NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 11 - Binomial Theorem

Question 1: 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]

Answer:

we need to prove,

$a^n-b^n=k(a-b)$ where k is some natural number.

Now let's add and subtract b from a so that we can prove the above result,

$a=a-b+b$

$a^n=(a-b+b)^n=[(a-b)+b]^n$

$=^nC_0(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_nb^n$

$=(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_{n-1}(a-b)b^{n-1}+b^n$$\Rightarrow a^n-b^n=(a-b)[(a-b)^{n-1}+^nC_2(a-b)^{n-2}+........+^nC_{n-1}b^{n-1}]$

$\Rightarrow a^n-b^n=k(a-b)$

Hence,$a - b$ is a factor of $a^n - b^n$.

Question 2: Evaluate $\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6$.

Answer:

First let's simplify the expression $(a+b)^6-(a-b)^6$ using binomial theorem,

So,

$(a+b)^6=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6b^6$$(a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$

And

$(a-b)^6=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6b^6$

$(a+b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$

Now,

$(a+b)^6-(a-b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$-a^6+6a^5b-15a^4b^2+20a^3b^3-15a^2b^4+6ab^5-b^6$

$(a+b)^6-(a-b)^6=2[6a^5b+20a^3b^3+6ab^5]$

Now, Putting $a=\sqrt{3}\:and\:b=\sqrt{2},$ we get

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2\times198\sqrt{6}$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=396\sqrt{6}$

Question 3: Find the value of $\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4$

Answer:

First, lets simplify the expression $(x+y)^4-(x-y)^4$ using binomial expansion,

$(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4$

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

And

$(x-y)^4=^4C_0x^4-^4C_1x^3y+^4C_2x^2y^2-^4C_3xy^3+^4C_4y^4$

$(x-y)^4=x^4-4x^3y+6x^2y^2-4xy^3+y^4$

Now,

$(x+y)^4-(x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4-$$x^4+4x^3y-6x^2y^2+4xy^3-y^4$

$(x+y)^4-(x-y)^4=2(x^4+6x^2y^2+y^4)$

Now, Putting $x=a^2 and\:y=\sqrt{a^2-1}$ we get,

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[a^8+6a^4(a^2-1)+(a^2-1)^2]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-12a^4+2a^4-4a^2+2$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-10a^4-4a^2+2$

Question 4: Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Answer:

As we can write 0.99 as 1-0.01,

$(0.99)^5=(1-0.001)^5=^5C_0(1)^5-^5C_1(1)^4(0.01)+^5C_2(1)^3(0.01)^2$$+\:other \:negligible \:terms$

$\Rightarrow (0.99)^5=1-5(0.01)+10(0.01)^2$

$\Rightarrow (0.99)^5=1-0.05+0.001$

$\Rightarrow (0.99)^5=0.951$

Hence the value of $(0.99)^5$ is 0.951 approximately.

Question 5: Expand using Binomial Theorem $\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Answer:

Given the expression,

$\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Binomial expansion of this expression is

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4\\=\left ( \left (1 + \frac{x}{2} \right ) -\frac{2}{x}\right )^4=^4C_0\left ( 1+\frac{x}{2} \right )^4-^4C_1\left ( 1+\frac{x}{2} \right )^3\left ( \frac{2}{x} \right )+$$^4C_2\left ( 1+\frac{x}{2} \right )^2\left ( \frac{2}{x} \right )^2$$-^4C_3\left ( 1+\frac{x}{2} \right )\left ( \frac{2}{x} \right )^3+^4C_4\left ( \frac{2}{x} \right )^4$

$\Rightarrow \left ( 1+\frac{x}{2} \right )^4-\frac{8}{x}\left ( 1+\frac{x}{2} \right )^3+$$\frac{24}{x^2}+\frac{24}{x}+6$$-\frac{32}{x^3}+\frac{16}{x^4}..........(1)$

Now Applying Binomial Theorem again,

$\left ( 1+\frac{x}{2} \right )^4=^4C_0(1)^4+^4C_1(1)^3\left ( \frac{x}{2} \right )+^4C_2(1)^2\left ( \frac{x}{2} \right )^2+^4C_3(1)\left ( \frac{x}{2} \right )^3$$+^4C_4\left ( \frac{x}{2} \right )^4$

$= 1+ 4\left ( \frac{x}{2} \right )+6\left ( \frac{x^2}{4} \right )+4\left ( \frac{x^3}{8} \right )+\frac{x^4}{16}$

$=1+2x+\frac{3x^2}{2}+\frac{x^3}{3}+\frac{x^4}{16}..............(2)$

And

$\left ( 1+\frac{x}{2} \right )^3=^3C_0(1)^3+^3C_1(1)^2\left ( \frac{x}{2} \right )+^3C_2(1)\left ( \frac{x}{2} \right )^2+^3C_3\left ( \frac{x}{2} \right )^3$

$\left ( 1+\frac{x}{2} \right )^3= 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}..........(3)$

Now, From (1), (2) and (3) we get,

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$=\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5$

Question 6: Find the expansion of $(3x^2 - 2ax +3a^2)^3$ using binomial theorem.

Answer:

Given $(3x^2 - 2ax +3a^2)^3$

By Binomial Theorem It can also be written as

$(3x^2 - 2ax +3a^2)^3=((3x^2 - 2ax) +3a^2)^3$

$= ^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3$

$= (3x^2-2ax)^3+3(3x^2-2ax)^2(3a^2)+3(3x^2-2ax)(3a^2)^2+(3a^2)^3$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6...........(1)$

Now, Again By Binomial Theorem,

$(3x^2-2ax)^3=^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3$

$(3x^2-2ax)^3= 27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^2x^3$

$(3x^2-2ax)^3= 27x^6-54x^5+36a^2x^4-8a^3x^3............(2)$

From (1) and (2) we get,

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+36a^2x^3+81a^2x^4-108a^3x^3+117a^4x^2$$-54a^5x+27a^6$

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+117a^2x^3-116a^3x^3+117a^4x^2$$-54a^5x+27a^6$