Class 12 Mathematics Notes Chapter 5 (Continuity and Differentiability) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 5: Continuity and Differentiability. This is a foundational chapter for calculus and frequently tested in various government exams. Understanding these concepts thoroughly is crucial. We'll break it down systematically, keeping the NCERT Exemplar level in mind.

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. No breaks, jumps, or holes.
• Formal Definition (at a point): A function f is said to be continuous at a point x = c in its domain if:
  lim_(x→c) f(x) = f(c)
  This single condition implicitly means three things must hold:
  1. f(c) is defined (c is in the domain of f).
  2. lim_(x→c) f(x) exists. This requires the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) to be equal:
     lim_(x→c⁻) f(x) = lim_(x→c⁺) f(x)
  3. The limit equals the function's value: lim_(x→c) f(x) = f(c).
• Continuity in an Interval:
  • Open Interval (a, b): f is continuous in (a, b) if it is continuous at every point in the interval.
  • Closed Interval [a, b]: f is continuous in [a, b] if:
    1. It is continuous in the open interval (a, b).
    2. It is continuous from the right at a: lim_(x→a⁺) f(x) = f(a).
    3. It is continuous from the left at b: lim_(x→b⁻) f(x) = f(b).
• Algebra of Continuous Functions: If f and g are two real functions continuous at x = c, then:
  1. f + g is continuous at x = c.
  2. f - g is continuous at x = c.
  3. f * g is continuous at x = c.
  4. f / g is continuous at x = c, provided g(c) ≠ 0.
  5. k * f is continuous at x = c (where k is a constant).
  6. (f o g) (composition) is continuous at x = c, if g is continuous at c and f is continuous at g(c).
• Continuity of Standard Functions:
  • Polynomial functions (P(x) = a_n x^n + ... + a_1 x + a_0): Continuous everywhere (R).
  • Rational functions (P(x)/Q(x) where Q(x) ≠ 0): Continuous in their domain.
  • Trigonometric functions (sin x, cos x): Continuous everywhere (R).
  • tan x, sec x: Continuous in their domains (R - {(2n+1)π/2, n ∈ Z}).
  • cot x, csc x: Continuous in their domains (R - {nπ, n ∈ Z}).
  • Exponential functions (e^x, a^x where a > 0, a ≠ 1): Continuous everywhere (R).
  • Logarithmic functions (log x, log_a x where a > 0, a ≠ 1): Continuous in their domain ((0, ∞)).
  • Modulus function (|x|): Continuous everywhere (R).
  • Greatest Integer Function ([x]): Discontinuous at all integer points. Continuous on (n, n+1) for any integer n.
  • Inverse Trigonometric Functions: Continuous in their respective domains.
• Intermediate Value Theorem (Statement): If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b) (f(a) ≠ f(b)), then there exists at least one number c in (a, b) such that f(c) = k.

2. Differentiability

• Intuitive Idea: A function is differentiable at a point if its graph is "smooth" at that point, meaning no sharp corners, kinks, cusps, or vertical tangents.
• Formal Definition (at a point): A function f 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 is called the derivative of f at c.
  For the limit to exist, the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) must be equal and finite:
  • LHD at c: L f'(c) = lim_(h→0⁻) [f(c+h) - f(c)] / h
  • RHD at c: R f'(c) = lim_(h→0⁺) [f(c+h) - f(c)] / h
    So, f is differentiable at c if L f'(c) = R f'(c) and this value is finite.
• Relationship between Continuity and Differentiability:
  Theorem: If a function f is differentiable at a point c, then it is also continuous at that point.
  Converse is NOT true: A function can be continuous at a point but not differentiable. The classic example is f(x) = |x| at x = 0. It's continuous but has a sharp corner, so LHD = -1 and RHD = +1.
• Differentiability in an Interval:
  • Open Interval (a, b): f is differentiable in (a, b) if it is differentiable at every point in the interval.
  • Closed Interval [a, b]: f is differentiable in [a, b] if:
    1. It is differentiable in the open interval (a, b).
    2. The right-hand derivative exists at a.
    3. The left-hand derivative exists at b.
       (Note: Often, differentiability is primarily discussed for open intervals).
• Algebra of Derivatives: If u and v are differentiable functions of x, then:
  1. Sum/Difference Rule: d/dx (u ± v) = du/dx ± dv/dx
  2. Product Rule: d/dx (u * v) = u * (dv/dx) + v * (du/dx)
  3. Quotient Rule: d/dx (u / v) = [v * (du/dx) - u * (dv/dx)] / v², provided v ≠ 0.
  4. Constant Multiple Rule: d/dx (k * u) = k * (du/dx)
• Chain Rule (Crucial!): 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 used extensively for composite 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
  • d/dx (log_e x) = 1/x (for x > 0)
  • d/dx (log_a x) = 1 / (x * log_e a) (for x > 0)
  • d/dx (sin⁻¹ x) = 1 / √(1 - x²) (for -1 < x < 1)
  • d/dx (cos⁻¹ x) = -1 / √(1 - x²) (for -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)) (for |x| > 1)
  • d/dx (csc⁻¹ x) = -1 / (|x| * √(x² - 1)) (for |x| > 1)
  • d/dx (constant) = 0
• Derivatives of Implicit Functions: When y cannot be easily expressed explicitly as a function of x (e.g., x² + xy + y³ = 5).
  • Method: Differentiate both sides of the equation with respect to x, treating y as a function of x. Use the chain rule whenever differentiating a term involving y (e.g., d/dx (y³) = 3y² * dy/dx). Then, algebraically solve for dy/dx.
• Derivatives of Parametric Functions: When x and y are given as functions of a third variable (parameter), say t (e.g., x = f(t), y = g(t)).
  • Method: Find dx/dt and dy/dt. Then, provided dx/dt ≠ 0:
    dy/dx = (dy/dt) / (dx/dt)
• Logarithmic Differentiation: Useful in two main cases:
  1. When the function is of the form [f(x)]^[g(x)] (function raised to the power of another function).
  2. When the function is a complicated product/quotient of several terms.
  • Method:
    1. Take the natural logarithm (log_e or ln) of both sides: log y = log [f(x)].
    2. Use logarithm properties to simplify the right side (e.g., log(a^b) = b log a, log(ab) = log a + log b, log(a/b) = log a - log b).
    3. Differentiate both sides implicitly with respect to x. Remember d/dx (log y) = (1/y) * dy/dx.
    4. Solve for dy/dx and substitute back the original expression for y.
• Second Order Derivatives: The derivative of the first derivative.
  • Notation: f''(x), y'', d²y/dx², D²y.
  • d²y/dx² = d/dx (dy/dx)
  • For parametric functions: d²y/dx² = d/dx (dy/dx) = d/dt (dy/dx) * dt/dx = [d/dt {(dy/dt)/(dx/dt)}] / (dx/dt). Be careful with this formula.

3. Mean Value Theorems

• 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.
  • Geometric Interpretation: There is at least one point on the curve between x=a and x=b where the tangent is horizontal (parallel to the x-axis).
• 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)
  • Geometric Interpretation: There is at least one point on the curve between x=a and x=b where the tangent is parallel to the secant line joining the points (a, f(a)) and (b, f(b)).
  • Rolle's Theorem is a special case of LMVT when f(a) = f(b).

Key Exam Points & Tips:

1. Check Continuity First: Before checking differentiability at a point, always check for continuity. If a function is discontinuous at a point, it cannot be differentiable there.
2. Master Chain Rule: This is arguably the most important differentiation technique. Practice it extensively with various composite functions.
3. LHD/RHD for Differentiability: For piecewise functions or functions involving modulus/greatest integer, always check differentiability using the LHD = RHD definition at the points where the function definition changes or might have issues.
4. Logarithmic Differentiation: Know when and how to apply it correctly. Don't forget the (1/y) * dy/dx term when differentiating log y.
5. Implicit Differentiation: Be systematic. Differentiate term by term, use product/chain rules correctly, and carefully isolate dy/dx.
6. Parametric Differentiation: Remember the formula dy/dx = (dy/dt) / (dx/dt). For the second derivative, use the correct formula involving d/dt(dy/dx) and dt/dx.
7. Standard Formulas: Derivatives of all standard functions (including inverse trig) must be on your fingertips.
8. Rolle's & LMVT: Understand the conditions and conclusions precisely. Problems often involve verifying the theorem or finding the value of 'c'.
9. Practice Exemplar Problems: NCERT Exemplar problems often test deeper understanding and involve more complex calculations or conceptual twists than the basic textbook exercises.

Multiple Choice Questions (MCQs)

1. The function f(x) = { (sin x / x) if x < 0, (x + 1) if x ≥ 0 } at x = 0 is:
   (A) Continuous but not differentiable
   (B) Differentiable but not continuous
   (C) Continuous and differentiable
   (D) Neither continuous nor differentiable

2. If f(x) = |cos x|, then at x = π/2, f(x) is:
   (A) Continuous and differentiable
   (B) Continuous but not differentiable
   (C) Not continuous but differentiable
   (D) Neither continuous nor differentiable

3. If y = log(log x), then dy/dx is:
   (A) 1 / log x
   (B) 1 / (x log x)
   (C) x / log x
   (D) log x / x

4. If x = a(θ - sin θ) and y = a(1 - cos θ), then dy/dx at θ = π/2 is:
   (A) 1
   (B) -1
   (C) 0
   (D) Undefined

5. The derivative of sin⁻¹(2x √(1 - x²)) with respect to x, for 1/√2 < x < 1, is:
   (A) 2 / √(1 - x²)
   (B) -2 / √(1 - x²)
   (C) 1 / √(1 - x²)
   (D) -1 / √(1 - x²)
   (Hint: Use substitution x = cos θ)

6. If x^y = e^(x-y), then dy/dx is:
   (A) log x / (1 + log x)
   (B) (x - y) / (y + x log x)
   (C) log x / (1 + log x)²
   (D) (log x - 1) / (log x + 1)²

7. Let f(x) = x³ - 6x² + ax + b be defined on [1, 3]. If Rolle's theorem holds for f(x) with c = 2 + 1/√3, find the value of a.
   (A) 11
   (B) -11
   (C) 6
   (D) -6

8. The function f(x) = [x] (Greatest Integer Function) is:
   (A) Continuous at x = 2.5
   (B) Differentiable at x = 3
   (C) Continuous at x = -1
   (D) Differentiable at x = 1.5

9. If y = tan⁻¹((cos x - sin x) / (cos x + sin x)), then dy/dx is:
   (A) 1
   (B) -1
   (C) 1 / (1 + x²)
   (D) 2 / (1 + x²)

10. If f(x) = e^x * g(x), g(0) = 2, g'(0) = 1, then f'(0) is:
    (A) 1
    (B) 2
    (C) 3
    (D) 0

Answer Key:

1. (A) LHL = lim_(x→0⁻) sin x / x = 1. RHL = lim_(x→0⁺) (x + 1) = 1. f(0) = 0 + 1 = 1. So, continuous. LHD = d/dx(sin x / x) is complex, but RHD = d/dx(x+1) = 1. LHD requires L'Hopital or series expansion, derivative of sin x / x at 0 is 0. Since LHD ≠ RHD, it's not differentiable.
2. (B) |cos x| is continuous everywhere. At x = π/2, cos x changes sign. Graph has a sharp corner. LHD = d/dx(-cos x) = sin x at π/2 is 1. RHD = d/dx(cos x) = -sin x at π/2 is -1. Not differentiable.
3. (B) Chain rule: Let u = log x. y = log u. dy/dx = (dy/du) * (du/dx) = (1/u) * (1/x) = (1 / log x) * (1/x) = 1 / (x log x).
4. (A) dx/dθ = a(1 - cos θ). dy/dθ = a(sin θ). dy/dx = (dy/dθ) / (dx/dθ) = (a sin θ) / (a(1 - cos θ)) = sin θ / (1 - cos θ). At θ = π/2, dy/dx = sin(π/2) / (1 - cos(π/2)) = 1 / (1 - 0) = 1.
5. (B) Let x = cos θ. Then θ = cos⁻¹ x. Since 1/√2 < x < 1, we have 0 < θ < π/4. sin⁻¹(2cos θ √(1 - cos² θ)) = sin⁻¹(2cos θ sin θ) = sin⁻¹(sin 2θ). Since 0 < θ < π/4, 0 < 2θ < π/2, so sin⁻¹(sin 2θ) = 2θ = 2 cos⁻¹ x. d/dx (2 cos⁻¹ x) = -2 / √(1 - x²).
6. (C) Take log: y log x = (x - y) log e = x - y. Differentiate implicitly: y * (1/x) + log x * (dy/dx) = 1 - dy/dx. dy/dx (log x + 1) = 1 - y/x = (x - y)/x. From y log x = x - y, y(log x + 1) = x, so y = x / (1 + log x). Substitute y: dy/dx (log x + 1) = (x - x/(1+log x))/x = (1 - 1/(1+log x)) = log x / (1 + log x). So, dy/dx = log x / (1 + log x)².
7. (A) Rolle's theorem requires f(1) = f(3). f'(x) = 3x² - 12x + a. f'(c) = 0. f'(2 + 1/√3) = 3(2 + 1/√3)² - 12(2 + 1/√3) + a = 0. 3(4 + 1/3 + 4/√3) - 24 - 12/√3 + a = 0. 3(13/3 + 4/√3) - 24 - 12/√3 + a = 0. 13 + 12/√3 - 24 - 12/√3 + a = 0. -11 + a = 0, so a = 11.
8. (A) [x] is continuous at non-integer points. At x = 2.5, lim_(x→2.5) [x] = 2 and f(2.5) = [2.5] = 2. It's discontinuous at integers (like 3 and -1) and hence not differentiable there. It is differentiable at non-integer points (like 1.5), but option (A) is the most accurate description among the choices.
9. (B) Divide numerator and denominator by cos x: y = tan⁻¹((1 - tan x) / (1 + tan x)) = tan⁻¹(tan(π/4 - x)). Assuming π/4 - x is in the principal range (-π/2, π/2), y = π/4 - x. Then dy/dx = -1.
10. (C) Use product rule: f'(x) = d/dx(e^x) * g(x) + e^x * d/dx(g(x)) = e^x * g(x) + e^x * g'(x). f'(0) = e^0 * g(0) + e^0 * g'(0) = 1 * 2 + 1 * 1 = 2 + 1 = 3.

Study these notes carefully and practice a variety of problems, especially from the Exemplar book. Good luck!