Continuity and Differentiability

Continuity

Chapter 5: Continuity and Differentiability

Page 1 of 7: Continuity

Concept Introduction

What does it mean for something to be continuous? In everyday language, it means uninterrupted, without a break. Think about the flow of time or the growth of a plant. In mathematics, the concept is remarkably similar. A function is continuous if you can draw its graph without lifting your pen from the paper. There are no sudden jumps, gaps, or holes.

This simple idea is the bedrock of calculus. Why? Because calculus is the study of change, and to study change at a specific instant (which is what a derivative does), the function must be "well-behaved" at that point. It can't suddenly vanish or leap to another value. Continuity is the formal mathematical way of guaranteeing this well-behaved nature.

In your CBSE board exams, the chapter on Continuity and Differentiability is crucial, carrying a weightage of approximately 8-9 marks. Questions are often application-based, asking you to find unknown constants to make a function continuous or to check the continuity of piecewise functions. Mastering this concept is the first essential step toward understanding derivatives and the rest of calculus.

{{FORMULA: expr=lim (x→c⁻) f(x) = lim (x→c⁺) f(x) = f(c) | symbols=lim:limit, x→c⁻:x approaches c from the left, x→c⁺:x approaches c from the right, f(c):value of the function at c}}

Definitions & Formulas

Understanding the formal definitions is key to solving problems rigorously.

{{VISUAL: diagram: A graph showing three functions. The first is a smooth, continuous curve. The second has a single point missing (removable discontinuity). The third has a jump (jump discontinuity).}}

Derivation / Theorem Proof

Let's prove one of the fundamental theorems of the algebra of continuous functions. This demonstrates how the definition of continuity is used in formal proofs.

Theorem: If f and g are two real functions continuous at a real number c, then their product f × g is also continuous at x=c.

Proof:

1. State the Given Information: Since f and g are continuous at x=c, we know from the definition of continuity that: lim (x→c) f(x) = f(c) lim (x→c) g(x) = g(c)

2. Define the New Function: Let the product function be h(x). So, h(x) = (f × g)(x) = f(x) × g(x).

3. State the Goal: To prove that h(x) is continuous at x=c, we must show that lim (x→c) h(x) = h(c).

4. Calculate the Value of the Function at c: First, let's find the value of h(x) at x=c. h(c) = f(c) × g(c)

5. Calculate the Limit of the Function as x approaches c: Now, we find the limit of h(x) as x→c. We will use the property of limits that states the limit of a product is the product of the limits. lim (x→c) h(x) = lim (x→c) [f(x) × g(x)]

6. Apply the Limit Property: lim (x→c) [f(x) × g(x)] = [lim (x→c) f(x)] × [lim (x→c) g(x)]

7. Substitute the Given Information: Using the information from Step 1, we can substitute f(c) and g(c) into the equation. ... = f(c) × g(c)

8. Conclusion: From our calculations, we see that lim (x→c) h(x) = f(c) × g(c) and h(c) = f(c) × g(c). Since lim (x→c) h(x) = h(c), the function h(x) = f(x) × g(x) is continuous at x=c. Hence, proved.

Solved Examples

Let's work through some problems, starting from easy and moving to more challenging ones.

Example 1: Easy (Checking a Polynomial Function)

Given: The function f(x) = 5x - 3. To Find: Whether the function is continuous at x=0, x=-3, and x=5. Approach: For a polynomial function, simply evaluate the limit and the function value at the given point and check if they are equal.

Solution:

A polynomial function is continuous everywhere in its domain (-∞, ∞). Let's verify this.

Case 1: At x = 0

• f(0) = 5(0) - 3 = -3
• lim (x→0) f(x) = lim (x→0) (5x - 3) = 5(0) - 3 = -3
• Since lim (x→0) f(x) = f(0), the function is continuous at x=0.

Case 2: At x = -3

• f(-3) = 5(-3) - 3 = -15 - 3 = -18
• lim (x→-3) f(x) = lim (x→-3) (5x - 3) = 5(-3) - 3 = -18
• Since lim (x→-3) f(x) = f(-3), the function is continuous at x=-3.

Case 3: At x = 5

• f(5) = 5(5) - 3 = 25 - 3 = 22
• lim (x→5) f(x) = lim (x→5) (5x - 3) = 5(5) - 3 = 22
• Since lim (x→5) f(x) = f(5), the function is continuous at x=5.

Final Answer:

The function f(x) = 5x - 3 is continuous at x=0, x=-3, and x=5.

Example 2: Medium (Piecewise Function)

Given: The function f(x) defined as: f(x) = { x + 2, if x ≤ 1 f(x) = { x - 2, if x > 1 To Find: Check if the function f(x) is continuous at x=1. Approach: Calculate the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and f(1) at the break-point x=1 and check if all three are equal.

{{KEY: type=concept | title=The Three-Point Check for Continuity | text=For any function f(x) at a point x=c, you must check three things: 1. f(c) is defined. 2. lim (x→c) f(x) exists (meaning LHL = RHL). 3. The limit equals the function value: lim (x→c) f(x) = f(c). If any one of these fails, the function is discontinuous.}}

Solution:

The point we need to investigate is x=1, as the function definition changes here.

1. Calculate f(1): For x=1, we use the first piece of the function (x ≤ 1): f(1) = 1 + 2 = 3

2. Calculate LHL (Left-Hand Limit): As x approaches 1 from the left, x < 1, so we use the first piece f(x) = x + 2. LHL = lim (x→1⁻) f(x) = lim (x→1⁻) (x + 2) Substitute x=1: LHL = 1 + 2 = 3

3. Calculate RHL (Right-Hand Limit): As x approaches 1 from the right, x > 1, so we use the second piece f(x) = x - 2. RHL = lim (x→1⁺) f(x) = lim (x→1⁺) (x - 2) Substitute x=1: RHL = 1 - 2 = -1

4. Compare LHL, RHL, and f(1): We have LHL = 3 and RHL = -1. Since LHL ≠ RHL, the limit lim (x→1) f(x) does not exist.

Final Answer:

Since LHL ≠ RHL at x=1, the function f(x) is not continuous (discontinuous) at x=1.

Example 3: Hard (Finding an Unknown Constant)

Given: The function f(x) is continuous at x=0. f(x) = { (1 - cos(4x)) / x² , if x ≠ 0 f(x) = { k , if x = 0 To Find: The value of the constant k. Approach: Since the function is continuous at x=0, we must have lim (x→0) f(x) = f(0). We will calculate the limit and equate it to f(0).

Solution:

1. Identify f(0): From the function definition, f(0) = k.

2. Calculate the Limit at x=0: We need to find lim (x→0) f(x). For x ≠ 0, f(x) = (1 - cos(4x)) / x². lim (x→0) (1 - cos(4x)) / x²

3. Use Trigonometric Identities: This is a 0/0 indeterminate form. We use the identity 1 - cos(2θ) = 2sin²(θ). Here, 2θ = 4x, so θ = 2x. Thus, 1 - cos(4x) = 2sin²(2x). The limit becomes: lim (x→0) (2sin²(2x)) / x²

4. Simplify and Use the Standard Limit lim (θ→0) sin(θ)/θ = 1: lim (x→0) 2 × (sin(2x)/x)² To use the standard limit, we need 2x in the denominator. We multiply and divide by 2² = 4. lim (x→0) 2 × (sin(2x)/(2x) × 2)² = lim (x→0) 2 × (sin(2x)/(2x))² × 4 = 8 × [lim (x→0) sin(2x)/(2x)]² As x→0, 2x→0. So, the limit in the bracket is 1. = 8 × (1)² = 8

5. Equate the Limit and f(0): For continuity, lim (x→0) f(x) = f(0). 8 = k

Final Answer:

The value of k for which the function is continuous at x=0 is 8.

Example 4: JEE-tricky (Greatest Integer Function)

Given: The function f(x) = x - [x], where [x] is the greatest integer function. To Find: Discuss the continuity of f(x) at x=3. Approach: Check the LHL and RHL at the integer point x=3. The greatest integer function behaves differently just to the left and right of an integer.

Solution:

The function f(x) = x - [x] is the fractional part function, usually denoted as {x}.

1. Calculate f(3): f(3) = 3 - [3] = 3 - 3 = 0

2. Calculate LHL (Left-Hand Limit): LHL = lim (x→3⁻) (x - [x]) Let x = 3 - h where h is a small positive number (h→0). LHL = lim (h→0) ((3 - h) - [3 - h]) For a small h > 0, 3 - h is slightly less than 3 (e.g., 2.999). So, [3 - h] = 2. LHL = lim (h→0) (3 - h - 2) = lim (h→0) (1 - h) = 1

3. Calculate RHL (Right-Hand Limit): RHL = lim (x→3⁺) (x - [x]) Let x = 3 + h where h→0. RHL = lim (h→0) ((3 + h) - [3 + h]) For a small h > 0, 3 + h is slightly more than 3 (e.g., 3.001). So, [3 + h] = 3. RHL = lim (h→0) (3 + h - 3) = lim (h→0) h = 0

4. Compare: We have LHL = 1 and RHL = 0. Since LHL ≠ RHL, the function is discontinuous at x=3.

Final Answer:

The function f(x) = x - [x] is discontinuous at x=3. In fact, it is discontinuous at every integer point.

Example 5: NEET-style (Oscillating Function)

Given: The function f(x) defined as: f(x) = { x² sin(1/x) , if x ≠ 0 f(x) = { 0 , if x = 0 To Find: Determine if f(x) is continuous at x=0. Approach: Use the Squeeze Theorem (or Sandwich Theorem) to evaluate the limit at x=0, as sin(1/x) oscillates rapidly.

Solution:

1. Identify f(0): The function is defined to be f(0) = 0.

2. Evaluate the Limit lim (x→0) f(x): We need to find lim (x→0) x² sin(1/x). We know that the sine function is always bounded between -1 and 1. -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.

3. Apply the Squeeze Theorem: Multiply the inequality by x². Since x² ≥ 0, the inequality signs do not change. -x² ≤ x² sin(1/x) ≤ x²

4. Take the Limit of the Bounding Functions: Now, take the limit as x→0 for all parts of the inequality. lim (x→0) (-x²) ≤ lim (x→0) (x² sin(1/x)) ≤ lim (x→0) (x²) 0 ≤ lim (x→0) (x² sin(1/x)) ≤ 0

5. Conclude the Limit: Since the limit of our function is "squeezed" between 0 and 0, it must be 0. lim (x→0) x² sin(1/x) = 0

6. Compare with f(0): We found that lim (x→0) f(x) = 0 and it is given that f(0) = 0. Since lim (x→0) f(x) = f(0), the function is continuous.

Final Answer:

The function f(x) is continuous at x=0.

Tips & Tricks (Shortcuts)

Common Mistakes / Sign-Errors

Concept-Trap Questions

Test your understanding with these tricky questions that often appear in exams.

1. Is the function f(x) = (x² - 4) / (x - 2) continuous at x=2?

💡 Answer & Trap: The function is not continuous at x=2. The Trap: Many students simplify the function to f(x) = x + 2 and substitute x=2 to get 4, concluding it's continuous. However, the original function f(x) is not defined at x=2 because the denominator becomes zero. Since f(2) is not defined, the first condition of continuity fails. This is a removable discontinuity. For it to be continuous, it would have to be defined as a piecewise function, like f(2) = 4.

2. A function is defined as f(x) = 1 if x is rational, and f(x) = 0 if x is irrational. Is this function (known as Dirichlet's function) continuous anywhere?

💡 Answer & Trap: This function is nowhere continuous. The Trap: It's tempting to think it might be continuous at the points where the value is defined. However, for any real number c you choose (rational or irrational), any tiny interval around c contains both rational and irrational numbers. This means that as x approaches c, the value of f(x) jumps wildly between 0 and 1. The limit lim (x→c) f(x) never exists. Therefore, it's discontinuous everywhere.

3. If f(x) and g(x) are two functions that are discontinuous at x=c, can their sum f(x) + g(x) be continuous at x=c?

💡 Answer & Trap: Yes, it is possible. The Trap: The algebra of continuous functions states that the sum of two continuous functions is continuous. It does not say anything about the sum of two discontinuous functions. Example: Let f(x) = [x] and g(x) = -[x]. Both are discontinuous at every integer, say c=1. f(x) + g(x) = [x] - [x] = 0. The sum function h(x) = 0 is a constant function, which is continuous everywhere, including at x=1.

Mini Cheatsheet

Differentiability

Page 2: Differentiability

Concept Introduction

On the previous page, we explored continuity, the idea of a function's graph being an unbroken curve. Now, we ask a more demanding question: Is the curve smooth? Imagine zooming into a point on a graph. If the curve starts looking more and more like a straight line, it's smooth at that point. This "smoothness" is the core idea of differentiability.

Geometrically, differentiability at a point means that a unique, non-vertical tangent line can be drawn to the curve at that point. The slope of this tangent line is the derivative of the function at that point, representing the instantaneous rate of change. Think of it this way:

• Continuity: Can you draw the graph without lifting your pen?
• Differentiability: Is the path you're drawing free of sharp corners or cusps?

In your CBSE exams, questions on differentiability often come combined with continuity. This chapter forms the bedrock of calculus, and this specific topic contributes significantly to the ~8-10 marks allocated to the "Continuity and Differentiability" unit. Mastering this concept is crucial for understanding applications of derivatives later on.

{{FORMULA: expr=f'(c) = lim(h→0) [f(c+h) - f(c)] / h | symbols=f'(c):Derivative at point c, h:infinitesimal change in x}}

Definitions & Formulas

The concept of differentiability is formally defined using limits. A function f(x) is said to be differentiable at a point x = c if the following limit exists finitely.

Derivation / Theorem Proof

One of the most important theorems in this chapter connects the concepts of continuity and differentiability.

Theorem: If a function f is differentiable at a point c, then it is also continuous at that point.

Let's prove this foundational result.

Given: A function f(x) is differentiable at x = c. This means that the limit f'(c) = lim(x→c) [f(x) - f(c)] / (x - c) exists and is finite.

To Prove: The function f(x) is continuous at x = c. This means we need to show that lim(x→c) f(x) = f(c), or equivalently, lim(x→c) [f(x) - f(c)] = 0.

Proof:

1. Start with the expression f(x) - f(c). We want to show this approaches 0 as x → c.
2. We can cleverly rewrite this expression by multiplying and dividing by (x - c), assuming x ≠ c. f(x) - f(c) = ([f(x) - f(c)] / (x - c)) × (x - c)
3. Now, let's take the limit of both sides as x → c. lim(x→c) [f(x) - f(c)] = lim(x→c) {([f(x) - f(c)] / (x - c)) × (x - c)}
4. Using the algebra of limits (limit of a product is the product of limits), we can separate the terms: lim(x→c) [f(x) - f(c)] = (lim(x→c) [f(x) - f(c)] / (x - c)) × (lim(x→c) (x - c))
5. From our "Given" information, we know that the first limit is the definition of the derivative, f'(c). lim(x→c) [f(x) - f(c)] / (x - c) = f'(c)
6. The second limit is straightforward. As x approaches c, (x - c) approaches 0. lim(x→c) (x - c) = c - c = 0
7. Substituting these results back into our equation from step 4: lim(x→c) [f(x) - f(c)] = f'(c) × 0
8. Since f'(c) is a finite value (as f is differentiable), the product is 0. lim(x→c) [f(x) - f(c)] = 0 This implies lim(x→c) f(x) = f(c). This is precisely the definition of continuity at x = c.

Conclusion: We have successfully proven that any function differentiable at a point must be continuous at that point.

{{KEY: type=concept | title=Differentiability implies Continuity | text=Every differentiable function is continuous. However, the converse is not true! A function can be continuous at a point but not differentiable there. The classic example is f(x) = |x| at x = 0. Its graph has a sharp corner, making it non-differentiable, but it is perfectly continuous.}}

Solved Examples

Example 1: Easy (Checking a Polynomial)

Given: The function f(x) = 3x² + 5. To Find: Show that f(x) is differentiable at x = 1. Approach: Calculate the LHD and RHD at x = 1 using the first principle and show they are equal.

Step-by-step Solution: We need to check the differentiability at c = 1. First, f(1) = 3(1)² + 5 = 8.

1. Calculate RHD (Right-Hand Derivative): R f'(1) = lim(h→0⁺) [f(1+h) - f(1)] / h = lim(h→0⁺) [ (3(1+h)² + 5) - 8 ] / h = lim(h→0⁺) [ 3(1 + 2h + h²) + 5 - 8 ] / h = lim(h→0⁺) [ 3 + 6h + 3h² - 3 ] / h = lim(h→0⁺) [ 6h + 3h² ] / h = lim(h→0⁺) h(6 + 3h) / h = lim(h→0⁺) (6 + 3h) = 6 + 3(0) = 6

2. Calculate LHD (Left-Hand Derivative): L f'(1) = lim(h→0⁺) [f(1-h) - f(1)] / (-h) = lim(h→0⁺) [ (3(1-h)² + 5) - 8 ] / (-h) = lim(h→0⁺) [ 3(1 - 2h + h²) + 5 - 8 ] / (-h) = lim(h→0⁺) [ 3 - 6h + 3h² - 3 ] / (-h) = lim(h→0⁺) [ -6h + 3h² ] / (-h) = lim(h→0⁺) -h(6 - 3h) / (-h) = lim(h→0⁺) (6 - 3h) = 6 - 3(0) = 6

3. Compare LHD and RHD: Since LHD = RHD = 6, the function is differentiable at x = 1.

Final Answer:

LHD = 6 and RHD = 6. Since LHD = RHD, the function f(x) = 3x² + 5 is differentiable at x = 1, and f'(1) = 6.

Example 2: Medium (Modulus Function - The Classic Case)

Given: The function f(x) = |x - 2|. To Find: Check the differentiability of f(x) at x = 2. Approach: Define the piecewise nature of the modulus function around x=2 and then calculate LHD and RHD.

Step-by-step Solution: First, let's write f(x) as a piecewise function: f(x) = { (x - 2), if x ≥ 2 { -(x - 2), if x < 2 Also, f(2) = |2 - 2| = 0.

{{VISUAL: diagram: A graph showing the V-shape of the function f(x) = |x-2|, with the sharp corner (vertex) located at the point (2, 0) on the x-axis.}}

1. Calculate RHD at x = 2: For x > 2, we use f(x) = x - 2. R f'(2) = lim(h→0⁺) [f(2+h) - f(2)] / h = lim(h→0⁺) [ |(2+h) - 2| - 0 ] / h = lim(h→0⁺) [ |h| ] / h Since h → 0⁺, h is positive, so |h| = h. = lim(h→0⁺) h / h = 1

2. Calculate LHD at x = 2: For x < 2, we use f(x) = -(x - 2) = 2 - x. L f'(2) = lim(h→0⁺) [f(2-h) - f(2)] / (-h) = lim(h→0⁺) [ |(2-h) - 2| - 0 ] / (-h) = lim(h→0⁺) [ |-h| ] / (-h) Since h is positive, -h is negative, but |-h| = h. = lim(h→0⁺) h / (-h) = -1

3. Compare LHD and RHD: Here, LHD = -1 and RHD = 1. Since LHD ≠ RHD, the function is not differentiable at x = 2.

Final Answer:

LHD = -1 and RHD = 1. Since LHD ≠ RHD, the function f(x) = |x - 2| is not differentiable at x = 2.

Example 3: Hard (Piecewise Function)

Given: The function f(x) = { ax + 1, if x ≤ 3 { bx² - 3, if x > 3 To Find: The values of a and b so that f(x) is differentiable at x = 3. Approach: For differentiability, the function must first be continuous. Use the continuity condition to get one equation, and the LHD = RHD condition to get a second equation. Solve the system.

Step-by-step Solution:

1. Condition for Continuity at x = 3: LHL = RHL = f(3) LHL = lim(x→3⁻) f(x) = lim(x→3⁻) (ax + 1) = 3a + 1 RHL = lim(x→3⁺) f(x) = lim(x→3⁺) (bx² - 3) = b(3)² - 3 = 9b - 3 f(3) = a(3) + 1 = 3a + 1 So, 3a + 1 = 9b - 3 3a - 9b = -4 ...(Equation 1)

2. Condition for Differentiability at x = 3: LHD = RHD Calculate LHD: L f'(3) = lim(h→0⁺) [f(3-h) - f(3)] / (-h) = lim(h→0⁺) [ (a(3-h) + 1) - (3a + 1) ] / (-h) = lim(h→0⁺) [ 3a - ah + 1 - 3a - 1 ] / (-h) = lim(h→0⁺) [ -ah ] / (-h) = a So, LHD = a.

   Calculate RHD: R f'(3) = lim(h→0⁺) [f(3+h) - f(3)] / h Note: we must use the value of f(3) from the continuity condition. f(3)=3a+1 = lim(h→0⁺) [ (b(3+h)² - 3) - (3a + 1) ] / h Since we know the function is continuous, we can substitute 3a+1 with 9b-3. = lim(h→0⁺) [ (b(9 + 6h + h²) - 3) - (9b - 3) ] / h = lim(h→0⁺) [ 9b + 6bh + bh² - 3 - 9b + 3 ] / h = lim(h→0⁺) [ 6bh + bh² ] / h = lim(h→0⁺) h(6b + bh) / h = 6b So, RHD = 6b.

   For differentiability, LHD = RHD, which means a = 6b ...(Equation 2)

3. Solve the System of Equations: Substitute a = 6b from Equation 2 into Equation 1: 3(6b) - 9b = -4 18b - 9b = -4 9b = -4 b = -4/9

   Now find a: a = 6b = 6(-4/9) = -24/9 = -8/3

Final Answer:

For f(x) to be differentiable at x = 3, the values must be a = -8/3 and b = -4/9.

Example 4: JEE-tricky (Greatest Integer Function)

Given: The function f(x) = x [x], where [x] is the greatest integer function. To Find: Discuss the differentiability of f(x) at x = 2. Approach: Check for continuity first. If continuous, then calculate LHD and RHD at x=2, paying close attention to how [x] behaves on either side of 2.

Step-by-step Solution:

1. Check for Continuity at x = 2: f(2) = 2[2] = 2 × 2 = 4 LHL = lim(x→2⁻) x[x]. As x approaches 2 from the left, x is like 1.999, so [x] = 1. LHL = lim(x→2⁻) x(1) = 2 RHL = lim(x→2⁺) x[x]. As x approaches 2 from the right, x is like 2.001, so [x] = 2. RHL = lim(x→2⁺) x(2) = 4 Since LHL ≠ RHL, the function is not continuous at x = 2.

2. Conclusion about Differentiability: We know that if a function is not continuous at a point, it cannot be differentiable at that point. There is no need to calculate LHD and RHD.

Final Answer:

The function f(x) = x[x] is not continuous at x = 2. Since differentiability implies continuity, the function is not differentiable at x = 2.

Example 5: CBSE-HOTS Style (Advanced Piecewise)

Given: f(x) = { |x-1| + a, if x ≤ 1 { b(x²-1), if x > 1 To Find: Find a and b if f(x) is differentiable at x = 1. Approach: Rewrite the modulus function, then apply continuity and differentiability conditions to form two equations and solve for a and b.

Step-by-step Solution:

1. Rewrite the function: For x ≤ 1, x-1 ≤ 0, so |x-1| = -(x-1) = 1-x. So, f(x) = { 1 - x + a, if x ≤ 1 { b(x²-1), if x > 1

2. Apply Continuity at x = 1: LHL = RHL = f(1) f(1) = 1 - 1 + a = a LHL = lim(x→1⁻) (1 - x + a) = 1 - 1 + a = a RHL = lim(x→1⁺) b(x² - 1) = b(1² - 1) = 0 For continuity, LHL = RHL, so a = 0.

3. Apply Differentiability at x = 1 (with a = 0): Our function is now: f(x) = { 1 - x, if x ≤ 1 { b(x²-1), if x > 1 And we know f(1) = a = 0. LHD = RHD

   Calculate LHD: L f'(1) = lim(h→0⁺) [f(1-h) - f(1)] / (-h) = lim(h→0⁺) [ (1 - (1-h)) - 0 ] / (-h) = lim(h→0⁺) [ h ] / (-h) = -1

   Calculate RHD: R f'(1) = lim(h→0⁺) [f(1+h) - f(1)] / h = lim(h→0⁺) [ b((1+h)² - 1) - 0 ] / h = lim(h→0⁺) [ b(1 + 2h + h² - 1) ] / h = lim(h→0⁺) [ b(2h + h²) ] / h = lim(h→0⁺) bh(2 + h) / h = lim(h→0⁺) b(2+h) = 2b

   For differentiability, LHD = RHD, so -1 = 2b. This gives b = -1/2.

Final Answer:

For f(x) to be differentiable at x = 1, the values must be a = 0 and b = -1/2.

Tips & Tricks (Shortcuts)

Common Mistakes / Sign-Errors

Concept-Trap Questions

Here are a few exam-style questions designed to test the subtleties of differentiability.

Question 1: A function f(x) is defined as f(x) = (x² - 4) / (x - 2) for x ≠ 2. Can f(x) be differentiable at x = 2?

💡 Answer & Trap: The function is not defined at x=2, so it's not continuous at x=2. Since it's not continuous, it cannot be differentiable at x=2. The Trap: Students might simplify f(x) to x+2 and conclude its derivative is 1 everywhere. However, the original function has a hole at x=2. For differentiability to even be considered, the function must be defined and continuous at the point. You could make it continuous by defining f(2)=4, and then it would be differentiable.

Question 2: Consider the function f(x) = x |x|. Is it differentiable at x = 0?

💡 Answer & Trap: Yes, it is differentiable at x = 0. Let's check: f(x) = { x², if x ≥ 0 and f(x) = { -x², if x < 0. f(0)=0. RHD at x=0: lim(h→0⁺) [f(h) - f(0)] / h = lim(h→0⁺) [h² - 0] / h = lim h = 0. LHD at x=0: lim(h→0⁺) [f(-h) - f(0)] / (-h) = lim(h→0⁺) [-(-h)² - 0] / (-h) = lim(h→0⁺) [-h²] / (-h) = lim h = 0. Since LHD = RHD = 0, the function is differentiable at x=0 and f'(0)=0. The Trap: Seeing the modulus function |x| makes students instinctively think "not differentiable at 0". But when multiplied by x, the function becomes smooth enough at the origin to be differentiable.

Question 3: If the derivative of a function f(x) exists at x=c, does the derivative f'(x) have to be continuous at x=c?

💡 Answer & Trap: No, not necessarily. A function can be differentiable at a point, but its derivative might not be continuous at that same point. A standard (though advanced for CBSE) example is f(x) = x² sin(1/x) for x≠0 and f(0)=0. This function is differentiable at x=0 with f'(0)=0, but f'(x) for x≠0 is 2x sin(1/x) - cos(1/x), which oscillates infinitely near x=0 and does not have a limit. The Trap: The common assumption is that if a function is "nice" enough to be differentiable, its derivative must also be a "nice" continuous function. This is usually true for polynomials and simple functions but is not a universal rule.

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