Unit 4: Calculus - Free Study Resources | Boundless Maths

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Topics Covered in Unit 4

Master these 4 important topics

1. Functions — Introduction & Types

Domain, range; polynomial, rational, composite, logarithm, exponential, modulus, greatest integer, signum functions

2. Graphical Representation of Functions

Sketching and interpreting graphs of standard functions; nature of function at various points

3. Limits & Continuity

Concept of limits, limit laws, standard limit results, continuity of a function at a point

4. Differentiation

Derivatives of algebraic functions using power rule, product rule, quotient rule, and chain rule

MCQs (30) Short Answers (10) Long Answers (10) Case Studies (3)

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Practice MCQs with Answers (27 MCQs + 3 Assertion-Reason)

Click "Show Answer" to reveal explanations

Question 1

The domain of f(x) = √(x − 3) is:

(a) x < 3

(b) 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

(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

(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

(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

(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

(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)

(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

lim(x→2) (x² − 4)/(x − 2) equals:

(a) 0

(b) 2

(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

lim(x→0) (sin x)/x equals:

(a) 0

(b) ∞

(d) −1

✓ Correct Answer: (c) 1

Standard result: lim(x→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) lim(x→a) f(x) exists only

(d) f(a) = 0

✓ Correct Answer: (c) lim(x→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

lim(x→3) (x² − 9)/(x − 3) is:

(a) 3

(b) 6

(d) 0

✓ Correct Answer: (b) 6

(x²−9)/(x−3) = x+3. As x→3, limit = 6.

Question 12

lim(x→∞) (3x² + 2x)/(5x² − 1) equals:

(a) 0

(b) 2/5

(d) ∞

✓ Correct Answer: (c) 3/5

Divide by x²: (3 + 2/x)/(5 − 1/x²) → 3/5 as x→∞.

Question 13

lim(x→0) (e^x − 1)/x equals:

(a) 0

(b) e

(d) ∞

✓ Correct Answer: (c) 1

Standard result: lim(x→0) (eˣ − 1)/x = 1.

Question 14

If f(x) = x⁵, then f'(x) is:

(a) 5x⁴

(b) x⁴

(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

(d) x

✓ Correct Answer: (c) 0

A constant does not change, so its derivative is 0.

Question 16

d/dx(e^x) is:

(a) xe^(x−1)

(b) e^x

(d) 1/x

✓ Correct Answer: (b) e^x

The exponential function e^x is its own derivative.

Question 17

d/dx(ln x) is:

(a) e^x

(b) 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

(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)

(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 · e^x) using product rule is:

(a) 2x · e^x

(b) x²e^x + 2xe^x

(d) 2xe^x − x²e^x

✓ Correct Answer: (b) x²e^x + 2xe^x

d/dx(x²·e^x) = x²·e^x + e^x·2x = e^x(x² + 2x).

Question 21

For y = √x, the derivative dy/dx is:

(a) 1/(2√x)

(b) 2√x

(d) √x/2

✓ Correct Answer: (a) 1/(2√x)

y = x^(1/2). dy/dx = (1/2)x^(−1/2) = 1/(2√x).

Question 22

If y = log(3x), then dy/dx is:

(a) 3/x

(b) 1/(3x)

(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

(d) 5

✓ Correct Answer: (b) 11

dy/dx = 3x² + 10x − 2. At x = 1: 3 + 10 − 2 = 11.

Question 24

If f(x) = e^(2x), then f'(x) using chain rule is:

(a) e^(2x)

(b) 2e^x

(d) e^(2x)/2

✓ Correct Answer: (c) 2e^(2x)

Chain rule: d/dx[e^(2x)] = e^(2x) · 2 = 2e^(2x).

Question 25

Using quotient rule, d/dx[x²/(x+1)] is:

(a) (x² + 2x)/(x+1)²

(b) 2x/(x+1)

(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)

(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

(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

(d) A is False but R is True

✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of 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·x^(n−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

(d) A is False but R is True

✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A

The power rule (R) directly gives d/dx(x⁴) = 4x³ (A).

Assertion-Reason 3

Assertion (A): lim(x→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

(d) A is False but R is True

✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A

Example: f(x) = (x²−4)/(x−2) is undefined at x=2 but lim(x→2) = 4. R explains exactly why.

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Short Answer Questions with Step-by-Step Solutions

Practice 2-mark and 3-mark questions

Question 1

Find the domain and range of f(x) = 1/(x − 4).

Solution:

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?

Solution:

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: lim(x→2) (x³ − 8)/(x − 2)

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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).

Solution:

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 = e^(x³)

Solution:

(i) dy/dx = 1/(x²+5) × 2x = 2x/(x²+5)

(ii) dy/dx = e^(x³) × 3x² = 3x²e^(x³)

(i) 2x/(x²+5) | (ii) 3x²e^(x³)

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Long Answer Questions with Complete Solutions

Practice 5-mark questions

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.

Complete Solution:

(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.

Complete Solution:

lim(x→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.

Complete Solution:

(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.

Complete Solution:

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 · e^(0.03t). Find: (i) P(0), (ii) P'(t), (iii) P'(10), use e^0.3 ≈ 1.35.

Complete Solution:

(i) P(0) = 500 × e⁰ = 500

(ii) P'(t) = 500 × 0.03 × e^(0.03t) = 15e^(0.03t)

(iii) P'(10) = 15 × 1.35 ≈ 20.25

(i) 500 | (ii) 15e^(0.03t) | (iii) ≈ 20.25

Question 6

Evaluate: (i) lim(x→0) [(1+x)^n − 1]/x (ii) lim(x→1) (x^n − 1)/(x − 1)

Complete Solution:

(i) Binomial expansion: (1+x)^n = 1+nx+... → [(1+x)^n−1]/x → n as x→0

(ii) x^n−1 = (x−1)(x^(n−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

Complete Solution:

(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.

Complete Solution:

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 · e^x · ln(x) stepwise using product rule.

Complete Solution:

Step 1 — Let u = x²·e^x: u' = 2x·e^x + x²·e^x = e^x(2x + x²)

Step 2 — Apply product rule to u · ln x: f'(x) = u'·ln x + u·(1/x)

= e^x(2x+x²)·ln x + x²·e^x·(1/x) = e^x[(x²+2x)ln x + x]

f'(x) = e^x[(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.

Complete Solution:

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

📊

Case Studies

Real-world application based questions

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·e^(0.1t) where t = months.

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

(a) Exponential

(b) Linear (polynomial)

(d) Rational

✓ Correct Answer: (b) Linear (polynomial)

C(x) = 50 + 2x is of the form mx + c — a degree 1 polynomial (linear function).

(ii) Is C(x) continuous at x = 100?

✓ 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) What type of function is G(t) = 1000·e^(0.1t)?

(a) Polynomial

(b) Modulus

(d) Signum

✓ Correct Answer: (c) Exponential

G(t) = 1000·e^(0.1t) is of the form a·e^(kt) — an exponential function.

(iii)(b) Find G'(t) using chain rule.

✓ G'(t) = 100e^(0.1t)

G'(t) = 1000 × 0.1 × e^(0.1t) = 100e^(0.1t) 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 lim(x→5) C(x).

(a) 0

(b) 5

(d) 25

✓ Correct Answer: (c) 10

(x²−25)/(x−5) = (x+5). As x→5, limit = 10.

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

(a) 5

(b) 10

(d) 0

✓ Correct Answer: (b) k = 10

For continuity, C(5) = limit = 10. So k = 10.

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

✓ 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·e^(0.05t), where t = weeks since launch.

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

(a) x > 0

(b) x > −1

(d) All real numbers

✓ Correct Answer: (b) x > −1

ln requires argument > 0, so x+1 > 0 → x > −1.

(ii) Find dE/dx. What happens as x → ∞?

✓ dE/dx = 1/(x+1) → 0 as x → ∞

Each additional content piece adds less and less to engagement — diminishing returns.

(iii)(a) Find U'(t) for U(t) = 200·e^(0.05t).

(a) 200e^(0.05t)

(b) 10e^(0.05t)

(d) 200 × 0.05

✓ Correct Answer: (b) 10e^(0.05t)

U'(t) = 200 × 0.05 × e^(0.05t) = 10e^(0.05t).

(iii)(b) Compare the growth of E(x) = ln(x+1) vs U(t) = 200e^(0.05t). Which grows faster?

✓ 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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∫ Unit 4: Calculus

Topics Covered in Unit 4

1. Functions — Introduction & Types

2. Graphical Representation of Functions

3. Limits & Continuity

4. Differentiation

Practice MCQs with Answers (27 MCQs + 3 Assertion-Reason)

📋 Assertion-Reason Questions

📐 All Calculus Formulas in One Place

Short Answer Questions with Step-by-Step Solutions

📐 Revise Smarter Before Your Exam

Long Answer Questions with Complete Solutions

Case Studies

🏆 Ace Your Class 11 Applied Maths Exams

📐 Class 11 Applied Maths Formula Deck — ₹199

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