Class 11 Maths Important Questions with Solutions

Class 11 Maths Important Questions provide chapter-wise practice across algebra, trigonometry, coordinate geometry, calculus, statistics and probability. These questions help students revise important concepts, apply formulas and prepare for school examinations using the current NCERT syllabus.

Class 11 Maths introduces several concepts that form the foundation for Class 12 Mathematics and higher-level entrance examinations. The syllabus includes Sets, Relations and Functions, Trigonometric Functions, Algebra, Coordinate Geometry, Limits, Derivatives, Statistics and Probability.

The Class 11 Maths Important Questions provided here help students revise these concepts chapter-wise. Students can use the questions to check their understanding, practise calculations and identify chapters that require more revision.

Key Takeaways

• The current NCERT Class 11 Maths textbook contains 14 chapters.
• Algebra includes Complex Numbers, Linear Inequalities, Permutations and Combinations, Binomial Theorem, and Sequences and Series.
• Coordinate Geometry covers Straight Lines, Conic Sections and introductory Three-Dimensional Geometry.
• Limits and Derivatives introduces the basic ideas required for Class 12 Calculus.
• Regular chapter-wise practice is more useful than attempting unrelated questions from the complete syllabus at once.

Important Questions Class 11 Maths Chapter-Wise

Important Class 11 Maths Formulas

Sets

Number of elements in the union of two finite sets:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

De Morgan’s laws:

(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′

Trigonometric Functions

sin²x + cos²x = 1

1 + tan²x = sec²x

1 + cot²x = cosec²x

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) = cos A cos B − sin A sin B

Complex Numbers

i² = −1

For z = a + ib:

|z| = √(a² + b²)

Conjugate of z:

z̄ = a − ib

Quadratic Equations

For ax² + bx + c = 0:

x = [−b ± √(b² − 4ac)]/(2a)

Discriminant:

D = b² − 4ac

Permutations and Combinations

n! = n(n − 1)(n − 2)...3 × 2 × 1

ⁿPᵣ = n!/(n − r)!

ⁿCᵣ = n!/[r!(n − r)!]

Binomial Theorem

(a + b)ⁿ = Σ ⁿCᵣaⁿ⁻ʳbʳ

General term:

Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ

Geometric Progression

nth term:

aₙ = arⁿ⁻¹

Sum of first n terms:

Sₙ = a(rⁿ − 1)/(r − 1), where r ≠ 1

Straight Lines

Slope of a line through two points:

m = (y₂ − y₁)/(x₂ − x₁)

Point-slope form:

y − y₁ = m(x − x₁)

Distance of point (x₁, y₁) from Ax + By + C = 0:

Distance = |Ax₁ + By₁ + C|/√(A² + B²)

Three-Dimensional Geometry

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):

PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]

Statistics

Variance:

σ² = Σfᵢ(xᵢ − x̄)²/Σfᵢ

Standard deviation:

σ = √σ²

Probability

P(E) = Number of favourable outcomes/Total number of equally likely outcomes

0 ≤ P(E) ≤ 1

P(E′) = 1 − P(E)

Access Class 11 Maths Important Questions in 30 Minutes

• First 10 minutes: Revise formulas from Sets, Trigonometric Functions, Algebra and Sequences and Series.
• Next 10 minutes: Practise one question each from Straight Lines, Conic Sections and Three-Dimensional Geometry.
• Final 10 minutes: Solve one question each from Limits and Derivatives, Statistics and Probability.

Need chapter-wise practice for the complete Class 11 Maths syllabus?

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Important Questions for Class 11 Maths

Q1. Write the set A = {1, 4, 9, 16, 25} in set-builder form.

Solution:

The elements are squares of the first five natural numbers.

Therefore:

A = {x : x = n², where n ∈ N and 1 ≤ n ≤ 5}

Q2. If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find A ∪ B and A ∩ B.

Solution:

The union contains all elements belonging to A or B.

A ∪ B = {1, 2, 3, 4, 5, 6}

The intersection contains the common elements.

A ∩ B = {3, 4}

Q3. Determine whether the relation R = {(1, 2), (2, 3), (3, 4)} is a function.

Solution:

Each first element has exactly one corresponding second element.

Therefore, R is a function.

Q4. Find the domain of f(x) = 1/(x − 3).

Solution:

The denominator cannot be zero.

x − 3 ≠ 0

x ≠ 3

Therefore, the domain is:

R − {3}

Q5. Prove that sin²x + cos²x = 1.

Solution:

In a right-angled triangle:

sin x = Perpendicular/Hypotenuse

cos x = Base/Hypotenuse

Therefore:

sin²x + cos²x

= Perpendicular²/Hypotenuse² + Base²/Hypotenuse²

= (Perpendicular² + Base²)/Hypotenuse²

Using the Pythagoras theorem:

Perpendicular² + Base² = Hypotenuse²

Therefore:

sin²x + cos²x = 1

Q6. Find the exact value of sin 75°.

Solution:

75° = 45° + 30°

Using:

sin(A + B) = sin A cos B + cos A sin B

sin 75°

= sin 45° cos 30° + cos 45° sin 30°

= (1/√2)(√3/2) + (1/√2)(1/2)

= (√3 + 1)/(2√2)

Q7. Simplify i³⁵.

Solution:

The powers of i repeat after every four powers.

35 = 4(8) + 3

Therefore:

i³⁵ = i³

i³⁵ = −i

Q8. Solve the quadratic equation x² − 5x + 6 = 0.

Solution:

Factorising:

x² − 5x + 6 = 0

(x − 2)(x − 3) = 0

Therefore:

x = 2 or x = 3

Q9. Solve the inequality 3x − 5 < 7.

Solution:

3x − 5 < 7

3x < 12

x < 4

Therefore, the solution set is:

(−∞, 4)

Q10. Solve 2x + 3 ≥ 5x − 6.

Solution:

2x + 3 ≥ 5x − 6

3 + 6 ≥ 5x − 2x

9 ≥ 3x

x ≤ 3

Therefore, the solution set is:

(−∞, 3]

Q11. How many three-digit numbers can be formed using 1, 2, 3, 4 and 5 without repetition?

Solution:

There are five choices for the hundreds place.

After selecting one digit, four choices remain for the tens place and three for the units place.

Number of three-digit numbers:

5 × 4 × 3 = 60

Alternatively:

⁵P₃ = 5!/(5 − 3)!

⁵P₃ = 5!/2!

⁵P₃ = 60

Q12. In how many ways can a committee of three students be selected from eight students?

Solution:

Since order does not matter, use combinations.

Number of committees:

⁸C₃ = 8!/[3!5!]

⁸C₃ = (8 × 7 × 6)/(3 × 2 × 1)

⁸C₃ = 56

Q13. Find the fourth term in the expansion of (x + 2)⁶.

Solution:

The general term is:

Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ

For the fourth term:

r + 1 = 4

r = 3

Therefore:

T₄ = ⁶C₃x⁶⁻³(2)³

T₄ = 20x³ × 8

T₄ = 160x³

Q14. Find the middle term in the expansion of (x + 1)⁸.

Solution:

The expansion contains:

8 + 1 = 9 terms

Therefore, the middle term is the fifth term.

For T₅:

r = 4

T₅ = ⁸C₄x⁸⁻⁴(1)⁴

T₅ = 70x⁴

Q15. Find the eighth term of the GP 3, 6, 12, 24, ...

Solution:

First term:

a = 3

Common ratio:

r = 6/3 = 2

The nth term is:

aₙ = arⁿ⁻¹

Therefore:

a₈ = 3(2)⁷

a₈ = 3 × 128

a₈ = 384

Q16. Find the sum of the first six terms of the GP 2, 6, 18, ...

Solution:

a = 2

r = 3

n = 6

Using:

Sₙ = a(rⁿ − 1)/(r − 1)

S₆ = 2(3⁶ − 1)/(3 − 1)

S₆ = 2(729 − 1)/2

S₆ = 728

Q17. Find the slope of the line joining (2, 3) and (6, 11).

Solution:

m = (y₂ − y₁)/(x₂ − x₁)

m = (11 − 3)/(6 − 2)

m = 8/4

m = 2

Q18. Find the equation of the line passing through (1, 2) with slope 3.

Solution:

Using the point-slope form:

y − y₁ = m(x − x₁)

y − 2 = 3(x − 1)

y − 2 = 3x − 3

Therefore:

y = 3x − 1

Q19. Find the distance of the point (2, −1) from the line 3x + 4y − 5 = 0.

Solution:

Distance:

= |Ax₁ + By₁ + C|/√(A² + B²)

= |3(2) + 4(−1) − 5|/√(3² + 4²)

= |6 − 4 − 5|/5

= 3/5

Therefore, the distance is 3/5 units.

Q20. Find the equation of the circle with centre at the origin and radius 5.

Solution:

The standard equation is:

x² + y² = r²

Here:

r = 5

Therefore:

x² + y² = 25

Q21. Identify the conic represented by y² = 8x and find its focus.

Solution:

The standard equation of a parabola is:

y² = 4ax

Comparing:

4a = 8

a = 2

Therefore, the conic is a parabola and its focus is:

(2, 0)

Q22. Find the distance between P(1, 2, 3) and Q(4, 6, 3).

Solution:

PQ = √[(4 − 1)² + (6 − 2)² + (3 − 3)²]

PQ = √(3² + 4² + 0²)

PQ = √25

PQ = 5 units

Q23. Evaluate lim x→2 (x² − 4)/(x − 2).

Solution:

Factorise the numerator:

x² − 4 = (x − 2)(x + 2)

Therefore:

lim x→2 (x² − 4)/(x − 2)

= lim x→2 (x + 2)

= 2 + 2

= 4

Q24. Find the derivative of f(x) = x³ from first principles.

Solution:

Using:

f′(x) = lim h→0 [f(x + h) − f(x)]/h

f′(x) = lim h→0 [(x + h)³ − x³]/h

Expanding:

f′(x) = lim h→0 [x³ + 3x²h + 3xh² + h³ − x³]/h

f′(x) = lim h→0 [3x²h + 3xh² + h³]/h

f′(x) = lim h→0 [3x² + 3xh + h²]

Therefore:

f′(x) = 3x²

Q25. Find the range of the data 7, 11, 15, 18, 21.

Solution:

Range = Largest value − Smallest value

Range = 21 − 7

Range = 14

Q26. Find the mean deviation about the mean for the observations 2, 4 and 6.

Solution:

Mean:

x̄ = (2 + 4 + 6)/3

x̄ = 4

Absolute deviations:

|2 − 4| = 2

|4 − 4| = 0

|6 − 4| = 2

Mean deviation:

MD = (2 + 0 + 2)/3

MD = 4/3

Q27. A die is thrown once. Find the probability of obtaining an even number.

Solution:

Sample space:

S = {1, 2, 3, 4, 5, 6}

Even-number outcomes:

E = {2, 4, 6}

Therefore:

P(E) = 3/6

P(E) = 1/2

Q28. Two coins are tossed together. Find the probability of obtaining exactly one head.

Solution:

Sample space:

S = {HH, HT, TH, TT}

Favourable outcomes:

E = {HT, TH}

Therefore:

P(E) = 2/4

P(E) = 1/2

Class 11 Maths Exam Preparation Tips

• Revise formulas before beginning a question set.
• Complete NCERT exercises before moving to extra questions.
• Practise graphs in Sets, Linear Inequalities and Coordinate Geometry.
• Write restrictions on variables wherever required.
• Check whether order matters before choosing between permutations and combinations.
• Draw a rough figure for Straight Lines, Conic Sections and Three-Dimensional Geometry.
• Show the limit and derivative steps instead of writing only the final answer.
• Recheck calculations in Statistics and Probability.