Unit 3: Calculus — Class 11 Applied Maths | Free MCQs & Solved Examples | Boundless Maths
Unit 3 of 7 12 Marks CBSE 2026–27

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Unit 3 — Free Study Resources

CBSE Class 11 Applied Mathematics · Free MCQs, Solved Examples & Case Studies

Unit 3 carries 12 marks in the CBSE Class 11 annual exam. Everything you need to master it: 30 practice MCQs (including 3 Assertion-Reason), 10 short-answer and 10 long-answer solved examples, and 3 board-pattern case studies. Covering Functions & Their Types, Graphical Representation, Limits & Continuity, and Differentiation. All content is aligned to the latest CBSE syllabus.

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Case Studies

Class 11 Applied Maths Unit 3 — Calculus Study Material

This page covers all topics in Unit 3 of CBSE Class 11 Applied Mathematics — carrying 12 marks in the CBSE Class 11 annual exam. You'll find 30 MCQs with solutions (including 3 Assertion-Reason), 10 short-answer questions and 10 long-answer questions with complete step-by-step workings, and 3 real-world case studies. Topics include Functions & Their Types (polynomial, rational, composite, logarithm, exponential, modulus, greatest integer, signum), Graphical Representation, Limits & Continuity, and Differentiation using power, product, quotient, and chain rules. Free CBSE 2026-27 aligned practice on how to find the domain and range of a function, limits by factorisation, continuity of piecewise functions, and chain rule differentiation with solved examples. All content is aligned to the latest CBSE syllabus.

Functions & Types Limits & Continuity Differentiation Chain Rule Product Rule Quotient Rule MCQs with Answers Case Studies
Unit 3 · 12 Marks

Topics & Key Formulas

Four topics examined across MCQs, short answers, long answers, and case studies.

3.1 Functions — Types & Graphs

Mapping between dependent and independent variables. Domain, codomain and range. Graphical representation.

  • Dependent & independent variables: y = f(x); x is input (independent), y is output (dependent)
  • Domain: valid inputs  ·  Codomain: target set  ·  Range ⊆ Codomain
  • Types: Polynomial, Rational, Logarithmic, Exponential, Modulus |x|, Greatest Integer ⌊x⌋, Signum, Composite
  • Graphical representation on coordinate plane — observe domain/range visually
f: A→B  ·  Range ⊆ Codomain  ·  y = f(x)

3.2 Limits & Continuity

Behaviour of a function as the input approaches a value. Continuity at a point.

  • Limit: limx→a f(x) = L — value f approaches (may ≠ f(a))
  • Left-Hand Limit = Right-Hand Limit → limit exists
  • Continuity at x=a: limit exists, f(a) defined, and limx→a f(x) = f(a)
  • Standard limits: limx→0 (sin x)/x = 1  ·  limx→0 (eˣ−1)/x = 1
limx→a f(x) = f(a) → f is continuous at a

3.3 Differentiation

Instantaneous rate of change. Derivative as the slope of the tangent to a curve at a point.

  • Instantaneous rate of change: how fast f(x) changes at a specific point
  • Derivative = slope of tangent to the curve y = f(x) at point (a, f(a))
  • Power rule: d/dx(xⁿ) = n·xⁿ⁻¹
  • d/dx(eˣ) = eˣ  ·  d/dx(ln x) = 1/x  ·  d/dx(c) = 0
d/dx(xⁿ) = n·xⁿ⁻¹  ·  derivative = slope of tangent

3.4 Algebra of Derivatives

Rules for differentiating combinations of functions — sum, difference, product, quotient, composite.

  • Sum/Difference: d/dx[f±g] = f′±g′
  • Product rule: d/dx[fg] = f′g + fg′
  • Quotient rule: d/dx[f/g] = (f′g − fg′)/g²
  • Chain rule (function of a function): d/dx[f(g(x))] = f′(g(x))·g′(x)
Chain rule: dy/dx = (dy/du)·(du/dx)
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Practice Questions

Practice MCQs — Unit 3 Calculus

27 MCQs + 3 Assertion-Reason questions. Click Show Answer to reveal the correct option and full explanation.

Question 1
The domain of f(x) = √(x − 3) is:
(a) x < 3
(b) x > 3
(c) x ≥ 3
(d) x ≤ 3
✓ Correct Answer: (c) x ≥ 3
For √(x − 3) to be defined, x − 3 ≥ 0, so x ≥ 3. Domain = [3, ∞).
Question 2
If f(x) = x² + 1 and g(x) = 2x, then f(g(x)) is:
(a) 4x² + 1
(b) 2x² + 2
(c) 4x + 1
(d) 2x² + 1
✓ Correct Answer: (a) 4x² + 1
f(g(x)) = f(2x) = (2x)² + 1 = 4x² + 1.
Question 3
The range of f(x) = |x| is:
(a) All real numbers
(b) x > 0
(c) x ≥ 0
(d) x ≤ 0
✓ Correct Answer: (c) x ≥ 0
The modulus function always returns non-negative values. Range = [0, ∞).
Question 4
⌊−2.4⌋ (greatest integer function) equals:
(a) −2
(b) −3
(c) 2
(d) 3
✓ Correct Answer: (b) −3
⌊x⌋ is the largest integer ≤ x. Since −3 < −2.4 < −2, the largest integer ≤ −2.4 is −3.
Question 5
The domain of f(x) = log(x − 2) is:
(a) x > 2
(b) x ≥ 2
(c) x < 2
(d) All real numbers
✓ Correct Answer: (a) x > 2
Logarithm requires argument strictly > 0, so x − 2 > 0 → x > 2.
Question 6
The signum function sgn(x) equals −1 when:
(a) x = 0
(b) x > 0
(c) x < 0
(d) x ≥ 0
✓ Correct Answer: (c) x < 0
sgn(x) = 1 if x > 0, 0 if x = 0, and −1 if x < 0.
Question 7
Which of the following is a rational function?
(a) f(x) = √x
(b) f(x) = (x² + 1)/(x − 3)
(c) f(x) = eˣ
(d) f(x) = log x
✓ Correct Answer: (b) f(x) = (x² + 1)/(x − 3)
A rational function is a ratio of two polynomials: p(x)/q(x) where q(x) ≠ 0.
Question 8
limx→2 (x² − 4)/(x − 2) equals:
(a) 0
(b) 2
(c) 4
(d) Undefined
✓ Correct Answer: (c) 4
(x²−4)/(x−2) = (x+2)(x−2)/(x−2) = x+2. As x→2, limit = 4.
Question 9
limx→0 (sin x)/x equals:
(a) 0
(b) ∞
(c) 1
(d) −1
✓ Correct Answer: (c) 1
Standard result: limx→0 (sin x)/x = 1 (x in radians).
Question 10
A function f(x) is continuous at x = a if:
(a) f(a) is defined only
(b) limx→a f(x) exists only
(c) limx→a f(x) = f(a)
(d) f(a) = 0
✓ Correct Answer: (c) limx→a f(x) = f(a)
For continuity: (1) f(a) must be defined, (2) the limit must exist, and (3) they must be equal.
Question 11
limx→3 (x² − 9)/(x − 3) is:
(a) 3
(b) 6
(c) 9
(d) 0
✓ Correct Answer: (b) 6
(x²−9)/(x−3) = x+3. As x→3, limit = 6.
Question 12
limx→∞ (3x² + 2x)/(5x² − 1) equals:
(a) 0
(b) 2/5
(c) 3/5
(d) ∞
✓ Correct Answer: (c) 3/5
Divide by x²: (3 + 2/x)/(5 − 1/x²) → 3/5 as x→∞.
Question 13
limx→0 (eˣ − 1)/x equals:
(a) 0
(b) e
(c) 1
(d) ∞
✓ Correct Answer: (c) 1
Standard result: limx→0 (eˣ − 1)/x = 1.
Question 14
If f(x) = x⁵, then f′(x) is:
(a) 5x⁴
(b) x⁴
(c) 5x⁶
(d) 5x
✓ Correct Answer: (a) 5x⁴
Power rule: d/dx(xn) = nxⁿ⁻¹. So d/dx(x⁵) = 5x⁴.
Question 15
The derivative of a constant function is:
(a) 1
(b) The constant itself
(c) 0
(d) x
✓ Correct Answer: (c) 0
A constant does not change, so its derivative is 0.
Question 16
d/dx(eˣ) is:
(a) xeˣ⁻¹
(b) eˣ
(c) ln x
(d) 1/x
✓ Correct Answer: (b) eˣ
The exponential function eˣ is its own derivative.
Question 17
d/dx(ln x) is:
(a) eˣ
(b) x
(c) 1/x
(d) ln x
✓ Correct Answer: (c) 1/x
d/dx(ln x) = 1/x for x > 0.
Question 18
If y = (3x² + 5)⁴, then dy/dx using chain rule is:
(a) 4(3x² + 5)³
(b) 24x(3x² + 5)³
(c) 4(6x)
(d) 12x(3x² + 5)⁴
✓ Correct Answer: (b) 24x(3x² + 5)³
Chain rule: dy/dx = 4(3x²+5)³ × 6x = 24x(3x²+5)³.
Question 19
The product rule states that d/dx[f(x)·g(x)] =
(a) f′(x)·g′(x)
(b) f(x)·g′(x) + g(x)·f′(x)
(c) f(x)·g′(x) − g(x)·f′(x)
(d) f′(x)/g′(x)
✓ Correct Answer: (b) f(x)·g′(x) + g(x)·f′(x)
Product rule (uv)' = uv' + vu'.
Question 20
d/dx(x²·eˣ) using product rule is:
(a) 2x·eˣ
(b) x²eˣ + 2xeˣ
(c) x²eˣ
(d) 2xeˣ − x²eˣ
✓ Correct Answer: (b) x²eˣ + 2xeˣ
d/dx(x²·eˣ) = x²·eˣ + eˣ·2x = eˣ(x² + 2x).
Question 21
For y = √x, the derivative dy/dx is:
(a) 1/(2√x)
(b) 2√x
(c) 1/x
(d) √x/2
✓ Correct Answer: (a) 1/(2√x)
y = x½. dy/dx = (1/2)x−½ = 1/(2√x).
Question 22
If y = log(3x), then dy/dx is:
(a) 3/x
(b) 1/(3x)
(c) 1/x
(d) 3 ln x
✓ Correct Answer: (c) 1/x
d/dx[log(3x)] = d/dx[log 3 + log x] = 1/x. (Via chain rule: (1/3x)·3 = 1/x.)
Question 23
d/dx(x³ + 5x² − 2x + 10) at x = 1 is:
(a) 8
(b) 11
(c) 14
(d) 5
✓ Correct Answer: (b) 11
dy/dx = 3x² + 10x − 2. At x = 1: 3 + 10 − 2 = 11.
Question 24
If f(x) = e²ˣ, then f′(x) using chain rule is:
(a) e²ˣ
(b) 2eˣ
(c) 2e²ˣ
(d) e²ˣ/2
✓ Correct Answer: (c) 2e²ˣ
Chain rule: d/dx[e2x] = e2x · 2 = 2e2x.
Question 25
Using quotient rule, d/dx[x²/(x+1)] is:
(a) (x² + 2x)/(x+1)²
(b) 2x/(x+1)
(c) x²/(x+1)²
(d) (2x+1)/(x+1)²
✓ Correct Answer: (a) (x² + 2x)/(x+1)²
Quotient rule: [2x(x+1) − x²·1]/(x+1)² = (2x²+2x−x²)/(x+1)² = (x²+2x)/(x+1)².
Question 26
d/dx[ln(x² + 1)] is:
(a) 1/(x²+1)
(b) 2x·ln(x²+1)
(c) 2x/(x²+1)
(d) x/(x²+1)
✓ Correct Answer: (c) 2x/(x²+1)
Chain rule: d/dx[ln(u)] = (1/u)·u'. Here u = x²+1, u′ = 2x. Result = 2x/(x²+1).
Question 27
If f(x) = x·ln x, then f′(x) is:
(a) ln x
(b) 1 + ln x
(c) 1/x
(d) x + ln x
✓ Correct Answer: (b) 1 + ln x
Product rule: f′(x) = 1·ln x + x·(1/x) = ln x + 1.

📋 Assertion-Reason Questions

Statement I is Assertion (A) and Statement II is Reason (R). Choose the correct option:

  • (a) Both A and R are True and R is the correct explanation of A
  • (b) Both A and R are True but R is not the correct explanation of A
  • (c) A is True but R is False
  • (d) A is False but R is True
Assertion-Reason 1
Assertion (A): The function f(x) = 1/x is not continuous at x = 0.
Reason (R): A function is discontinuous at any point where it is not defined.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a)
f(x) = 1/x is undefined at x = 0, hence discontinuous there. R correctly explains A.
Assertion-Reason 2
Assertion (A): The derivative of f(x) = x⁴ is 4x³.
Reason (R): For any constant n, d/dx(xn) = n·xn−1.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a)
The power rule (R) directly gives d/dx(x⁴) = 4x³ (A).
Assertion-Reason 3
Assertion (A): limx→2 f(x) can exist even if f(2) is not defined.
Reason (R): The limit of a function at a point depends on values near that point, not at the point itself.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a)
Example: f(x) = (x²−4)/(x−2) is undefined at x=2 but limx→2 = 4. R explains exactly why.

📐 All Calculus Formulas in One Place

Formula Deck covers all units of Class 11 Applied Maths — derivatives, limits, function types — crisp & printable.

  • All formulas organised chapter-wise
  • Function types, derivative rules, limit formulas and more
  • Perfect for daily revision & last-minute exam prep
  • Instant PDF — print & pin up anytime
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Step-by-Step Solutions

Short Answer Questions

2-mark and 3-mark questions with complete working. Click Show Solution to reveal each answer.

Question 1
Find the domain and range of f(x) = 1/(x − 4).
f(x) undefined when x − 4 = 0, i.e., x = 4.
Domain = ℝ − {4}
Since 1/(x−4) = 0 has no solution, 0 is excluded from the range.
Domain: ℝ − {4}  |  Range: ℝ − {0}
Question 2
If f(x) = 2x + 3 and g(x) = x², find g(f(x)) and f(g(x)). Are they equal?
g(f(x)) = g(2x+3) = (2x+3)² = 4x² + 12x + 9
f(g(x)) = f(x²) = 2x² + 3
4x² + 12x + 9 ≠ 2x² + 3 — composition is generally not commutative.
g(f(x)) = (2x+3)²  |  f(g(x)) = 2x² + 3  |  Not equal.
Question 3
Evaluate: limx→2 (x³ − 8)/(x − 2)
x³ − 8 = (x − 2)(x² + 2x + 4)
So (x³−8)/(x−2) = x² + 2x + 4, for x ≠ 2
As x→2: 4 + 4 + 4 = 12
Limit = 12
Question 4
Check continuity of f(x) = {2x + 1, x < 1 ; 4, x = 1 ; 3x − 1, x > 1} at x = 1.
LHL (x→1⁻): 2(1) + 1 = 3
RHL (x→1⁺): 3(1) − 1 = 2
LHL ≠ RHL — limit does not exist at x = 1.
NOT continuous at x = 1
Question 5
Differentiate f(x) = x⁴ − 3x³ + 5x − 7.
Apply power rule term by term.
d/dx(x⁴) = 4x³  |  d/dx(−3x³) = −9x²  |  d/dx(5x) = 5  |  d/dx(−7) = 0
f′(x) = 4x³ − 9x² + 5
Question 6
Find dy/dx for y = (x² + 3)⁵ using chain rule.
Let u = x² + 3, so y = u⁵
dy/du = 5u⁴  |  du/dx = 2x
dy/dx = 5u⁴ × 2x = 10x(x²+3)⁴
dy/dx = 10x(x² + 3)⁴
Question 7
Differentiate y = x · ln(x) using product rule.
u = x, v = ln x → u′ = 1, v′ = 1/x
dy/dx = u'v + uv′ = ln x + x·(1/x) = ln x + 1
dy/dx = ln x + 1
Question 8
Find k so that f(x) = {kx + 1, x ≤ 5 ; 3x − 5, x > 5} is continuous at x = 5.
LHL = 5k + 1  |  RHL = 3(5) − 5 = 10
For continuity: 5k + 1 = 10 → k = 9/5
k = 9/5
Question 9
Using quotient rule, differentiate f(x) = (x² + 1)/(x − 1).
u = x²+1, v = x−1 → u′ = 2x, v′ = 1
f′(x) = [(2x)(x−1) − (x²+1)(1)]/(x−1)²
= (2x²−2x−x²−1)/(x−1)² = (x²−2x−1)/(x−1)²
f′(x) = (x² − 2x − 1)/(x − 1)²
Question 10
Differentiate using chain rule: (i) y = ln(x² + 5)   (ii) y = ex³
(i) dy/dx = 1/(x²+5) × 2x = 2x/(x²+5)
(ii) dy/dx = ex³ × 3x² = 3x²ex³
(i) 2x/(x²+5)  |  (ii) 3x²ex³

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5-Mark Questions

Long Answer Questions

5-mark practice questions with complete working. Click Show Solution to reveal each answer.

Question 1
Given f(x) = x³ − 6x² + 11x − 6, find: (i) f′(x), (ii) f′(1), f′(2), f′(3), (iii) values of x where f′(x) = 0.
(i) f′(x) = 3x² − 12x + 11
(ii) f′(1) = 3−12+11 = 2; f′(2) = 12−24+11 = −1; f′(3) = 27−36+11 = 2
(iii) 3x²−12x+11 = 0 → x = (12 ± √12)/6 = 2 ± 1/√3
(i) f′(x) = 3x²−12x+11  |  (ii) 2, −1, 2  |  (iii) x = 2 ± 1/√3
Question 2
Determine continuity of f(x) at x = 2, where f(x) = {(x²−4)/(x−2), x ≠ 2 ; 5, x = 2}. If discontinuous, redefine f(2) to make it continuous.
limx→2 (x²−4)/(x−2) = lim(x+2) = 4
f(2) = 5 ≠ 4 → Discontinuous at x = 2
Redefine f(2) = 4 to make it continuous.
Discontinuous at x = 2; redefine f(2) = 4.
Question 3
Differentiate y = (x + 1)(x² − x + 1) in two ways: (i) by expanding, (ii) by product rule. Verify both give the same answer.
(i) Expand: y = x³ + 1. dy/dx = 3x²
(ii) u = x+1, v = x²−x+1; u'=1, v'=2x−1
dy/dx = (x²−x+1) + (x+1)(2x−1) = x²−x+1+2x²+x−1 = 3x²
Both give dy/dx = 3x² ✓
Question 4
Find a and b so that f(x) = {ax + b, x < −1 ; 2, x = −1 ; bx² − a, x > −1} is continuous at x = −1.
LHL = −a + b; RHL = b − a; f(−1) = 2
Both LHL and RHL give the same expression: b − a
For continuity: b − a = 2
Any a, b satisfying b − a = 2 (e.g., a = 0, b = 2)
Question 5
A population is modelled by P(t) = 500 · e0.03t. Find: (i) P(0), (ii) P′(t), (iii) P′(10), use e0.3 ≈ 1.35.
(i) P(0) = 500 × e⁰ = 500
(ii) P′(t) = 500 × 0.03 × e0.03t = 15e0.03t
(iii) P′(10) = 15 × 1.35 ≈ 20.25
(i) 500  |  (ii) 15e0.03t  |  (iii) ≈ 20.25
Question 6
Evaluate: (i) limx→0 [(1+x)n − 1]/x    (ii) limx→1 (xn − 1)/(x − 1)
(i) Binomial expansion: (1+x)n = 1+nx+... → [(1+x)n−1]/x → n as x→0
(ii) xⁿ−1 = (x−1)(xn−1+…+1). Dividing by (x−1): sum of n terms, each →1 as x→1.
Both limits equal n
Question 7
Using chain rule, differentiate: (i) y = √(3x−2)   (ii) y = ln(x²+5)   (iii) y = (2x³+1)⁶
(i) dy/dx = 3/[2√(3x−2)]
(ii) dy/dx = 2x/(x²+5)
(iii) dy/dx = 6(2x³+1)⁵ × 6x² = 36x²(2x³+1)⁵
(i) 3/[2√(3x−2)]  |  (ii) 2x/(x²+5)  |  (iii) 36x²(2x³+1)⁵
Question 8
Find points of discontinuity of f(x) = (x²−5x+6)/(x²−3x+2) and simplify wherever possible.
Numerator = (x−2)(x−3); Denominator = (x−1)(x−2)
Discontinuous at x = 1 and x = 2 (denominator = 0)
For x ≠ 1, 2: simplified form = (x−3)/(x−1)
At x=2: lim = −1 (removable discontinuity); at x=1: infinite discontinuity
Discontinuous at x = 1 (infinite) and x = 2 (removable); simplified: (x−3)/(x−1)
Question 9
Differentiate f(x) = x² · eˣ · ln(x) stepwise using product rule.
Step 1 — Let u = x²·eˣ: u′ = 2x·eˣ + x²·eˣ = eˣ(2x + x²)
Step 2 — Apply product rule to u · ln x: f′(x) = u′·ln x + u·(1/x)
= eˣ(2x+x²)·ln x + x²·eˣ·(1/x) = eˣ[(x²+2x)ln x + x]
f′(x) = eˣ[(x² + 2x)ln x + x]
Question 10
Discuss continuity of f(x) = |x − 2| at x = 2. Also find f′(x) for x > 2 and x < 2.
f(x) = x−2 for x ≥ 2;   f(x) = 2−x for x < 2
LHL at x=2: lim(2−x) = 0; RHL: lim(x−2) = 0; f(2) = 0 → Continuous
For x > 2: f′(x) = 1
For x < 2: f′(x) = −1
Continuous at x = 2; f′(x) = 1 for x > 2, f′(x) = −1 for x < 2
Real-World Applications

Case Study Questions

Board-pattern case studies. Read the context carefully, then click Show Answers for each case.

📱

Case Study 1: Mobile Recharge Plans — Function Types & Continuity

A telecom company models cost (in ₹) as C(x) = 50 + 2x for x ≤ 100 calls, and C(x) = 250 + 1.5(x−100) for x > 100 calls. A data analyst models user growth as G(t) = 1000·e0.1t where t = months.
(i)

What type of function is C(x) for x ≤ 100?

(a) Exponential
(b) Linear (polynomial)
(c) Logarithmic
(d) Rational
(ii)

Is C(x) continuous at x = 100?

(iii)(a)

What type of function is G(t) = 1000·e0.1t?

(a) Polynomial
(b) Modulus
(c) Exponential
(d) Signum
(iii)(b)

Find G′(t) using chain rule.

(i)(b) Linear (polynomial) — C(x) = 50 + 2x is of the form mx + c — a degree 1 polynomial (linear function).
(ii)Yes, C(x) is continuous at x = 100. LHL: 50+2(100) = 250. RHL: 250+1.5(0) = 250. f(100) = 250. All three equal → Continuous.
(iii)(a)(c) Exponential — G(t) = 1000·e0.1t is of the form a·ekt — an exponential function.
(iii)(b)G′(t) = 100e0.1t — G′(t) = 1000 × 0.1 × e0.1t = 100e0.1t by chain rule.
🏭

Case Study 2: Limits & Continuity in Manufacturing

A machine's cost per item (₹) as quantity x increases is C(x) = (x²−25)/(x−5) for x ≠ 5. At x = 5 (retooling), the cost is set to ₹k per item.
(i)

Find limx→5 C(x).

(a) 0
(b) 5
(c) 10
(d) 25
(ii)

What value of k makes C(x) continuous at x = 5?

(a) 5
(b) 10
(c) 25
(d) 0
(iii)

If C(x) = x² + 3x for all x, find dC/dx and evaluate at x = 4.

(i)(c) 10 — (x²−25)/(x−5) = (x+5). As x→5, limit = 10.
(ii)(b) k = 10 — For continuity, C(5) = limit = 10. So k = 10.
(iii)dC/dx = 2x + 3; At x = 4: dC/dx = 11 — d/dx(x²+3x) = 2x+3. At x=4: 8+3 = 11.
💻

Case Study 3: Online Learning Platform — Differentiation in Action

A startup models user engagement as E(x) = ln(x + 1), where x = content pieces uploaded, and daily active users as U(t) = 200·e0.05t, where t = weeks since launch.
(i)

What is the domain of E(x) = ln(x + 1)?

(a) x > 0
(b) x > −1
(c) x ≥ 0
(d) All real numbers
(ii)

Find dE/dx. What happens as x → ∞?

(iii)(a)

Find U′(t) for U(t) = 200·e0.05t.

(a) 200e0.05t
(b) 10e0.05t
(c) 0.05e0.05t
(d) 200 × 0.05
(iii)(b)

Compare the growth of E(x) = ln(x+1) vs U(t) = 200e0.05t. Which grows faster?

(i)(b) x > −1 — ln requires argument > 0, so x+1 > 0 → x > −1.
(ii)dE/dx = 1/(x+1) → 0 as x → ∞ — Each additional content piece adds less and less to engagement — diminishing returns.
(iii)(a)(b) 10e0.05t — U′(t) = 200 × 0.05 × e0.05t = 10e0.05t.
(iii)(b)U(t) grows much faster — exponential growth outpaces logarithmic growth. Logarithmic functions grow very slowly; exponential functions grow without bound. The startup's user base grows rapidly, but engagement per content piece grows only logarithmically — quality matters more than quantity.

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Exam Tips — Unit 3

Common mistakes students make in the exam — and how to avoid them.

✓ Tip 1 — Write All Three Conditions for Continuity

Never just compute the limit. To prove continuity at x = a, explicitly show all three: (1) f(a) is defined, (2) limx→a f(x) exists, and (3) the limit equals f(a). Writing only the limit calculation without these steps costs marks in every short-answer and long-answer question.

✓ Tip 2 — State the Rule Before Applying It

Always name the differentiation rule — power rule, product rule, quotient rule, or chain rule — before applying it. Define u and v clearly when using the product or quotient rule. Examiners award method marks for this setup even if arithmetic slips later.

❌ Common Mistakes to Avoid in Unit 3

  • Confusing the domain of log(x−a) — it is x > a, not x ≥ a
  • Forgetting that ⌊x⌋ rounds down, not toward zero (⌊−2.4⌋ = −3, not −2)
  • Not checking LHL = RHL = f(a) separately when testing continuity
  • Applying the chain rule but forgetting to multiply by the inner derivative
  • Mixing up the quotient rule sign: it is (vu' − uv')/v², not (uv' − vu')/v²
  • Writing f(g(x)) and g(f(x)) without clearly substituting step by step
Common Questions

Frequently Asked Questions

Questions students frequently ask about Unit 3 — topics, formulas, and exam strategy.

Unit 3 — Calculus carries 12 marks in the CBSE Class 11 Applied Mathematics annual exam. Questions appear as 1-mark MCQs (including Assertion-Reason), 2–3 mark short answers, and 5-mark long answers.
Four topics: (1) Functions — Introduction & Types (polynomial, rational, composite, logarithm, exponential, modulus, greatest integer, signum), (2) Graphical Representation of Functions, (3) Limits & Continuity (limit laws, standard results, continuity at a point), and (4) Differentiation (power rule, product rule, quotient rule, chain rule).
The two most frequently tested standard limits are: limx→0 (sin x)/x = 1 and limx→0 (eˣ − 1)/x = 1. Also memorise limx→0 (ln(1+x))/x = 1 and limx→a (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹. These appear as direct 1-mark MCQs almost every year.
A function f(x) is continuous at x = a if and only if: (1) f(a) is defined, (2) limx→a f(x) exists (meaning LHL = RHL), and (3) limx→a f(x) = f(a). All three must be stated explicitly in your answer to earn full marks.
The chain rule: dy/dx = (dy/du) × (du/dx). Use it when differentiating a composite function — a function inside another function. Examples: y = (3x²+5)⁴ (polynomial inside power), y = e2x (linear inside exponential), y = ln(x²+1) (polynomial inside log). Always identify the inner function u first.
Modulus function f(x) = |x|: returns the non-negative value of x. f(−3) = 3, f(3) = 3. Range = [0, ∞). Greatest integer function ⌊x⌋: returns the largest integer ≤ x. ⌊2.7⌋ = 2, ⌊−2.3⌋ = −3 (not −2). This floor/ceiling distinction for negative numbers is a very common MCQ trap.
Based on past CBSE papers: Domain and range of functions — almost always an MCQ. Composition of functions — short answer. Limits by factorisation — short answer. Continuity of piecewise functions — long answer most years. Chain rule differentiation — short or long answer. Product or quotient rule — long answer.
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