Chapter-7 Binomial Theorem — Online MCQ Test
MATHS · CLASS 11 FIRST PUC · Karnataka State Board
Practice Chapter-7 Binomial Theorem with a free chapter-wise online MCQ test.
This chapter covers: Binomial expansion - Pascal's Triangle - Binomial coefficients (nCk) - General term (Tr+1) - Middle term.
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Chapter-7 Binomial Theorem — Important Questions & Answers
In Pascal's Triangle, which row represents the binomial coefficients for (a+b)³?
- A. Row 2
- B. Row 3
- C. Row 4
- D. Row 5
Answer: B. Row 3
Pascal's Triangle row n contains the binomial coefficients for (a+b)ⁿ. Row 3 corresponds to (a+b)³.
Pascal's Triangle row n contains the binomial coefficients for (a+b)ⁿ. Row 3 corresponds to (a+b)³.
What is the general formula for the (r+1)th term in the binomial expansion of (a+b)ⁿ?
- A. Tr+1 = nCr × aⁿ⁻ʳ × bʳ
- B. Tr+1 = nCr × aʳ × bⁿ⁻ʳ
- C. Tr+1 = (n-r)C(n-r) × aⁿ × bʳ
- D. Tr+1 = nCr × aⁿ × bʳ
Answer: A. Tr+1 = nCr × aⁿ⁻ʳ × bʳ
The general term Tr+1 = nCr × aⁿ⁻ʳ × bʳ is the standard formula where the power of a decreases and power of b increases.
The general term Tr+1 = nCr × aⁿ⁻ʳ × bʳ is the standard formula where the power of a decreases and power of b increases.
What is the relationship between nCr and nC(n-r)?
- A. nCr > nC(n-r)
- B. nCr = nC(n-r)
- C. nCr < nC(n-r)
- D. nCr + nC(n-r) = 2n
Answer: B. nCr = nC(n-r)
By the symmetry property of binomial coefficients: nCr = nC(n-r) = n!/(r!(n-r)!).
By the symmetry property of binomial coefficients: nCr = nC(n-r) = n!/(r!(n-r)!).
How many terms are there in the binomial expansion of (x+y)⁷?
- A. 6
- B. 7
- C. 8
- D. 9
Answer: C. 8
The binomial expansion of (x+y)ⁿ contains (n+1) terms. For n=7, there are 8 terms.
The binomial expansion of (x+y)ⁿ contains (n+1) terms. For n=7, there are 8 terms.
Find the middle term in the binomial expansion of (2x+3y)⁶.
- A. T₃ and T₄
- B. T₄
- C. T₃
- D. T₂ and T₃
Answer: B. T₄
For even n=6, there are 7 terms total, so the middle term is the 4th term (T₄) which occurs at r=3.
For even n=6, there are 7 terms total, so the middle term is the 4th term (T₄) which occurs at r=3.
In the expansion of (1+x)¹⁰, what is the coefficient of x³?
- A. 100
- B. 110
- C. 120
- D. 140
Answer: C. 120
The coefficient of x³ is 10C3 = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120.
The coefficient of x³ is 10C3 = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120.
In the expansion of (1+x)ⁿ, if the coefficient of x² is 45, what is n?
- A. 9
- B. 10
- C. 8
- D. 12
Answer: B. 10
Coefficient of x² is nC2 = 45. So n(n-1)/2 = 45, giving n(n-1) = 90. Solving: n = 10.
Coefficient of x² is nC2 = 45. So n(n-1)/2 = 45, giving n(n-1) = 90. Solving: n = 10.
In (2x + 3y)⁵, the coefficient of x²y³ is:
- A. 540
- B. 1080
- C. 2160
- D. 4320
Answer: B. 1080
General term: 5Cr × (2x)⁵⁻ʳ × (3y)ʳ. For x²y³: r=3, so T₄ = 5C3 × 2² × 3³ × x² × y³ = 10 × 4 × 27 = 1080.
General term: 5Cr × (2x)⁵⁻ʳ × (3y)ʳ. For x²y³: r=3, so T₄ = 5C3 × 2² × 3³ × x² × y³ = 10 × 4 × 27 = 1080.
In the binomial expansion, if Tr = 5C2 × a³ × b², then Tr equals:
- A. T₂
- B. T₃
- C. T₄
- D. T₅
Answer: B. T₃
Since the term is nCr × aⁿ⁻ʳ × bʳ and here n=5, r=2, this is the (r+1)th = 3rd term = T₃.
Since the term is nCr × aⁿ⁻ʳ × bʳ and here n=5, r=2, this is the (r+1)th = 3rd term = T₃.
In (x+y)ⁿ, if 8 times the 3rd term equals 9 times the 2nd term, and both contain equal powers of x and y, then n is:
- A. 16
- B. 15
- C. 14
- D. 17
Answer: C. 14
Given: 8×T₃ = 9×T₂ and equal powers of x,y means we're at the middle. This requires solving 8×nC2/9×nC1 = 1 with middle term condition, yielding n = 14.
Given: 8×T₃ = 9×T₂ and equal powers of x,y means we're at the middle. This requires solving 8×nC2/9×nC1 = 1 with middle term condition, yielding n = 14.