3 Marks Question

Question 13 Marks

Find the middle term $($terms$)$ in the expansion of:
$\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)^{10}$

Answer

Given expression is $\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)^{10}$
Here index$, n = 10\ ($even$).$
So$,$ there is one middle term which is $\Big(\frac{10}{2}+1\Big)$ th term, i.e., $6^{th}$ term.
$\therefore\text{T}_6=\text{T}_{5+1}=\ ^{10}\text{C}_5\Big(\frac{\text{x}}{\text{a}}\Big)^{10-5}\Big(\frac{-\text{a}}{\text{x}}\Big)^5$
$=-\ ^{10}\text{C}_5\Big(\frac{\text{x}}{\text{a}}\Big)^5\Big(\frac{\text{a}}{\text{x}}\Big)^5$
$=-\frac{10\times9\times8\times7\times6\times5!}{5!\times5\times4\times3\times2\times1}\Big(\frac{\text{x}}{\text{a}}\Big)^5\Big(\frac{\text{x}}{\text{a}}\Big)^{-5}=-252$

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Question 23 Marks

Show that the middle term in the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}}$ is $\frac{1\times3\times5\times....(2\text{n}-1)}{\text{n}!}\times(-2)^\text{n}.$

Answer

Given, expression is $\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}}.$
Since the index is 2n, which is even. So, there is only one middle term, i.e., $\Big(\frac{2\text{n}}{2}+1\Big)$th term = (n + 1)th term
$\text{T}_{\text{n}+1}=\ ^{2\text{n}}\text{C}_\text{n}(\text{x})^{2\text{n}-\text{n}}\Big(-\frac{1}{\text{x}}\Big)^\text{n}=^{2\text{n}}\text{C}_\text{n}(-1)^\text{n}=(-1)^\text{n}\frac{(2\text{n}!)}{\text{n}!.\text{n}!}$
$=(-1)^\text{n}\frac{1\cdot2\cdot3\cdot4\cdot5...(2\text{n}-1)\cdot(2\text{n})}{\text{n}!\cdot\text{n}!}=(-1)^\text{n}\frac{[1\cdot3\cdot5..(2\text{n}-1)].[2\cdot4\cdot6...(2\text{n})]}{(1\cdot2\cdot3...\text{n})\cdot\text{n}!}$
$=(-1)^\text{n}\frac{[1\cdot3\cdot5...(2\text{n}-1)]\cdot2^\text{n}[1\cdot2\cdot3...\text{n}]}{(1\cdot2\cdot3...\text{n})\cdot\text{n}!}=(-2)^\text{n}\frac{[1\cdot3\cdot5..(2\text{n}-1)]\cdot2^\text{n}}{\text{n}!}$

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Question 33 Marks

Find the coefficient of $\frac{1}{\text{x}^{17}}$ in the expansion of $\Big(\text{x}^4-\frac{1}{\text{x}^3}\Big)^{15}.$

Answer

Given expression is $\Big(\text{x}^4-\frac{1}{\text{x}^3}\Big)^{15}$
$\therefore\text{T}_{\text{r}+1}=\ ^{15}\text{C}_\text{r}(\text{x}^4)^{15-\text{r}}\Big(-\frac{1}{\text{x}^3}\Big)^\text{r}$ $=\ ^{15}\text{C}_\text{r}\text{x}^{60-4\text{r}}(-1)^\text{r}\text{x}^{-3\text{r}}=\ ^{15}\text{C}_\text{r}\text{x}^{60-7\text{r}(-1)^\text{r}}$
For the coefficient $x^{-17},$ we have
$60-7\text{r}=-17\Rightarrow7\text{r}=77\Rightarrow\text{r}=11$
$\therefore\text{T}_{11+1}=\ ^{15}\text{C}_{11}\text{x}^{60-77}(-1)^{11}$
$\therefore$ Coefficient of $\text{x}^{-17}=\frac{-15\times14\times13\times12\times11!}{11!\times4\times3\times2\times1}$
$=-15\times7\times13=-1365$

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Question 43 Marks

Find n in the binomial $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}.$

Answer

Give expression is $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$
Now, $7^{th}$ term from beginning$,$ $\text{T}_7=\text{T}_{6+1}=^\text{n}\text{C}_6(3\sqrt{2})^{\text{n}-6}\Big(\frac{1}{3\sqrt{3}}\Big)^6\ ...(\text{i})$
And $7^{th}$ term from end is same as $7^{th}$ term from the beginning of $\Big(\frac{1}{3\sqrt{3}}+3\sqrt{2}\Big)^\text{n}$
i.e., $\text{T}_7=^\text{n}\text{C}_6\Big(\frac{1}{3\sqrt{3}}\Big)^{\text{n}-6}(3\sqrt{2})^6\ \ \ ....(\text{ii})$
Given that, $\frac{^\text{n}\text{C}_6(3\sqrt{2})^{\text{n}-6}\Big(\frac{1}{3\sqrt{3}}\Big)^6}{^\text{n}\text{C}_6\Big(\frac{1}{3\sqrt{3}}\Big)^{\text{n}-6}(3\sqrt{2})^6}=\frac{1}{6}$
$\Rightarrow\frac{(3\sqrt{2})^{\text{n}-12}}{\Big(\frac{1}{3\sqrt{3}}\Big)^{\text{n}-12}}=\frac{1}{6}$
$\Rightarrow(3\sqrt{2}\ \ 3\sqrt{3})^{\text{n}-12}$ $=6^{-1}$
$\Rightarrow6^\frac{\text{n}-12}{3}=6^{-1}$
$\Rightarrow\frac{\text{n}-12}{3}=-1$
$\Rightarrow\text{n}=9$

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Question 53 Marks

Find the coefficient of $x^{15}$ in the expansion of $(x - x^2)^{10}.$

Answer

The given expression is $(\text{x}-\text{x}^2)^{10}$
General Term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$=\ ^{10}\text{C}_\text{r}(\text{x})^{10-\text{r}}(-\text{x}^2)^\text{r}=\ ^{10}\text{C}_\text{r}(\text{x})^{10-\text{r}}(-1)^\text{r}.(\text{x}^2)^\text{r}$
$=(-1)^\text{r}.^{10}\text{C}_\text{r}(\text{x})^{10-\text{r}+2\text{r}}=(-1)^\text{r}.^{10}\text{C}_\text{r}(\text{x})^{10+\text{r}}$
To find the coefficient of $\text{x}^{15},\text{Put}\ 10+\text{r}=15$
$\Rightarrow\text{r}=5$
$\therefore$ Hence$,$ the required coefficient $= -252$

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Question 63 Marks

If the coefficient of second, third and fourth terms in the expansion of $(1 + x)^{2n}$ are in $A.P. $Show that $2n^2 - 9n + 7 = 0.$

Answer

Given expression is $(1 + \text{x})^\text{2n}$
Now$,$ coefficient of $2^{\text{nd}}, 3^\text{rd}$ and $4^{th}$ terms are $^\text{2n}\text{C}_1,\ ^\text{2n}\text{C}_2$ and $^\text{2n}\text{C}_3,$ respectively.
Given that, $^\text{2n}\text{C}_2,\ ^\text{2n}\text{C}_2$ and $^\text{2n}\text{C}_3$ are in $A.P.$
Then, $2.^{2\text{n}}\text{C}_2=\ ^{2\text{n}}\text{C}_1+\ ^{2\text{n}}\text{C}_3$
$\Rightarrow2\Big[\frac{2\text{n}(2\text{n}-1)(2\text{n}-2)!}{2\times1\times(2\text{n}-2)!}\Big]=2\text{n}+\frac{2\text{n}(2\text{n}-1)(2\text{n}-2)(2\text{n}-3)!}{3!(2\text{n}-3)!}$
$\Rightarrow\text{n}(2\text{n}-1)=\text{n}+\frac{\text{n}(2\text{n}-1)(\text{n}-1)}{3}$
$\Rightarrow3(2\text{n}-1)=3+(2\text{n}^2-3\text{n}+1)$
$\Rightarrow6\text{n}-3=2\text{n}^2-3\text{n}+4\Rightarrow2\text{n}^2-9\text{n}+7=0$

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Question 73 Marks

If $p$ is a real number and if the middle term in the expansion of $\Big(\frac{\text{p}}{2}+2\Big)^8$ is $1120,$ find $P.$

Answer

Give expression is $\Big(\frac{\text{P}}{2}+2\Big)^8$
Number of term $= 8 + 1 = 9 ($odd$)$
$\therefore$ Middle term $=\frac{9+1}{2}\text{th}$ term $= 5^{th}$ term
$\therefore\text{T}_5=\text{T}_{4+1}=\ ^8\text{C}_\text{4}\Big(\frac{\text{P}}{2}\Big)^{8-4}(2)^4$
$=\ ^8\text{C}_4\frac{\text{P}^4}{2^4}\times2^4=\ ^8\text{C}_4\text{P}^4$
Now $^8\text{C}_4\text{P}^4=1120\Rightarrow\frac{8\times7\times6\times5}{4\times3\times2\times1}\text{P}^4=1120$
$\Rightarrow70\text{P}^4=1120$
$\Rightarrow\text{P}^4=\frac{1120}{70}=16$
$\Rightarrow\text{P}^4=2^2$
$\Rightarrow\text{P}=\pm2$
Hence$,$ the required value of $\text{P}=\pm2.$

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Question 83 Marks

Find the term independent of $\text{x},\text{x}\neq0,$, in the expansion of $\Big(\frac{3\text{x}^2}{2}-\frac{1}{3\text{x}}\Big)^{15}$

Answer

Given expansion is $\Big(\frac{3\text{x}^2}{2}-\frac{1}{3\text{x}}\Big)^{15}$
$\therefore\text{T}_{\text{r}+1}=\ ^{15}\text{C}_\text{r}\Big(\frac{3\text{x}^2}{2}\Big)^{15-\text{r}}\Big(-\frac{1}{3\text{x}}\Big)^\text{r}$
or $\text{T}_{\text{r}+1}=\ ^{15}\text{C}_\text{r}(-1)^\text{r}3^{15-\text{r}}\ 2^{\text{r}-15}\ \text{x}^{30-3\text{r}}\ \ ....(\text{i})$
For the term independent of $\text{x},30-3\text{r=0}\Rightarrow\text{r}=10$
$\therefore$ The term independent of x is
$\text{T}_{10+1}=\ ^{15}\text{C}_{10}\ 3^{-5}\ 2^{-5}$ (Putting r = 10 in (i))
$=\ ^{15}\text{C}_{10}\Big(\frac{1}{6}\Big)^5$

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Question 93 Marks

Find the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3 )^{11}.$

Answer

Given expression is $(1+\text{x}+\text{x}^2+\text{x}^3)^{11}$
$=[(1+\text{x})+\text{x}^2(1+\text{x})]^{11}=[(1+\text{x})(1+\text{x}^2)]^{11}$
$=(1+\text{x})^{11}.(1+\text{x}^2)^{11}$
Expanding the above expression$,$ we get$,$
$(\ ^{11}\text{C}_0+\ ^{11}\text{C}_1\text{x}+\ ^{11}\text{C}_2\text{x}^2+\ ^{11}\text{C}_3\text{x}^3+\ ^{11}\text{C}_4\text{x}^4+\ ...).$
$(\ ^{11}\text{C}_0+\ ^{11}\text{C}_1\text{x}^2+\ ^{11}\text{C}_1\text{x}^4+)$
$=(1+11\text{x}+55\text{x}^2+165\text{x}^3+330\text{x}^4...).(1+11\text{x}^2+55\text{x}^4+...)$
$(55+605+330)\text{x}^4=990\text{x}^4$
Hence$,$ the coefficient of $x^4 = 990.$

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Question 103 Marks

Find the value of $r,$ if the coefficients of$ (2r + 4)^{th}$ and $(r - 2)^{th}$ terms in the expansion of $(1 + x)^{18}$ are equal.

Answer

Given expression is $(1 + x)^{18}$
Now$, (2r + 4)^{th}$ term, i.e., $\text{T}_{(2\text{r}+3)+1}$
$\therefore\ \text{T}_{(2\text{r}+3)+1}=\ ^{18}\text{C}_{2\text{r}+3}(\text{x})^{2\text{r}+3}$
And $(r - 2)^{th}$ term, i.e., $\text{T}_{(\text{r}-3)+1}$
$\therefore\ \text{T}_{(\text{r}-3)+1}=\ ^{18}\text{C}_{\text{r}-3}\text{x}^{\text{r}-3}$
According to the question,
$^{18}\text{C}_{2\text{r}+3}=\ ^{18}\text{C}_{\text{r}-3}$
$\Rightarrow2\text{r}+3+\text{r}-3=18$
$[\because\ ^\text{n}\text{C}_\text{x}=\ ^\text{n}\text{C}_\text{y}\Rightarrow\text{x}+\text{y}=\text{n}]$
$\Rightarrow3\text{r}=18$
$\therefore\text{r}=6$

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