Class 12 Mathematics Notes Chapter 5 (Continuity and differentiability) – Mathematics Part-I Book

Alright class, let's delve into Chapter 5: Continuity and Differentiability. This is a foundational chapter for calculus and frequently tested in various government examinations. Pay close attention to the definitions, standard results, and techniques.

Chapter 5: Continuity and Differentiability - Detailed Notes

1. Continuity

• Intuitive Idea: A function is continuous if you can draw its graph without lifting your pen. There are no breaks, jumps, or holes in the graph.
• Formal Definition (Continuity at a Point): A function f(x) is said to be continuous at a point x = c in its domain if:
  1. f(c) is defined (the function value exists at c).
  2. The limit of the function as x approaches c exists, i.e., lim (x→c) f(x) exists.
  3. The limit equals the function value, i.e., lim (x→c) f(x) = f(c).
• Limit Existence (LHL & RHL): For the limit lim (x→c) f(x) to exist, the Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL).
  • LHL: lim (x→c⁻) f(x) = lim (h→0) f(c-h)
  • RHL: lim (x→c⁺) f(x) = lim (h→0) f(c+h)
  • Thus, for continuity at x=c, we need: lim (h→0) f(c-h) = lim (h→0) f(c+h) = f(c).
• Continuity in an Interval:
  • Open Interval (a, b): f(x) is continuous in (a, b) if it is continuous at every point in the interval.
  • Closed Interval [a, b]: f(x) is continuous in [a, b] if:
    • It is continuous in the open interval (a, b).
    • It is continuous from the right at a, i.e., lim (x→a⁺) f(x) = f(a).
    • It is continuous from the left at b, i.e., lim (x→b⁻) f(x) = f(b).
• Algebra of Continuous Functions: If f and g are two real functions continuous at a real number c, then:
  • f + g is continuous at x = c.
  • f - g is continuous at x = c.
  • f * g is continuous at x = c.
  • f / g is continuous at x = c, provided g(c) ≠ 0.
  • k * f is continuous at x = c, where k is a constant.
  • Composition: If f is continuous at c and g is continuous at f(c), then the composite function g(f(x)) is continuous at c.
• Standard Continuous Functions:
  • Polynomial functions (P(x) = a_n x^n + ... + a_1 x + a_0) are continuous everywhere (ℝ).
  • Rational functions (P(x)/Q(x), where P, Q are polynomials) are continuous everywhere except where the denominator Q(x) = 0.
  • Trigonometric functions (sin x, cos x) are continuous everywhere. tan x, cot x, sec x, csc x are continuous in their respective domains.
  • Inverse trigonometric functions are continuous in their respective domains.
  • Exponential function (a^x, a > 0, a ≠ 1, especially e^x) is continuous everywhere.
  • Logarithmic function (log_a x, a > 0, a ≠ 1, especially ln x) is continuous in its domain (x > 0).
  • Modulus function (|x|) is continuous everywhere.
• Discontinuity: A function is discontinuous at x=c if any of the three conditions for continuity fail.

2. Differentiability

• Intuitive Idea: A function is differentiable at a point if its graph is "smooth" at that point, meaning there are no sharp corners or cusps, and there's a unique tangent line.
• Formal Definition (Differentiability at a Point): A function f(x) is said to be differentiable (or derivable) at a point x = c in its domain if the following limit exists:
  f'(c) = lim (h→0) [f(c+h) - f(c)] / h
  This limit, if it exists, is called the derivative of f at c.
• Existence of Derivative (LHD & RHD): For the derivative f'(c) to exist, the Left-Hand Derivative (LHD) must equal the Right-Hand Derivative (RHD).
  • LHD at c: L f'(c) = lim (h→0⁻) [f(c+h) - f(c)] / h = lim (h→0⁺) [f(c-h) - f(c)] / (-h)
  • RHD at c: R f'(c) = lim (h→0⁺) [f(c+h) - f(c)] / h
  • Thus, for differentiability at x=c, we need: L f'(c) = R f'(c).
• Key Theorem: Differentiability implies Continuity: If a function f is differentiable at a point c, then it is also continuous at that point c.
• Important Note: The converse is NOT true. A function can be continuous at a point but not differentiable there.
  • Standard Example: f(x) = |x| is continuous at x = 0 but not differentiable at x = 0 (LHD = -1, RHD = +1).
• Algebra of Derivatives: If u and v are differentiable functions of x, then:
  • Sum/Difference Rule: d/dx (u ± v) = du/dx ± dv/dx
  • Product Rule: d/dx (u * v) = u * (dv/dx) + v * (du/dx)
  • Quotient Rule: d/dx (u / v) = [v * (du/dx) - u * (dv/dx)] / v², provided v ≠ 0.
  • Constant Multiple Rule: d/dx (k * u) = k * (du/dx), where k is a constant.
• Chain Rule (Derivative of Composite Functions): If y = f(t) and t = g(x) are differentiable functions, then y = f(g(x)) is differentiable with respect to x, and:
  dy/dx = dy/dt * dt/dx
  Alternatively, if y = f(u) where u is a function of x, then dy/dx = f'(u) * du/dx. This is extremely important for differentiating complex functions.
• Derivatives of Standard Functions: (Memorize these!)
  • d/dx (x^n) = n * x^(n-1)
  • d/dx (sin x) = cos x
  • d/dx (cos x) = -sin x
  • d/dx (tan x) = sec² x
  • d/dx (cot x) = -csc² x
  • d/dx (sec x) = sec x * tan x
  • d/dx (csc x) = -csc x * cot x
  • d/dx (e^x) = e^x
  • d/dx (a^x) = a^x * log_e a (or a^x * ln a)
  • d/dx (log_e x) = d/dx (ln x) = 1/x, x > 0
  • d/dx (log_a x) = 1 / (x * ln a), x > 0
  • d/dx (sin⁻¹ x) = 1 / √(1 - x²), -1 < x < 1
  • d/dx (cos⁻¹ x) = -1 / √(1 - x²), -1 < x < 1
  • d/dx (tan⁻¹ x) = 1 / (1 + x²)
  • d/dx (cot⁻¹ x) = -1 / (1 + x²)
  • d/dx (sec⁻¹ x) = 1 / (|x| * √(x² - 1)), |x| > 1
  • d/dx (csc⁻¹ x) = -1 / (|x| * √(x² - 1)), |x| > 1
  • d/dx (constant) = 0
• Differentiation Techniques:
  • Implicit Differentiation: Used when y cannot be easily expressed explicitly as a function of x (e.g., x² + xy + y² = 1). Differentiate both sides of the equation with respect to x, treating y as a function of x and using the chain rule (d/dx(y^n) = n*y^(n-1) * dy/dx). Then, algebraically solve for dy/dx.
  • Logarithmic Differentiation: Useful for:
    1. Functions of the form y = [f(x)]^[g(x)]. Take the natural logarithm (ln) of both sides, use log properties to simplify, then differentiate implicitly w.r.t x.
    2. Functions involving complicated products, quotients, or powers. Taking ln simplifies the expression before differentiation.
  • Parametric Differentiation: When x and y are given as functions of a third variable (parameter), say t (x = f(t), y = g(t)). Then:
    dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
• Second Order Derivative: The derivative of the first derivative (dy/dx or f'(x)) is called the second order derivative.
  • Notation: d²y/dx² or f''(x) or y''.
  • d²y/dx² = d/dx (dy/dx)

3. Mean Value Theorems (Often relevant in competitive exams)

• Rolle's Theorem: Let f be a function such that:
  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).
  3. f(a) = f(b).
     Then, there exists at least one point c in the open interval (a, b) such that f'(c) = 0. (Geometrically: there's a point where the tangent is horizontal).
• Lagrange's Mean Value Theorem (LMVT): Let f be a function such that:
  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).
     Then, there exists at least one point c in the open interval (a, b) such that:
     f'(c) = [f(b) - f(a)] / (b - a)
     (Geometrically: there's a point where the tangent is parallel to the secant line joining the endpoints (a, f(a)) and (b, f(b))).

Key Takeaways for Exams:

• Master the definitions of continuity and differentiability (LHL=RHL=f(c) and LHD=RHD).
• Know the relationship: Differentiability implies Continuity, but not vice-versa (|x| is the key example).
• Memorize all standard derivative formulas.
• Be proficient in applying the Product Rule, Quotient Rule, and especially the Chain Rule.
• Understand when and how to use Implicit Differentiation, Logarithmic Differentiation, and Parametric Differentiation.
• Be aware of the conditions and conclusions of Rolle's Theorem and LMVT.

Now, let's test your understanding with some multiple-choice questions.

Multiple Choice Questions (MCQs)

1. The function f(x) = |x - 3| is:
   (a) Continuous everywhere but not differentiable at x = 3
   (b) Differentiable everywhere
   (c) Not continuous at x = 3
   (d) Neither continuous nor differentiable at x = 3

2. If f(x) = kx + 5 is continuous at x = 2 and f(2) = 11, what is the value of k?
   (a) 2
   (b) 3
   (c) 5
   (d) 6

3. The derivative of sin(x²) with respect to x is:
   (a) cos(x²)
   (b) 2x sin(x²)
   (c) 2x cos(x²)
   (d) -cos(2x)

4. If y = log(sin x), then dy/dx is:
   (a) cos x
   (b) cot x
   (c) tan x
   (d) 1 / sin x

5. If x = a t² and y = 2at, then dy/dx is:
   (a) t
   (b) 1/t
   (c) a
   (d) 2a

6. The function f(x) = { (sin x / x, if x ≠ 0), (1, if x = 0) } at x = 0 is:
   (a) Continuous and differentiable
   (b) Continuous but not differentiable
   (c) Differentiable but not continuous
   (d) Neither continuous nor differentiable

7. If x³ + y³ = 3axy, then dy/dx is:
   (a) (ax - y²) / (x² - ay)
   (b) (ay - x²) / (y² - ax)
   (c) (x² + ay) / (y² + ax)
   (d) (ax + y²) / (x² + ay)

8. The derivative of e^(tan⁻¹ x) with respect to x is:
   (a) e^(tan⁻¹ x) / (1 + x²)
   (b) e^(tan⁻¹ x) * (1 + x²)
   (c) e^(tan⁻¹ x) / √(1 - x²)
   (d) e^(tan⁻¹ x)

9. The second derivative of log x (base e) is:
   (a) 1/x
   (b) -1/x²
   (c) 1/x²
   (d) -2/x³

10. For the function f(x) = x² in the interval [-1, 1], the value 'c' satisfying Rolle's Theorem is:
    (a) 1
    (b) -1
    (c) 0
    (d) 1/2

Answer Key:

1. (a) - The modulus function creates a sharp corner at the point where the inside becomes zero (x=3), making it non-differentiable there, but it's continuous everywhere.
2. (b) - For continuity, lim (x→2) f(x) = f(2). So, k(2) + 5 = 11 => 2k = 6 => k = 3.
3. (c) - Use Chain Rule: d/dx(sin(u)) = cos(u) * du/dx. Here u = x², du/dx = 2x. So, cos(x²) * 2x.
4. (b) - Use Chain Rule: d/dx(log(u)) = (1/u) * du/dx. Here u = sin x, du/dx = cos x. So, (1/sin x) * cos x = cot x.
5. (b) - Parametric differentiation: dx/dt = 2at, dy/dt = 2a. dy/dx = (dy/dt) / (dx/dt) = (2a) / (2at) = 1/t.
6. (b) - We know lim (x→0) sin x / x = 1, and f(0) = 1, so it's continuous. To check differentiability, you'd need LHD and RHD of sin x / x at 0, which involves the derivative of sin x / x. The derivative is (x cos x - sin x) / x². The limit of this as x->0 requires L'Hopital's rule or series expansion and shows the derivative doesn't exist finitely at 0 (or more simply, check LHD/RHD definition - it gets complex, but standard result is continuous not differentiable). Correction: While the derivative expression has issues at x=0, the definition of differentiability lim (h->0) [f(h)-f(0)]/h = lim (h->0) [(sin h / h) - 1] / h = lim (h->0) [sin h - h] / h². Using L'Hopital twice gives 0. Let me re-evaluate. f'(0) = lim h->0 (f(h)-f(0))/h = lim h->0 ((sin h / h) - 1)/h = lim h->0 (sin h - h)/h^2. L'Hopital: lim h->0 (cos h - 1)/(2h). L'Hopital again: lim h->0 (-sin h)/2 = 0. So the derivative exists and is 0. Thus it is differentiable. Let me correct the standard result understanding. The standard example of continuous but not diff is |x|. Let's re-evaluate sin x / x. It is differentiable at x=0 if defined as 1. The derivative is 0. So the answer should be (a). Self-Correction: Let's double check standard examples. x^2 sin(1/x) for x!=0, 0 for x=0 is differentiable. x sin(1/x) is continuous but not differentiable. sin x / x - its derivative is (x cos x - sin x)/x^2. The limit of this as x->0 is indeed 0 (using Taylor series (x(1-x^2/2!) - (x-x^3/3!))/x^2 = (x-x^3/2 - x+x^3/6)/x^2 = (-x^3/3)/x^2 -> 0). So, yes, it is differentiable. Answer is (a).
7. (b) - Implicit Differentiation: Differentiate x³ + y³ = 3axy w.r.t x. 3x² + 3y²(dy/dx) = 3a [x(dy/dx) + y(1)]. Divide by 3: x² + y²(dy/dx) = ax(dy/dx) + ay. Rearrange: y²(dy/dx) - ax(dy/dx) = ay - x². Factor out dy/dx: (dy/dx) (y² - ax) = ay - x². Solve: dy/dx = (ay - x²) / (y² - ax).
8. (a) - Chain Rule: d/dx(e^u) = e^u * du/dx. Here u = tan⁻¹ x, du/dx = 1 / (1 + x²). So, e^(tan⁻¹ x) * [1 / (1 + x²)].
9. (b) - First derivative: d/dx(ln x) = 1/x = x⁻¹. Second derivative: d/dx(x⁻¹) = -1 * x⁻² = -1/x².
10. (c) - Rolle's Theorem requires f'(c) = 0. f(x) = x², so f'(x) = 2x. Set f'(c) = 0: 2c = 0 => c = 0. Check conditions: f(x)=x² is continuous on [-1, 1] (polynomial), differentiable on (-1, 1). f(-1) = (-1)² = 1, f(1) = 1² = 1. So f(-1) = f(1). Conditions met. c=0 is in the interval (-1, 1).

Revised Answer Key based on correction for Q6:

1. (a)
2. (b)
3. (c)
4. (b)
5. (b)
6. (a)
7. (b)
8. (a)
9. (b)
10. (c)

Make sure you understand the reasoning behind each answer, especially the application of rules and definitions. Good luck with your preparation!