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,

$a=a-b+b$

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

${=}^n{C}_0(a-b{)}^n{+}^n{C}_1(a-b{)}^{n-1}b+.......{.}^n{C}_n{b}^n$

$=(a-b{)}^n{+}^n{C}_1(a-b{)}^{n-1}b+.......{.}^n{C}_{n-1}(a-b)b^{n-1}+b^n$$\Rightarrow a^n-b^n=(a-b)[(a-b{)}^{n-1}{+}^n{C}_2(a-b{)}^{n-2}+........{+}^n{C}_{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 ${(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6$.

Answer:

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

So,

$(a+b{)}^6{=}^6{C}_0{a}^6{+}^6{C}_1{a}^{5}b{+}^6{C}_2{a}^4{b}^2{+}^6{C}_3{a}^3{b}^3{+}^6{C}_4{a}^2{b}^4{+}^6{C}_{5}a{b}^5{+}^6{C}_6{b}^6$$(a+b{)}^6=a^6+6a^{5}b+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a{b}^5+b^6$

And

$(a-b{)}^6{=}^6{C}_0{a}^6{-}^6{C}_1{a}^{5}b{+}^6{C}_2{a}^4{b}^2{-}^6{C}_3{a}^3{b}^3{+}^6{C}_4{a}^2{b}^4{-}^6{C}_{5}a{b}^5{+}^6{C}_6{b}^6$

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

Now,

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

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

Now, Putting $a=\sqrt{3}andb=\sqrt{2},$ we get

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

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

${(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6=2\times 198\sqrt{6}$

${(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6=396\sqrt{6}$

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

Answer:

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

$(x+y{)}^4{=}^4{C}_0{x}^4{+}^4{C}_1{x}^{3}y{+}^4{C}_2{x}^2{y}^2{+}^4{C}_{3}x{y}^3{+}^4{C}_4{y}^4$

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

And

$(x-y{)}^4{=}^4{C}_0{x}^4{-}^4{C}_1{x}^{3}y{+}^4{C}_2{x}^2{y}^2{-}^4{C}_{3}x{y}^3{+}^4{C}_4{y}^4$

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

Now,

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

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

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

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

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

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

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

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{=}^5{C}_0(1{)}^5{-}^5{C}_1(1{)}^4(0.01){+}^5{C}_2(1{)}^3(0.01{)}^2$$+othernegligibleterms$

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

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

$\Rightarrow (0.99{)}^5=0.951$

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

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

Answer:

Given the expression,

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

Binomial expansion of this expression is

$$\begin{gathered} {(1+\frac{x}{2}-\frac{2}{x})}^4 \\ ={((1+\frac{x}{2})-\frac{2}{x})}^4{=}^4{C}_0{(1+\frac{x}{2})}^4{-}^4{C}_1{(1+\frac{x}{2})}^3\left(\frac{2}{x}\right)+ \end{gathered}$$${}^4{C}_2{(1+\frac{x}{2})}^2{\left(\frac{2}{x}\right)}^2$${-}^4{C}_3(1+\frac{x}{2}){\left(\frac{2}{x}\right)}^3{+}^4{C}_4{\left(\frac{2}{x}\right)}^4$

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

Now Applying Binomial Theorem again,

${(1+\frac{x}{2})}^4{=}^4{C}_0(1{)}^4{+}^4{C}_1(1{)}^3\left(\frac{x}{2}\right){+}^4{C}_2(1{)}^2{\left(\frac{x}{2}\right)}^2{+}^4{C}_3(1){\left(\frac{x}{2}\right)}^3$${+}^4{C}_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

${(1+\frac{x}{2})}^3{=}^3{C}_0(1{)}^3{+}^3{C}_1(1{)}^2\left(\frac{x}{2}\right){+}^3{C}_2(1){\left(\frac{x}{2}\right)}^2{+}^3{C}_3{\left(\frac{x}{2}\right)}^3$

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

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

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

${(1+\frac{x}{2}-\frac{2}{x})}^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}$

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

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

${=}^3{C}_0(3x^2-2ax{)}^3{+}^3{C}_1(3x^2-2ax{)}^2(3a^2){+}^3{C}_2(3x^2-2ax)(3a^2{)}^2{+}^3{C}_3(3a^2{)}^3$

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

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

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

Now, Again By Binomial Theorem,

$(3x^2-2ax{)}^3{=}^3{C}_0(3x^2{)}^3{-}^3{C}_1(3x^2{)}^2(2ax){+}^3{C}_2(3x^2)(2ax{)}^2{-}^3{C}_3(2ax{)}^3$

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

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

From (1) and (2) we get,

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

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