Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for the Chapter 7 Miscellaneous Exercise — all 6 questions solved, covering a general divisibility proof, binomial shortcuts for surd expressions, decimal approximations, and trinomial expansions.
Question 1 proves a genuinely useful general result — that a-b always divides a^n-b^n — by writing a=(a-b)+b and expanding. Questions 2 and 3 reuse the odd/even-term cancellation trick from Exercise 7.1 (Questions 11–12) on trickier surd expressions. Question 4 shows how just three terms of a binomial expansion already give an excellent decimal approximation. Questions 5 and 6 close the chapter with trinomial expansions — expressions with three terms rather than two — handled by grouping two of the three terms together so the Binomial Theorem still applies directly.
Following the hint, write a=(a-b)+b, so that a^n=[(a-b)+b]^n.
Expanding by the Binomial Theorem, with the first term (a-b) and second term b:
a^n={}^nC_0(a-b)^nb^0+{}^nC_1(a-b)^{n-1}b^1+{}^nC_2(a-b)^{n-2}b^2+\ldots+{}^nC_{n-1}(a-b)^1b^{n-1}+{}^nC_nb^n
Every term except the very last one, {}^nC_nb^n=b^n, carries at least one factor of (a-b). So:
a^n=(a-b)\left[{}^nC_0(a-b)^{n-1}+{}^nC_1(a-b)^{n-2}b+\ldots+{}^nC_{n-1}b^{n-1}\right]+b^n
Rearranging to isolate a^n-b^n:
a^n-b^n=(a-b)\left[{}^nC_0(a-b)^{n-1}+{}^nC_1(a-b)^{n-2}b+\ldots+{}^nC_{n-1}b^{n-1}\right]
This expresses a^n-b^n as (a-b) multiplied by an integer (since a and b are integers, every term in the bracket is an integer).
For any (x+y)^n-(x-y)^n, every even-power term in y cancels, leaving only twice the odd-power terms. With n=6, x=\sqrt3, y=\sqrt2, the surviving terms use {}^6C_1, {}^6C_3 and {}^6C_5:
(\sqrt3+\sqrt2)^6-(\sqrt3-\sqrt2)^6=2\left[{}^6C_1(\sqrt3)^5(\sqrt2)+{}^6C_3(\sqrt3)^3(\sqrt2)^3+{}^6C_5(\sqrt3)(\sqrt2)^5\right]
Computing each piece: (\sqrt3)^5=9\sqrt3, (\sqrt3)^3=3\sqrt3, (\sqrt2)^3=2\sqrt2, (\sqrt2)^5=4\sqrt2, and {}^6C_1=6, {}^6C_3=20, {}^6C_5=6.
=2\left[6(9\sqrt3)(\sqrt2)+20(3\sqrt3)(2\sqrt2)+6(\sqrt3)(4\sqrt2)\right]
=2\left[54\sqrt6+120\sqrt6+24\sqrt6\right]=2(198\sqrt6)=396\sqrt6
Let x=a^2 and y=\sqrt{a^2-1}. For (x+y)^4+(x-y)^4, every odd-power term in y cancels, leaving only twice the even-power terms, using {}^4C_0, {}^4C_2 and {}^4C_4:
(x+y)^4+(x-y)^4=2\left[{}^4C_0x^4+{}^4C_2x^2y^2+{}^4C_4y^4\right]=2\left[x^4+6x^2y^2+y^4\right]
Now substitute back x=a^2, so x^2=a^4, x^4=a^8; and y^2=a^2-1, so y^4=(a^2-1)^2=a^4-2a^2+1.
x^2y^2=a^4(a^2-1)=a^6-a^4
Substituting:
2\left[a^8+6(a^6-a^4)+(a^4-2a^2+1)\right]=2\left[a^8+6a^6-6a^4+a^4-2a^2+1\right]
=2\left[a^8+6a^6-5a^4-2a^2+1\right]=2a^8+12a^6-10a^4-4a^2+2
Write 0.99=1-0.01, so (0.99)^5=(1-0.01)^5.
By the Binomial Theorem:
(1-0.01)^5={}^5C_0(1)^5-{}^5C_1(1)^4(0.01)+{}^5C_2(1)^3(0.01)^2-\ldots
Taking only the first three terms as an approximation:
\approx1-5(0.01)+10(0.01)^2=1-0.05+10(0.0001)
=1-0.05+0.001=0.951
Treat this as a binomial by grouping the last two terms: a=1, b=\dfrac{x}{2}-\dfrac{2}{x}, n=4.
\left[1+\left(\dfrac{x}{2}-\dfrac{2}{x}\right)\right]^4={}^4C_0+{}^4C_1\left(\dfrac{x}{2}-\dfrac{2}{x}\right)+{}^4C_2\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^2+{}^4C_3\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^3+{}^4C_4\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^4
Now expand each power of \left(\dfrac{x}{2}-\dfrac{2}{x}\right) separately, using the Binomial Theorem again:
\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^1=\dfrac{x}{2}-\dfrac{2}{x}
\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^2=\dfrac{x^2}{4}-2+\dfrac{4}{x^2}
\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^3=\dfrac{x^3}{8}-\dfrac{3x}{2}+\dfrac{6}{x}-\dfrac{8}{x^3}
\left(\dfrac{x}{2}-\dfrac{2}{x}\right)^4=\dfrac{x^4}{16}-x^2+6-\dfrac{16}{x^2}+\dfrac{16}{x^4}
Multiplying each by its coefficient ({}^4C_1=4, {}^4C_2=6, {}^4C_3=4, {}^4C_4=1) and adding the constant term {}^4C_0=1:
1\ +\ 4\left(\dfrac{x}{2}-\dfrac{2}{x}\right)\ +\ 6\left(\dfrac{x^2}{4}-2+\dfrac{4}{x^2}\right)\ +\ 4\left(\dfrac{x^3}{8}-\dfrac{3x}{2}+\dfrac{6}{x}-\dfrac{8}{x^3}\right)\ +\ \left(\dfrac{x^4}{16}-x^2+6-\dfrac{16}{x^2}+\dfrac{16}{x^4}\right)
Collecting like powers of x term by term:
Constant: 1-12+6=-5 | x: 2x-6x=-4x | x^2: \tfrac{3}{2}x^2-x^2=\tfrac{1}{2}x^2 | x^3: \tfrac{1}{2}x^3 | x^4: \tfrac{1}{16}x^4
1/x: -8/x+24/x=16/x | 1/x^2: 24/x^2-16/x^2=8/x^2 | 1/x^3: -32/x^3 | 1/x^4: 16/x^4
Group the two square terms together so this becomes a true binomial: let A=3x^2+3a^2 and B=2ax, so the expression is (A-B)^3.
(A-B)^3=A^3-3A^2B+3AB^2-B^3
Finding A³: A=3(x^2+a^2), so A^3=27(x^2+a^2)^3=27\left(x^6+3x^4a^2+3x^2a^4+a^6\right)
=27x^6+81a^2x^4+81a^4x^2+27a^6
Finding 3A²B: A^2=9(x^2+a^2)^2=9(x^4+2a^2x^2+a^4), so 3A^2B=3\times9(x^4+2a^2x^2+a^4)\times2ax=54ax(x^4+2a^2x^2+a^4)
=54ax^5+108a^3x^3+54a^5x
Finding 3AB²: B^2=4a^2x^2, so 3AB^2=3\times3(x^2+a^2)\times4a^2x^2=36a^2x^2(x^2+a^2)=36a^2x^4+36a^4x^2
Finding B³: B^3=8a^3x^3
Combining A^3-3A^2B+3AB^2-B^3 and collecting like terms:
x^6:\ 27x^6 | x^5:\ -54ax^5 | x^4:\ 81a^2x^4+36a^2x^4=117a^2x^4
x^3:\ -108a^3x^3-8a^3x^3=-116a^3x^3 | x^2:\ 81a^4x^2+36a^4x^2=117a^4x^2
x^1:\ -54a^5x | a^6:\ 27a^6
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