Class 11 Maths Chapter 7 Binomial Theorem Worksheet with Answers, free PDF (CBSE Class 11 and JEE Main; based on NCERT)

A 20-mark worksheet on Binomial Theorem for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

The number of terms in the expansion of (3x − 5y)⁹ is

(a) 9(b) 10(c) 18(d) 8

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Answer: (b) 10

(a + b)ⁿ has n + 1 terms, so (3x − 5y)⁹ has 9 + 1 = 10 terms. 9 is the index itself, not the count of terms.

Question 2

The value of ¹⁰C₀ − ¹⁰C₁ + ¹⁰C₂ − ¹⁰C₃ + ... + ¹⁰C₁₀ is

(a) 0(b) 1(c) 1024(d) −1

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Answer: (a) 0

Putting x = 1 in (1 − x)¹⁰ = ¹⁰C₀ − ¹⁰C₁x + ¹⁰C₂x² − ... + ¹⁰C₁₀x¹⁰ gives 0¹⁰ = 0. 1024 is the sum without the signs.

Question 3

The binomial coefficient of the third term in the expansion of (a + b)¹⁰ is

(a) 120(b) 10(c) 45(d) 90

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Answer: (c) 45

The third term is T₂₊₁, with coefficient ¹⁰C₂ = (10 × 9)/2 = 45. ¹⁰C₃ = 120 belongs to the fourth term.

Question 4

The fourth term of (y + 3)⁵ is k y². The value of k is

(a) 90(b) 405(c) 135(d) 270

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Answer: (d) 270

T₄ = T₃₊₁ = ⁵C₃ y² 3³ = 10 × 27 y² = 270y². 90 is the coefficient of the third term, which has y³.

Question 5

The row of Pascal's triangle for index 6 is 1 6 15 20 15 6 1. The fourth entry of the row for index 7 is

(a) 21(b) 70(c) 35(d) 30

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Answer: (c) 35

Each entry is the sum of the two entries above it; the fourth entry of the next row is 15 + 20 = 35 (that is, ⁷C₃). 21 = 6 + 15 is the third entry.

Question 6 · 2 marks

Find the sum of all the coefficients in the expansion of (3x − 2y)⁵.

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(3x − 2y)⁵ = Σ c_k x^5−ky^k; putting x = y = 1 adds all the coefficients: (3 − 2)⁵ = 1⁵ = 1.

puts x = 1 and y = 1 (1)
sum = 1 (1)

Question 7 · 2 marks

Find the 5th term in the expansion of (3 + x2)⁶.

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T₅ = T₄₊₁ = ⁶C₄ (3)² (x2)⁴ = 15 × 9 × x⁴16 = 13516x⁴, that is, 135/16 x⁴.

uses T_r+1 = ⁶C_r 3^6−r(x/2)^r with r = 4 (1)
T₅ = 135/16 x⁴ (1)

Question 8 · 2 marks

The row of Pascal's triangle for index 5 is 1 5 10 10 5 1. Write the row for index 6, and use it to find the coefficient of a²b⁴ in (a − b)⁶.

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Adding neighbours, with 1 at each end: 1, 1 + 5 = 6, 5 + 10 = 15, 10 + 10 = 20, 15, 6, 1. The term in a²b⁴ is the fifth, 15a²(−b)⁴ = 15a²b⁴, so the coefficient is 15.

row 1 6 15 20 15 6 1 (1)
coefficient 15, with the sign of (−b)⁴ positive (1)

Question 9 · 3 marks

Writing (1 + x + x²)⁴ as [(1 + x) + x²]⁴, find the coefficient of x³ in its expansion.

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[(1 + x) + x²]⁴ = ⁴C₀(1 + x)⁴ + ⁴C₁(1 + x)³x² + ⁴C₂(1 + x)²x⁴ + ...; only the first two terms reach x³. From (1 + x)⁴: ⁴C₃ = 4. From 4(1 + x)³x²: 4 × ³C₁ = 12. Coefficient of x³ = 4 + 12 = 16.

expands the grouped binomial (1)
finds 4 and 12 from the two terms (1)
16 (1)

Question 10 · 3 marks

Find the term independent of x in the expansion of (3x − 16x²)⁹.

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T_r+1 = ⁹C_r (3x)^9−r(−16x²)^r = ⁹C_r 3^9−r(−16)^r x^9−3r. For x⁰, r = 3: T₄ = ⁹C₃ × 3⁶ × (−1216) = −84 × 729216 = −567/2.

general term with power x^9−3r (1)
r = 3, the 4th term (1)
T₄ = −567/2 (1)

Question 11 · 3 marks

For the expansion of (x + 2y)⁹, find (a) the number of terms, (b) which term contains y⁴, (c) that term.

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(a) 9 + 1 = 10 terms. (b) The index of y in T_r+1 is r, so y⁴ is in T₅. (c) T₅ = ⁹C₄ x⁵(2y)⁴ = 126 × 16x⁵y⁴ = 2016x⁵y⁴.

10 terms (1)
the 5th term (1)
2016x⁵y⁴ (1)

Class 11 Maths Chapter 7: Binomial Theorem worksheet with answer key