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Maths Fill In The Blanks[1 Marks ]

Binomial Theorem · CBSE STD 11 Science (CBSE - English Medium). 9 questions.

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Fill In The Blanks[1 Marks ]

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Question 11 Mark
The coefficient of $a^{-6}b^4$ in the expansion of $\Big(\frac{1}{\text{a}}-\frac{2\text{b}}{3}\Big)^{10}$ is ___________.
$[$Hint: $\text{T}_5=\ ^{10}\text{C}_4\Big(\frac{1}{\text{a}}\Big)^\text{b}\Big(\frac{-2\text{b}}{3}\Big)^4=\frac{1120}{27}\text{a}^{-6}\text{b}^4]$
Answer
The coefficient of $a^{-6} b^4$ in the expansion of $\Big(\frac{1}{\text{a}}-\frac{2\text{b}}{3}\Big)^{10}$ is $=\frac{1120}{27}$.
The given expansion is $\Big(\frac{1}{\text{a}}-\frac{2\text{b}}{3}\Big)^{10}$
From a$^{-6}b^4,$ we can take $r = 4$
$\text{T}_5=\text{T}_{4+1}=\ ^{10}\text{C}_4\Big(\frac{1}{\text{a}}\Big)^{10-4}\Big(-\frac{2\text{b}}{3}\Big)^4=\ ^{10}\text{C}_4\Big(\frac{1}{\text{a}}\Big)^{6}\Big(-\frac{-2}{3}\Big)^4.\text{b}^4$
$=\frac{10\times9\times8\times7}{4\times3\times2\times1}\times\frac{16}{81}.\text{a}^{-6}\text{b}^4=210\times\frac{16}{81 }\text{a}^{-6}\text{b}^4$
$=\frac{1120}{27}\text{a}^{-6}\text{b}^4$
Hence$,$ the value of the filler $=\frac{1120}{27}$
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Question 21 Mark
The number of terms in the expansion of $(x + y + z)^n$ __________.
$[$Hint: $(x + y + z)^n = [x + (y + z)]^n]$
Answer
The number of terms in the expansion of $(x + y + z)^n =\frac{(\text{n}+1)(\text{n}+2)}{2}$.
$(\text{x}+\text{y}+\text{z})^\text{n}=[\text{x}+(\text{y}+\text{z})]^\text{n}$
$=\ ^\text{n}\text{C}_0\text{x}^\text{n}+\ ^\text{n}\text{C}_1\text{x}^{\text{n}-1}(\text{y}+\text{z})+\ ^\text{n}\text{C}_2\text{x}^{\text{n}-2}(\text{y}+\text{z})^2+...+\ ^\text{n}\text{C}_\text{n}(\text{y}+\text{z})^\text{n}$
$\therefore$ Number of terms in expansion $= 1 + 2 + 3 + ... + \text{n} + (\text{n} + 1)$
$=\frac{(\text{n}+1)(\text{n}+2)}{2}$
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Question 31 Mark
If the seventh terms from the beginning and the end in the expansion of $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$ are equal, then n equals _____________.
[Hint: $\text{T}_7=\text{T}_{\text{n}-7+2}\Rightarrow\ ^\text{n}\text{C}_6\Big(2^\frac{1}{3}\Big)^{\text{n}-6}\bigg(\frac{1}{3^\frac{1}{3}}\bigg)^6$ $=\ ^\text{n}\text{C}_{\text{n}-6}\Big(2^\frac{1}{3}\Big)^6\bigg(\frac{1}{3^\frac{1}{3}}\bigg)^{\text{n}-6}$
$\Rightarrow\Big(2^\frac{1}{3}\Big)^{\text{n}-12}=\bigg(\frac{1}{3^{\frac{1}{3}}}\bigg)^{\text{n}-12}\Rightarrow$ only problem when $\text{n}-12=0\Rightarrow\text{n}=12]$
Answer
If the seventh terms from the beginning and the end in the expansion of $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$ are equal, then n equals n = 12.
Solution:
Given expression is $\Big(3\sqrt{2}+\frac{1}{3\sqrt{3}}\Big)^\text{n}$
According to the question,
$\frac{^\text{n}\text{C}_6(3\sqrt{2})^{\text{n}-6}\Big(\frac{1}{3\sqrt{3}}\Big)^6}{^\text{n}\text{C}_6\Big(\frac{1}{3\sqrt{3}}\Big)^{\text{n}-6}(3\sqrt{2})^6}=6$
$\Rightarrow(3\sqrt{2}\ 3\sqrt{3})^{\text{n}-12}=1\Rightarrow6^{\frac{\text{n}-12}{3}}=6^0\Rightarrow\frac{\text{n}-12}{3}=0$
$\Rightarrow\text{n}=12$
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Question 41 Mark
The position of the term independent of x in the expansion of $\Big(\sqrt{\frac{\text{x}}{3}}+\frac{3}{2\text{x}^2}\Big)^{10}$ is __________.
Answer
The position of the term independent of x in the expansion of $\Big(\sqrt{\frac{\text{x}}{3}}+\frac{3}{2\text{x}^2}\Big)^{10}$ is r = 2.
Solution:
Given expression is $\Big(\sqrt{\frac{\text{x}}{3}}+\frac{3}{2\text{x}^2}\Big)^{10}$
$\therefore\text{T}_{\text{r}+1}=\ ^{10}\text{C}_\text{r}\Big(\sqrt{\frac{\text{x}}{3}}\Big)^{10-\text{r}}\Big(\frac{3}{2\text{x}^2}\Big)^\text{r}$
For constant term, $10-5\text{r}=0\Rightarrow\text{r}=2$
Hence, third term is independent of x.
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Question 51 Mark
Middle term in the expansion of $(a^3 + b^a)^{28}$ is _________.
Answer
Middle term in the expansion of $(a^3 + b^a)^{28}$ is $=\ ^{28}\text{C}_{14}\text{a}^{56}\text{b}^{14}$.
Given expression is $(a^3 + ba)^{28}.$
Since index is $n = 28 ($even$)$
So$,$ there is only one middle term which is $\Big(\frac{28}{2}+1\Big)^\text{th}$ term or $15^{th}$ term.
$\therefore$ Middle term, $\text{T}_{15}=\text{T}_{14+1}=\ ^{28}\text{C}_{14}(\text{a}^3)^{28-14}(\text{ba})^{14}$
$=\ ^{28}\text{C}_{14}\text{a}^{42}\text{b}^{14}\text{a}^{14}=\ ^{28}\text{C}_{14}\text{a}^{56}\text{b}^{14}$
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Question 61 Mark
The largest coefficient in the expansion of $(1 + x)^{30}$ is _________.
Answer
The largest coefficient in the expansion of $(1 + x)^{30}$ is $^{30}\text{C}_{15}.$
Here $n = 30$ which is even
$\therefore$ the largest coefficient in $(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_{\frac{\text{n}}{2}}$
So$,$ the largest coefficient in $(1 + \text{x})^{30} =\ ^{30}\text{C}_{15}$
Hence$,$ the value of the filler is $^{30}\text{C}_{15}.$
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Question 71 Mark
The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ is _________.
$[$Hint: $^{\text{p} + \text{q}}\text{C}_\text{p} =\ ^{\text{p} + \text{q}}\text{C}_\text{q}]$
Answer
The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ is $1 : 1$.
Given expression is $(1 + \text{x})^{\text{p} + \text{q}}.$
$\therefore$ Coefficient of $\text{x}^\text{p}=\ ^{\text{p + q}}\text{C}_\text{p}$
And coefficient of $\text{x}^\text{q}=\ ^{\text{p + q}}\text{C}_\text{q}$
$\therefore\frac{^\text{p + q}\text{C}_\text{p}}{^\text{p + q}\text{C}_\text{q}}=\frac{^\text{p + q}\text{C}_\text{p}}{^\text{p + q}\text{C}_\text{p}}=1:1$
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Question 81 Mark
If $25^{15} $is divided by $13,$ the reminder is _________.
Answer
If $25^{15}$ is divided by $13,$ the reminder is $12.$
$25^{15}=(26-1)^{15}$ $=\ ^{15}\text{C}_0{26}^{15}-\ ^{15}\text{C}_126^{14}+...+\ ^{15}\text{C}_{14}26-\ ^{15}\text{C}_{15}$
$=(^{15}\text{C}_026^{15}-\ ^{15}\text{C}_126^{14}+...+\ ^{15}\text{C}_{14}26-13)+12$
Clearly$,$ when $25^{15}$ is divided by $13,$ then remainder will be $12.$
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Question 91 Mark
In the expansion of $\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)^{16},$ the value of constant term is ____________.
Answer
In the expansion of $\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)^{16},$ the value of constant term is $=\ ^{16}\text{C}_8$.
Let $T_{r + 1}$ be the constant term in the expansion of $\Big(\text{x}^2-\frac{1}{\text{x}^{2\text{r}}}\Big)^{16}$
$\therefore\text{T}_{\text{r}+1}=\ ^{16}\text{C}_\text{r}(\text{x}^2)^{16-\text{r}}\Big(\frac{-1}{\text{x}^2}\Big)^\text{r}=\ ^{16}\text{C}_\text{r}(\text{x})^{32-2\text{r}}(-1)^\text{r}.\frac{1}{\text{x}^{2\text{r}}}$
$=(-1)^\text{r}.\ ^{16}\text{C}_\text{r}(\text{x})^{32-2\text{r}-2\text{r}}\Rightarrow(-1)^\text{r}.\ ^{16}\text{C}_\text{r}(\text{x})^{32-4\text{r}}$
For getting constant term$, 32 - 4r = 0$
$\Rightarrow\text{r}=8$
$\therefore\text{T}_{\text{r}+1}=(-1)^8.\ ^{16}\text{C}_8=\ ^{16}\text{C}_8.$
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