breaks at certain points are known as discontinuous functions. When we talk about continuity, it generally means that the graph of a function can be drawn without lifting your pen off the paper. In mathematical terms, for a function \( f(x) \) to be continuous at a point \( c \), three conditions must be met:

• \( f(c) \) is defined.
• The limit \( \lim_{{x \to c}} f(x) \) exists.
• The limit \( \lim_{{x \to c}} f(x) = f(c) \).

If any of these conditions fail, the function is discontinuous at that point.
In the given problem, the ceiling function part, \( \lceil 3+x \rceil \), creates a jump discontinuity at \( x = 3 \). At this point, the function suddenly shifts value instead of changing gradually. The discontinuity caused by the ceiling function renders \( f(x) = |3-x| + \lceil 3+x \rceil \) discontinuous at \( x = 3 \). Since a discontinuous function cannot be differentiable, it confirms why option (D) is correct.

Ceiling Function Properties

The ceiling function, denoted by \( \lceil x \rceil \), returns the smallest integer greater than or equal to a given number \( x \). This function has unique properties that can lead to sudden jumps in the output value, which contribute to potential discontinuities.
Notably, the ceiling function jumps by 1 at integer values. Hence, when \( x \) is an integer or makes \( 3+x \) an integer, as seen in this problem where \( x=3 \), it will cause \( \lceil 3+x \rceil \) to jump from one integer to the next. This results in a rectangular step-like graph.
Such behavior disrupts continuity, as suggested by the exercise, where the ceiling function part introduces a jump discontinuity at \( x=3 \). Therefore, while the absolute value function \(|3-x|\) is continuous everywhere, the ceiling component is not, making \( f(x) \) discontinuous at this specific point.

Absolute Value Functions

An absolute value function like \(|3-x|\) is defined as always taking the non-negative value of its input. This function is continuous everywhere on the real number line, meaning it has no breaks or jumps.
For an expression \(|a|\), it is defined as:

• \(a\), if \(a \geq 0\)
• -\(a\), if \(a < 0\)

The graph of an absolute value function resembles a "V" shape, smoothly joining parts without any disruptive behavior. As for our function \( |3-x| \), it remains continuous even at \( x = 3 \), simplifying to \( |3-3| = 0 \). It does not create any discontinuity itself.
However, despite its innate continuity, when combined with a discontinuous function like the ceiling function in this exercise, the overall function \( f(x) \) becomes discontinuous where the jump occurs in the ceiling part. This is crucial because for any function to be differentiable at a point, it needs to be continuous there first.