5 Marks Questions

Question 15 Marks

Find the sixth term of the expansion $\Big(\text{y}^\frac{1}{2}+\text{x}^\frac{1}{3}\Big)^\text{n},$ if the binomial coefficient of the third term from the end is $45.$
$[$Hint: Binomial coefficient of third term from the end $=$ Binomial coefficient of third term from beginning $= ^nC_2.]$

Answer

The given expression is $\Big(\text{y}^{\frac{1}{2}}+\text{x}^\frac{1}{3}\Big),$ since the binomial coefficient of third term from the end $=$ Binomial coefficient of third term from the beginning $= ^nC_2$
$\therefore\ ^\text{n}\text{C}_2=45$
$\Rightarrow\frac{\text{n}(\text{n}-1)}{2}=45\Rightarrow\text{n}^2-\text{n}=90$
$\Rightarrow\text{n}^2-\text{n}-90=0\Rightarrow\text{n}^2-10\text{n}+9\text{n}-90=0$
$\Rightarrow\text{n}(\text{n}-10)+9(\text{n}-10)\Rightarrow(\text{n}-10)(\text{n}+9)=0$
$\Rightarrow\text{n}=10,\text{n}=-9\Rightarrow\text{n}=10,\text{n}\neq-9$
So, the given expression becomes $\Big(\text{y}^{\frac{1}{2}}+\text{x}^\frac{1}{3}\Big)^{10}$
Sixth term is this expression
$\text{T}_6=\text{T}_{5+1}=\ ^{10}\text{C}_5\Big(\text{y}^\frac{1}{2}\Big)^{10-5}\Big(\text{x}^\frac{1}{3}\Big)^5=\ ^{10}\text{C}_5\text{y}^\frac{5}{2}.\text{x}^\frac{5}{3}$
$=252\ \text{y}^\frac{5}{2}\text{x}^\frac{5}{3}$
Hence$,$ the required term $=252\ \text{y}^\frac{5}{2}.\text{x}^\frac{5}{3}$

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Question 25 Marks

If the term free from x in the expansion of $\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^2}^{10}$ is 405, find the value of k.

Answer

The given expression is $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^2}\Big)^{10}$
General term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\ \text{y}^\text{r}$
$=\ ^{10}\text{C}_\text{r}(\sqrt{x})^{10-\text{r}}\Big(\frac{-\text{k}}{\text{x}^2}\Big)=\ ^{10}\text{C}_\text{r}(\text{x})^\frac{10-\text{r}}{2}(-\text{k})^\text{r}\Big(\frac{1}{\text{x}^{2\text{r}}}\Big)$
$=\ ^{10}\text{C}_\text{r}(\text{x})^{\frac{10-\text{r}}{2}-2\text{r}}(-\text{k})^\text{r}=\ ^{10}\text{C}_\text{r}(\text{x})^\frac{10-\text{r}-4\text{r}}{2}(-\text{k})^\text{r}$
$=\ ^{10}\text{C}_\text{r}(\text{x})^{\frac{10-5\text{r}}{2}}(-\text{k})^\text{r}$
For getting term free from $\text{x},\frac{10-5\text{r}}{2}=0$
$\Rightarrow\text{r}=2$
On putting the value of r in the above expression, we get $^{10}\text{C}_2(-\text{k})^2$
According to the condition of the queston, we have
$^{10}\text{C}_2\text{K}^2=405\Rightarrow\frac{10.9}{2.1}\text{K}^2=405$
$\Rightarrow45\text{K}^2=405$
$\Rightarrow\text{K}^2=\frac{405}{45}=9$
$\therefore\text{K}=\pm3$
Hence, the value of $\text{K}=\pm3$

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Question 35 Marks

Find the term independent of x in the expansion of $(1+\text{x}+2\text{x}^3)\Big(\frac{3}{2}\text{x}^2-\frac{1}{3\text{x}}\Big)^9.$

Answer

Given expansion is $(1+\text{x}+2\text{x}^3)\Big(\frac{3}{2}\text{x}^2-\frac{1}{3\text{x}}\Big)^9$
Now, consider $\Big(\frac{3}{2}\text{x}^2-\frac{1}{3\text{x}}\Big)^9$
$\text{T}_{\text{r}+1}=\ ^9\text{C}_\text{r}\Big(\frac{3}{2}\text{x}^2\Big)^{9-\text{r}}\Big(-\frac{1}{3\text{x}}\Big)=\ ^9\text{C}_\text{r}\Big(\frac{3}{2}\Big)^{9-\text{r}}\Big(-\frac{1}{3}\Big)^\text{r}\text{x}^{18-3\text{r}}$
Hence, the general term in the expansion of $(1+\text{x}+2\text{x}^3)\Big(\frac{3}{2}\text{x}^2-\frac{1}{3\text{x}}\Big)^9$ is:
$^9\text{C}_\text{r}\Big(\frac{3}{2}\Big)^{9-\text{r}}\Big(-\frac{1}{3}\Big)^\text{r}\text{x}^{18-3\text{r}}+\ ^9\text{C}_\text{r}\Big(\frac{3}{2}\Big)^{9-\text{r}}\Big(-\frac{1}{3}\Big)^\text{r}\text{x}^{19-3\text{r}}\\+2.\ ^9\text{C}_\text{r}\Big(\frac{3}{2}\Big)^{9-\text{r}}\Big(-\frac{1}{3}\Big)^\text{r}\text{x}^{21-3\text{r}}$
For term independent of x, putting 18 - 3r = 0, 19 - 3r = 0 and 21 - 3r = 0, we get,
r = 6, r = 7
Hence, second term is not independent of x. Therefore, term independent of x is:
$^9\text{C}_6\Big(\frac{3}{2}\Big)^{9-6}\Big(-\frac{1}{3}\Big)^6+2.\ ^9\text{C}_7\Big(\frac{3}{2}\Big)^{9-7}\Big(-\frac{1}3\Big)^7$
$=\frac{9\times8\times7\times6!}{6!\times3\times2}\cdot\frac{1}{2^3\cdot3^3}-2\cdot\frac{9\times8\times7!}{7!\times2\times1}\cdot\frac{3^2}{2^2}\cdot\frac{1}{3^7}$
$=\frac{84}{8}\cdot\frac{1}{3^3}-\frac{36}{4}\cdot\frac{2}{3^5}=\frac{21-4}{54}=\frac{17}{54}$

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Question 45 Marks

In the expansion of $(x + a)^n$ if the sum of odd terms is denoted by $O$ and the sum of even term by $E$.
Then prove that,

$\text{O}^2 – \text{E}^2 = (\text{x}^2 – \text{a}^2 )^\text{n}$
$4\text{OE} = (\text{x} + \text{a})^{2\text{n}} – (\text{x} – \text{a})^{2\text{n}}$

Answer

Given expression is $(x - a)^n$
$(\text{x}+\text{a})\text{n}=\ ^\text{n}\text{C}_0\text{x}^\text{n}\text{a}^0+\ ^\text{n}\text{C}_1\text{x}^{\text{n}-1}\text{a}+\ ^\text{n}\text{C}_2\text{x}^{\text{n}-2}\text{a}^2$
$+\ ^\text{n}\text{C}_3\text{x}^{\text{n}-3}\text{a}^3+...+\ ^\text{n}\text{C}_\text{n}\text{a}^\text{n}$Sum of odd terms,
$\text{O}=\ ^\text{n}\text{C}_0\text{x}^\text{n}+\ ^\text{n}\text{C}_2\text{x}^{\text{n}-2}\text{a}^2+\ ^\text{n}\text{C}_4\text{x}^{\text{n}-4}\text{a}^4+...$ and the sum of even terms,
$\text{E}=\ ^\text{n}\text{C}_1\text{x}^{\text{n}-1}.\text{a}+\ ^\text{n}\text{C}_3\text{x}^{\text{n}-3}\text{a}^3+\ ^\text{n}\text{C}_5\text{x}^{\text{n}-5}\text{a}^5+...$
Now $(x + a)^n = O + E ....(i)$
Similarly $(x - a)^n = O - E ....(ii)$
Multiplying eq. $(i)$ and rq. $(ii),$ we get$,$
$(\text{x}+\text{a})^\text{n}(\text{x}-\text{a})^\text{n}=(\text{O}+\text{E})(\text{O}-\text{E})$
$\Rightarrow(\text{x}^2-\text{a}^2)\text{n}=\text{O}^2-\text{E}^2$
Hence $\text{O}^2-\text{E}^2=(\text{x}^2-\text{a}^2)^\text{n}$
$4\text{OE}=(\text{O}+\text{E})^2-(\text{O}-\text{E})^2$$=[(\text{x}+\text{a})^\text{n}]^2-[(\text{x}-\text{a})^\text{n}]^2$
$=[(\text{x}+\text{a})]^{2\text{n}}-[\text{x}-\text{a}]^{2\text{n}}$
Hence$, 4\text{OE}=(\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}$

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Question 55 Marks

Find the term independent of x in the expansion of, $3\text{x}-\frac{2}{\text{x}^2}^{15}.$

Answer

Given expression is $\Big(3\text{x}-\frac{2}{\text{x}^2}\Big)^{15}$
General term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$=\ ^{15}\text{C}_\text{r}(3\text{x})^{15-\text{r}}\Big(-\frac{2}{\text{x}^2}\Big)^\text{r}=\ ^{15}\text{C}_\text{r}(3)^{15-\text{r}}.\text{x}^{15-\text{r}}(-2)^\text{r.}\frac{1}{\text{x}^{2\text{r}}}$
$=\ ^{15}\text{C}_\text{r}(3)^{15-\text{r}}.\text{x}^{15-\text{r}-2\text{r}}.(-2)^\text{r}=\ ^{15}\text{C}_\text{r}(3)^{15-\text{r}}.\text{x}^{15-3\text{r}}(-2)^\text{r}$
For getting a term independent of x, put $15-3\text{r}=0\Rightarrow\text{r}=5$
$\therefore$ The required term is $^{15}\text{C}_5(3)^{15-5}(-2)^5$
$=-\ ^{15}\text{C}_5(3)^{10}(2)^5=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}.(3)^{10}(2)^5$
$=-7\times13\times3\times11.(3)^{10}(2)^5=-3003.(3)^{10}(2)^5$
Hence, the required term $=-3003(3)^{10}(2)^5$

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Question 65 Marks

If $x^p$ occurs in the expansion of $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{2\text{n}},$ prove that its coefficient is $\frac{2\text{n}!}{\Big(\frac{4\text{n}-\text{p}}{3}\Big)!\Big(\frac{2\text{n}+\text{p}}{3}\Big)!}.$

Answer

Given expression is $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{2\text{n}}$
General terms$, \text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
$=\ ^{2\text{n}}\text{C}_\text{r}(\text{x}^2)^{2\text{n}-\text{r}}.\Big(\frac{1}{\text{x}}\Big)=\ ^{\text{2n}}\text{C}_\text{r}(\text{x})^{4\text{n}-2\text{r}}.\frac{1}{\text{x}^\text{r}}$
$=\ ^{2\text{n}}\text{C}_\text{r}(\text{x})^{4\text{n}-2\text{r}-\text{r}}=\ ^{2\text{n}}\text{C}_\text{r}(\text{x})^{4\text{n}-2\text{r}}$
If $x^p$ occurs in $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{2\text{n}}$
Then $4\text{n}-3\text{r}=\text{p}\Rightarrow4\text{n}-\text{p}$
$\Rightarrow\text{r}=\frac{4\text{n}-\text{p}}{3}$
$\therefore$ Coefficient of $\text{x}^\text{p}=\ ^{2\text{n}}\text{C}_\text{r}=\ ^{2\text{n}}\text{C}_{\frac{4\text{n}-\text{p}}{3}}$
$=\frac{(2\text{n})!}{\Big(\frac{4\text{n}-\text{p}}{3}\Big)!\Big(2\text{n}-\frac{4\text{n}-\text{p}}{3}\Big)!}=\frac{(2\text{n)!}}{\Big(\frac{4\text{n}-\text{p}}{3}\Big)!\Big(\frac{6\text{n}-4\text{n}+\text{p}}{3}\Big)!}$
$=\frac{(2\text{n})!}{\Big(\frac{4\text{n}-\text{p}}3\Big)!\Big(\frac{2\text{n}+\text{p}}{3}\Big)!}$
Hence$,$ the coefficient of $\text{x}^\text{p}=\frac{(2\text{n})!}{\Big(\frac{4\text{n}-\text{p}}{3}\Big)!\Big(\frac{2\text{n}+\text{p}}{3}\Big)!}$

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Question 75 Marks

Find the middle term $($terms$)$ in the expansion of:
$\Big(3\text{x}-\frac{\text{x}^3}{6}\Big)^{9}$

Answer

Given expression is $\Big(3\text{x}-\frac{\text{x}^3}{6}\Big)^{9}$
Here index$, n = 9\ ($odd$).$
So, there are two middle term which are $\Big(\frac{9+1}{2}\Big)$ th i.e., $5^{th}$ term and $\Big(\frac{9+1}{2}+1\Big)$ th i.e., $6^{th}$ term.
$\therefore\text{T}_5=\text{T}_{4+1}=\ ^{9}\text{C}_4(3\text{x})^{9-4}\Big(-\frac{\text{x}^3}{6}\Big)^4$
$=\frac{9\times8\times7\times6\times5!}{4\times3\times2\times1\times5!}3^5\text{x}^5\text{x}^{12}6^{-4}$
$=\frac{9\times8\times7\times6}{4\times3\times2\times1}\times\frac{3^5}{3^4\times2^4}\text{x}^{17}=\frac{189}{8}\text{x}^{17}$
And $\text{T}_6=\text{T}_{5+1}=\ ^{9}\text{C}_5(3\text{x})^{9-5}\Big(-\frac{\text{x}^3}{6}\Big)^5$
$=-\frac{9\times8\times7\times6\times5!}{5!\times4\times3\times2\times1}.3^4.\text{x}^4.\text{x}^{15}.6^{-5}=-\frac{-21}{16}\text{x}^{19}$

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