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Continuity and Differentiability – Class 12 Mathematics notes, important questions and MCQs
NCERT Class 12 Mathematics, chapter 5. Notes, key terms, important questions with answers and MCQs.
Chapter overview
A function is continuous if its graph has no breaks; it is differentiable if it has a well-defined tangent. Rules of differentiation handle composite, implicit, inverse, exponential and logarithmic functions.
Continuity and differentiability
- f is continuous at a if lim(x→a) f(x) = f(a).
- Differentiable ⇒ continuous, but not conversely (|x| at 0 is continuous but not differentiable).
- Polynomials, sin x, cos x, eˣ are continuous everywhere.
Rules
- Chain rule: d/dx f(g(x)) = f′(g(x))·g′(x).
- d/dx eˣ = eˣ; d/dx log x = 1/x; d/dx sin⁻¹x = 1/√(1 − x²); d/dx tan⁻¹x = 1/(1 + x²).
- Logarithmic differentiation for xˣ: d/dx xˣ = xˣ(1 + log x); parametric dy/dx = (dy/dt)/(dx/dt).
- Rolle's and Mean Value theorems link derivative values with function values.
Key terms
- Continuous function
- function whose limit equals its value at every point of the domain
- Derivative
- instantaneous rate of change, dy/dx
- Rolle's theorem
- if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f′(c) = 0 for some c in (a, b)
Formulas
- d/dx (uv) = u′v + uv′
- d/dx (u/v) = (u′v − uv′)/v²
- d/dx xˣ = xˣ(1 + log x)
Important questions with answers
Find k so that f(x) = kx + 1 for x ≤ π and cos x for x > π is continuous at x = π.2 marks
Left limit = kπ + 1, right limit = cos π = −1. kπ + 1 = −1 gives k = −2/π.
Find dy/dx if x² + xy + y² = 100.3 marks
2x + y + x dy/dx + 2y dy/dx = 0, so dy/dx = −(2x + y)/(x + 2y).
If y = (tan⁻¹x)², show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.5 marks
y₁ = 2 tan⁻¹x/(1 + x²), so (1 + x²) y₁ = 2 tan⁻¹x. Differentiate again: (1 + x²) y₂ + 2x y₁ = 2/(1 + x²). Multiply by (1 + x²): (1 + x²)² y₂ + 2x(1 + x²) y₁ = 2. Hence proved.
Quick revision
- Differentiable ⇒ continuous.
- |x| is not differentiable at 0.
- Chain rule for composite functions.
Exam tips
- Check left-hand and right-hand derivatives at corner points.
- For implicit differentiation, differentiate both sides with respect to x and treat y as a function of x (d/dx y² = 2y dy/dx).
Practice MCQs
d/dx (sin x) is
- −cos x
- −sin x
- sin x
- cos x
Show answer
Answer: (d) cos x. Standard derivative.
f(x) = |x| at x = 0 is
- not continuous
- differentiable
- continuous but not differentiable
- undefined
Show answer
Answer: (c) continuous but not differentiable. Corner point.
d/dx (e^(3x)) is
- e^(3x)
- 3e^(3x)
- 3x e^(3x−1)
- e³
Show answer
Answer: (b) 3e^(3x). Chain rule.
d/dx (log(sin x)) is
- 1/sin x
- tan x
- cot x
- cos x
Show answer
Answer: (c) cot x. cos x / sin x.
d/dx [sin⁻¹(2x√(1 − x²))] for −1/√2 < x < 1/√2 is
- 1/√(1 − x²)
- 2x
- −2/√(1 − x²)
- 2/√(1 − x²)
Show answer
Answer: (d) 2/√(1 − x²). Put x = sin θ.
If x = a cos t, y = a sin t, then dy/dx is
- tan t
- −cot t
- cot t
- −tan t
Show answer
Answer: (b) −cot t. (a cos t)/(−a sin t).
d/dx (xˣ) is
- x·xˣ⁻¹
- xˣ log x
- xˣ
- xˣ(1 + log x)
Show answer
Answer: (d) xˣ(1 + log x). Logarithmic differentiation.
For f(x) = x² on [1, 3], the c of the Mean Value Theorem is
- 1.5
- 2
- 2.5
- 3
Show answer
Answer: (b) 2. f′(c) = (9 − 1)/2 = 4 = 2c.
All chapters of Class 12 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.