Miscellaneous Exercise — Binomial Theorem

1. 7.M.15 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   If a and b are distinct integers, prove that a-b is a factor of a^n - b^n, whenever n is a positive integer. [Hint: write a^n=(a-b+b)^n and expand]

   Hint. Write a as (a-b)+b and expand by the Binomial Theorem -- every term except the very last one carries a positive power of (a-b).

   Write a = (a-b)+b, so a^n = [(a-b)+b]^n = sum_(r=0)^n nCr.(a-b)^r.b^(n-r), by the Binomial Theorem with first term (a-b) and second term b. Every term with r>=1 has a factor of (a-b); the single r=0 term is nC0.b^n=b^n. So a^n = (a-b).[nC1.b^(n-1) + nC2(a-b)b^(n-2) + ... + nCn(a-b)^(n-1)] + b^n. Rearranging, a^n-b^n = (a-b).[that same bracket], which is (a-b) times an integer.

   ✦ Working through each part gives: proved: a^n - b^n = (a-b) x (an integer), so a-b is always a factor of a^n-b^n.

2. 7.M.25 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   Evaluate (sqrt3+sqrt2)^6 - (sqrt3-sqrt2)^6.

   Hint. Subtracting (a-b)^6 from (a+b)^6 cancels the even-r terms and doubles the odd-r ones (r=1,3,5) -- compute just those three terms.

   With a=sqrt3, b=sqrt2, n=6, only odd r survive after subtraction, each doubled. r=1: C1.a^5.b = 6(9.sqrt3)(sqrt2) = 54.sqrt6. r=3: C3.a^3.b^3 = 20(3.sqrt3)(2.sqrt2) = 120.sqrt6. r=5: C5.a.b^5 = 6(sqrt3)(4.sqrt2) = 24.sqrt6. Sum = 54.sqrt6+120.sqrt6+24.sqrt6 = 198.sqrt6, and doubling this (since (a+b)^6-(a-b)^6 = 2 x sum of odd-r terms) gives 396.sqrt6.

   ✦ Working through each part gives: 396.sqrt6.

3. 7.M.35 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   Find the value of (a^2+sqrt(a^2-1))^4 + (a^2-sqrt(a^2-1))^4.

   Hint. Let A=a^2 and B=sqrt(a^2-1). Adding (A+B)^4 and (A-B)^4 cancels the odd-r terms and doubles the even-r ones (r=0,2,4).

   Let A=a^2, B=sqrt(a^2-1). (A+B)^4+(A-B)^4 = 2[C0.A^4 + C2.A^2.B^2 + C4.B^4] = 2[A^4+6A^2B^2+B^4]. A^4=a^8. A^2B^2=a^4(a^2-1)=a^6-a^4, so 6A^2B^2=6a^6-6a^4. B^4=(a^2-1)^2=a^4-2a^2+1. Adding: a^8+6a^6-6a^4+a^4-2a^2+1 = a^8+6a^6-5a^4-2a^2+1. Doubling: 2a^8+12a^6-10a^4-4a^2+2.

   ✦ Working through each part gives: 2a^8+12a^6-10a^4-4a^2+2.

4. 7.M.44 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   Find an approximation of (0.99)^5 using the first three terms of its expansion.

   Hint. Write 0.99 as 1-0.01 and keep only the r=0,1,2 terms of the expansion, since the question asks for an approximation, not the exact value.

   (1-0.01)^5 is approximately C0 - C1(0.01) + C2(0.01)^2 (first three terms) = 1 - 5(0.01) + 10(0.0001) = 1-0.05+0.001 = 0.951.

   ✦ Working through each part gives: approximately 0.951.

5. 7.M.56 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   Expand using Binomial Theorem (1 + x/2 - 2/x)^4, x != 0.

   Hint. Group the last two terms as a single quantity y=x/2-2/x, expand (1+y)^4 = 1+4y+6y^2+4y^3+y^4 using the Binomial Theorem, then substitute back and expand each power of y in x.

   Let y = x/2-2/x. Then (1+y)^4 = 1+4y+6y^2+4y^3+y^4. Computing each power: y^2 = x^2/4-2+4/x^2. y^3 = x^3/8-3x/2+6/x-8/x^3. y^4 = x^4/16-x^2-16/x^2+16/x^4+6. Substituting 1, 4y, 6y^2, 4y^3, y^4 and collecting term by term by power of x gives x^4/16 + x^3/2 + x^2/2 - 4x - 5 + 16/x + 8/x^2 - 32/x^3 + 16/x^4 (verified numerically at x=1 and x=2 against the original expression).

   ✦ Working through each part gives: x^4/16 + x^3/2 + x^2/2 - 4x - 5 + 16/x + 8/x^2 - 32/x^3 + 16/x^4.

6. 7.M.66 marksNCERT Class 11 Mathematics, Binomial Theorem, Reprint 2026-27

   Find the expansion of (3x^2 - 2ax + 3a^2)^3 using binomial theorem.

   Hint. Group the two like-signed terms as one quantity, P=3x^2+3a^2, and the middle term as Q=-2ax, then expand (P+Q)^3=P^3+3P^2Q+3PQ^2+Q^3.

   Let P=3x^2+3a^2=3(x^2+a^2), Q=-2ax. P^3 = 27(x^2+a^2)^3 = 27(x^6+3x^4a^2+3x^2a^4+a^6) = 27x^6+81a^2x^4+81a^4x^2+27a^6. 3P^2Q = -54ax(x^2+a^2)^2 = -54ax(x^4+2a^2x^2+a^4) = -54ax^5-108a^3x^3-54a^5x. 3PQ^2 = 36a^2x^2(x^2+a^2) = 36a^2x^4+36a^4x^2. Q^3 = -8a^3x^3. Adding all four and collecting by power of x (verified numerically at a=1,x=1 giving 64=4^3, and at a=1,x=2 giving 1331=11^3) gives the full expansion.

   ✦ Working through each part gives: 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh107.pdf) — a 9-page, single-exercise chapter (Exercise 7.1, 14 questions) plus a 6-question Miscellaneous Exercise, 20 questions total. Confirmed against the CBSE curriculum PDF that the general term and middle term of a binomial expansion are formative-only content for this chapter (not summative), matching the current book's complete absence of any such section or a second numbered exercise. Several stacked-fraction and trinomial questions (Ex 7.1 Q2, Q4, Q5; Misc Q3, Q5, Q6) were cross-checked via 300dpi page renders, and the two trinomial-expansion answers (Misc Q5, Q6) were independently verified by numeric substitution at multiple values before being finalised. Questions are referenced from the NCERT textbook for identification.