2 Marks Questions

Question 12 Marks

Find the coefficient of:
x in the expansion of $(1-2\text{x}^3+3\text{x}^5)\Big(1+\frac{1}{\text{x}}\Big)^8.$

Answer

$(1-2\text{x}^2+3\text{x}^3)\Big(1+\frac{1}{\text{x}}\Big)^2$
$=(1-2\text{x}^3+3\text{x}^5)\Big({^8\text{C}}_0+{^8\text{C}}_1\big(\frac{1}{\text{x}}\big)+{^8\text{C}}_2\big(\frac{1}{\text{x}}\big)^2+{^8\text{C}}_3\big(\frac{1}{\text{x}}\big)^3\\+{^8\text{C}}_4\big(\frac{1}{\text{x}}\big)^4+{^8\text{C}}_5\big(\frac{1}{\text{x}}\big)^5+{^8\text{C}}_6\big(\frac{1}{\text{x}}\big)^6\Big)$
$$x occurs in the above expression at $-2\text{x}^3.{^8\text{C}}_2\big(\frac{1}{\text{x}^2}\big)+3\text{x}^5.{^8\text{C}}_4\big(\frac{1}{\text{x}}\big)4.$
$\therefore\ \text{Coefficient}\ \text{of}\ \text{x}=-2\Big(\frac{8!}{2!6!}\Big)+3\Big(\frac{8!}{4!4!}\Big)=-56+210=154$

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Question 22 Marks

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

Answer

$\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)^{10}$
Here, n = 10 which is even, therefore it has 11 terms
$\therefore$ middle term is $\Big(\frac{\text{n}}{2}+1\Big)=6^\text{k}\ \text{term}$
$\text{T}_\text{k}=\text{T}_{\text{r}+1}=(-1)^\text{r}\ {^\text{k}\text{C}}_\text{r}\text{x}^{\text{k}-\text{r}}\text{y}^\text{r}$
$\text{T}_6=\text{T}_{\text{s}+1}=(-1)^\text{s}{^{10}\text{C}}_\text{s}\big(\frac{\text{x}}{\text{a}}\big)^{10-5}\big(\frac{\text{a}}{\text{x}}\big)^5$
$=-\frac{10!}{5!5!}\times\frac{\text{x}^5}{\text{a}^5}\times\text{a}^5\times\text{x}^{-5}$
$=-252$

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Question 32 Marks

Find the 11th term from the beginning and the 11th term from the end in the expansion of $\Big(2\text{x}-\frac{1}{\text{x}^2}\Big)^{25}$

Answer

$\text{T}_{\text{r}+1={\text{T}}_\text{n}}=(-1)^\text{r}\ {^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$\text{T}_{11}=\text{T}_{10+1}=(-1)^{10}\ {^{25}\text{C}}_{10}(2\text{x})^{15}\Big(\frac{1}{\text{x}^2}\Big)^{10}\\={^{25}\text{C}}_{10}\Big(\frac{2^{15}}{\text{x}^5}\Big)=\frac{25!}{10!5!}2^{15}\text{x}{15}\times\text{x}^{-20}$
11th term from the end = (26 - 11 + 1) = 16th from beginning.
$\Rightarrow\text{T}_{16}=\text{T}_{15+1}=(-1)^{15}\ {^{25}\text{C}}_{15}(2\text{x})^{10}\Big(\frac{1}{\text{x}^2}\Big)^{15}={^{-25}\text{C}}_{15}\frac{2^{10}}{\text{x}^{20}}$

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Question 42 Marks

Find the 4th term from the beginning and 4th term from the end in the expansion of $\Big(\text{x}+\frac{2}{\text{x}}\Big)^9.$

Answer

Term from the beginning
$\text{T}_\text{N}=\text{T}_{\text{r}+1}={^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}...(\text{i})$
$\text{N}=4\ \text{r}=3,\ \text{n}=9,\text{x}=\text{ x},\text{y}=\frac{2}{\text{x}}$
$\text{T}_4=\text{T}_{3+1}={^9\text{C}}_3\text{x}^6\big(\frac{2}{\text{x}}\big)^3$$=\frac{9\times7\times8}{3\times2}\text{x}^3\times8=672\text{x}^3$
4th term from the end = 7th term from beginning
Using (i)
$\text{N}=7,\ \text{r}=6,\ \text{n}=9,\ \text{x}=\text{x},\ \text{y}=\frac{2}{\text{x}}$
$\text{T}_7=\text{T}_{6+1}={^9\text{C}_6\text{x}}^3\Big(\frac{2}{\text{x}}\Big)^6=\frac{9\times8\times7}{3\times2}\times\frac{2^6}{\text{x}^3}=\frac{5376}{\text{x}^3}$

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Question 52 Marks

Find the 7th term in the expansion of $\Big(3\text{x}^2-\frac{1}{\text{x}^3}\Big)^{10}$

Answer

$\text{T}_\text{n}=\text{T}_{\text{t}+1}=(-1)^\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}\times{^{10}\text{C}}_\text{r}$
$\text{n}=7,\ \text{r}=6,\ \text{x}=3\text{x}^2,\ \text{y}=\frac{1}{\text{x}^3}$
$\text{T}_7=\text{T}_{6+1}(-1)^6{^{10}\text{C}}_6(3\text{x}^2)^4\Big(\frac{1}{\text{x}^3}\Big)^6\\={^{10}\text{C}}_63^4\text{x}^8\times\frac{1}{\text{x}^{18}}={^{10}\text{C}}_6\times\frac{81}{\text{x}^{10}}=\frac{210\times81}{\text{x}^{10}}=\frac{17010}{\text{x}^{10}}$

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Question 62 Marks

Does the expansion of $\Big(2\text{x}^2-\frac{1}{\text{x}}\Big)^{20}$ contain any term involving $x^9?$

Answer

$(-1)^\text{r}\ {^{20}\text{C}}_\text{r}(2\text{x}^2)^{20-\text{r}}\big(\frac{1}{\text{x}}\big)^\text{r}$
$\text{x}^{40-2\text{r}}\text{x}^{-\text{r}}=\text{x}^9$
$40-3\text{r}=9$
$31=3\text{r}$
$\text{r}=\frac{31}{3}$
$r$ can not be in fraction
$\therefore$ There is no term involving $x^9.$

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Question 72 Marks

Find the 7th term in the expansion of $\Big(\frac{4\text{x}}{5}+\frac{5}{2\text{x}}\Big).$

Answer

$\text{T}_\text{N}=\text{T}_{\text{r}+1}={^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$\text{N}=7,\ \text{r}=6,\ \text{n}=8,\ \text{x}=\frac{4\text{x}}{5},\ \text{y}=\frac{5}{2\text{x}}$
$\text{T}_7=\text{T}_{6+1}={^8\text{C}}_6\Big(\frac{4\text{x}}{5}\Big)^2\Big(\frac{5}{2\text{x}}\Big)=28\times\frac{4^2}{5^2}\times\text{x}^4\times\frac{5^6}{2^6\times\text{x}^6}\\=\frac{28}{4}\times\frac{5^4}{\text{x}^4}=\frac{7\times5\times125}{\text{x}^4}=\frac{4375}{\text{x}^4}$

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Question 82 Marks

Find the middle term in the expansion of:
$\Big(\text{x}^2-\frac{2}{\text{x}}\Big)^{10}$

Answer

$\Big(\text{x}^2-\frac{2}{\text{x}}\Big)^{10}$
Here, $n = 10$
$\therefore\Big(\frac{\text{n}}{2}+1\Big)^\text{th}=\Big(\frac{10}{2}+1\Big)^\text{th} = 6^{th}$ term is the middle term.
The term formula is
$\text{T}_\text{n}=\text{T}_{\text{r}+1}=({-1})^\text{r}\ {^\text{n}\text{C}}_\text{r}\text{x}^{\text{r}-\text{n}}\text{y}^\text{r}$
$\text{T}_6=\text{T}_{5+1}=(-1)^5\ {^{10}\text{C}}_5(\text{x}^2)^{10-5}\big(\frac{2}{\text{x}}\big)^5$
$=-{^{10}\text{C}}_5\text{x}^{20-10}\frac{2^5}{\text{x}^5}$
$=\frac{-10\times9\times8\times7\times6}{5\times4\times3\times2}\times2^5\text{x}^5$
$=-8064\text{x}^5$

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Question 92 Marks

Which term in the expansion of $\Bigg\{\Big(\frac{\text{x}}{\sqrt{\text{y}}}\Big)^\frac{1}{3}+\bigg(\frac{\text{y}}{\text{x}^\frac{1}{3}}\bigg)^\frac{1}{2}\Bigg\}^{21}$ contains x and y to one and the same power?

Answer

$\text{T}_\text{n}=\text{T}_{\text{r}+1}={^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$={^{21}\text{C}}_\text{r}\Bigg(\Big(\frac{\text{x}}{\sqrt{\text{y}}}\Big)^\frac{1}{3}\Bigg)^{21-\text{r}}\Bigg(\bigg(\frac{\text{y}}{\text{x}^\frac{1}{3}}\bigg)^\frac{1}{2}\Bigg)^\text{r}$
$={^{21}\text{C}}_\text{r}\Bigg(\frac{\text{x}^{7-\frac{\text{r}}{3}}}{\text{y}^{\frac{7}{2}-\frac{\text{r}}{6}}}\Bigg)\frac{\text{y}^\frac{\text{r}}{2}}{\text{x}^\frac{\text{r}}{6}}$
$\frac{\text{x}^{7-\frac{\text{r}}{3}-\frac{\text{r}}{6}}}{\text{y}^{\frac{7}{2}-\frac{\text{r}}{6}-\frac{\text{r}}{2}}}$
$\Rightarrow\text{x}\frac{42-2\text{r}-\text{r}}{6}=\text{y}\frac{21-\text{r}-3\text{r}}{6}$
Since x and y have same power
$\frac{42-3\text{r}}{6}=\frac{-(21-4\text{r})}{6}$
$42+21=4\text{r}+3\text{r}$
$63=7\text{r}$
$\text{r}=9$
Term is 10th
$(\text{t}_\text{n}=\text{t}_{\text{r}+1})$

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Question 102 Marks

Find the coefficient of:
$x^7$ in the expansion of $\Big(\text{x}-\frac{1}{\text{x}^2}\Big)^{40}.$

Answer

$\text{x}^{7}\ \text{in}\Big(\text{x}-\frac{1}{\text{x}^2}\Big)^{40}$
$\text{T}_\text{n}=\text{T}_{\text{r}+1}=(-1)^\text{r}\ {^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$(-1)^\text{r}\ {^{40}\text{C}}_\text{r}\text{x}^{40-\text{r}}\Big(\frac{1}{\text{x}^2}\Big)^\text{r}$
$=(-1)^\text{r}\ {^{40}\text{C}}_\text{r}\text{x}^{40-\text{r}-2\text{r}}$
$\Rightarrow\text{x}^7=\text{x}^{40-3\text{r}}$
$7=40-3\text{r}$
$3\text{r}=33$
$\text{r}=11$
$=(-1)^{11}{^{40}\text{C}}_{11}\ \text{is}\ \text{coefficient}\ \text{of}\ \text{x}^7$
$=-{^{40}\text{C}}_{11}$

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Question 112 Marks

Find the coefficient of:
$x^m$ in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{n}.$

Answer

$\text{x}^\text{m}\ \text{in}\ \text{expansion}\ \text{of}\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{n}$
$\text{T}_\text{n}={^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$={^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^\text{r}$
$\text{x}^{\text{n}-2\text{r}}=\text{x}$
$\text{n}-2\text{r}=\text{m}$
$\text{r}=\frac{\text{n}-\text{m}}{2}$
$\frac{{^\text{n}\text{C}_{\text{n}-\text{m}}}}{2}=\frac{\text{n}!}{\Big(\frac{\text{n}-\text{m}}{2}\Big)!\Big(\frac{\text{n}+\text{m}}{2}\Big)!}$

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Question 122 Marks

Find the coefficient of:
$x^{10}$ in the expansion of $\Big(2\text{x}^2-\frac{1}{\text{x}}\Big)^{20}.$

Answer

$\text{x}^{10}\ \text{in}\Big(2\text{x}^2-\frac{1}{\text{x}}\Big)^{20}$
$\text{T}_\text{n}=\text{T}_{\text{r}+1}=(-1)^\text{r}\ {^\text{n}\text{C}}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$(-1)^\text{r}\ {^{20}\text{C}}_\text{r}(2\text{x}^2)^{20-\text{r}}\big(\frac{1}{\text{x}}\big)^\text{r}$
Coefficient of $x^{10}$ is
$(-1)^\text{r}\ {^{20}\text{C}}_\text{r}(2)^{20-\text{r}}\text{x}^{40-2\text{r}}\text{x}^{-\text{r}}....(\text{i})$
$\Rightarrow\text{x}^{40-3\text{r}}=\text{x}^{10}$
$\Rightarrow10-3\text{r}=10$
$3\text{r}=30$
$\text{r}=10$
Substituting $r = 10$ in $...(i)$
$(-1)^{10}\ {^{20}\text{C}}_{10}2^{10}$
$={^{20}\text{C}}_{10}2^{10}$

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