1 Marks Question

Question 11 Mark

Write the number of terms in the expansion of $(1-3\text{x}+3\text{x}^{2}-\text{x}^{3})^{8}.$

Answer

We have,
$(1-3\text{x}+3\text{x}^{2}-\text{x}^{3})^{8}=\Big[(1+\text{x}^{3})\Big]^{8}=(1-\text{x})^{24}$
So, there are 25 terms in the expansion of $(1-3\text{x}+3\text{x}^{2}-\text{x}^{3})^{8}.$

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Question 21 Mark

Write the number of terms in the expansion of $(2+\sqrt{3}\text{x})^{10}+(2-\sqrt{3}\text{x})^{10}.$

Answer

Number of terms in the expansion $(\text{x}+\text{y})^{\text{n}}+(\text{x}-\text{y})^{\text{n}}$ where n is even $=\big(\frac{\text{n}}{2}+1\big)$
Thus, we have
Number of terms in the given expansion $=\Big(\frac{10}{2}+1\Big)=6$

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Question 31 Mark

Write the total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}.$

Answer

The total number of terms are 101 of which 50 terms get cancelled.
Hence, the total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}$ is 51.

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Question 41 Mark

Find the ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1+\text{x})^{\text{p}+\text{q}}.$

Answer

Coefficients of $x^p$ in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is ${^\text{q}}\text{C}_{\text{p}}.$
Coefficients of $x^q$ in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is ${^\text{q}}\text{C}_{\text{p}}.$
Now,
$\frac{{^\text{p+q}}\text{C}_{\text{p}}}{{^\text{p+q}}\text{C}_{\text{q}}}=\frac{\frac{(\text{p+q})!}{\text{p!}\text{q}!}}{{\frac{(\text{p+q})!}{\text{q!}\text{p}!}}}=1$
Hence, the ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is $1 : 1.$

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Question 51 Mark

Write last two digits of the number $3^{400}.$

Answer

$3^{400}=(9)^{200}$
$=(10-1)^{200}$
$={^\text{200}}\text{C}_{\text{0}}(10)^{200}+{^\text{200}}\text{C}_{\text{1}}(10)^{199}(-1)^{1}.....\\+{^\text{200}}\text{C}_{\text{108}}(10)^{2}(-1)^{198}+{^\text{200}}\text{C}_{\text{199}}(10)^{1}(-1)^{199}+{^\text{200}}\text{C}_{\text{200}}(-1)^{200}$
$=100\Big[(10)^{198}+{^\text{200}}\text{C}_{\text{1}}(10)^{197}(-1)^{1}+......+{^\text{200}}\text{C}_{\text{198}}(-1)^{198}\Big]+200(10)^{199}+(-1)^{200}$
$=100\Big[(10)^{198}-{^\text{200}}\text{C}_{\text{1}}(10)^{197}+......+{^\text{200}}\text{C}_{\text{198}}-2(10)\Big]+1$
$=100(\text{a natural number})+1$
Hence, last two digits of the number $3^{400}$ is $01.$

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Question 61 Mark

If $(1-\text{x}+\text{x}^{2})^{\text{n}}=\text{a}^{0}+\text{a}_{1}\text{x}+\text{a}_{2}\text{x}^{2}+...+\text{a}_{2\text{n}}\text{x}^{2\text{n}},$find the value of $\text{a}_{0}+\text{a}_{2}+\text{a}_{4}+...+\text{a}_{2\text{n}}.$

Answer

Putting x = 1 and -1 in $(1-\text{x}+\text{x}^{2})^{\text{n}}=\text{a}^{0}+\text{a}_{1}\text{x}+\text{a}_{2}\text{x}^{2}+...+\text{a}_{2\text{n}}\text{x}^{2\text{n}}$
we get,
$1=\text{a}_{0}+\text{a}_{1}+\text{a}_{2}+....+\text{a}_{2\text{n}}\ ...(\text{1})$ and,
$3^{\text{n}}=\text{a}_{0}-\text{a}_{1}+\text{a}_{2}-....+\text{a}_{2\text{n}}\ ...(\text{2})$
Adding (i) and (ii), we get
$3^{\text{n}}+1=2(\text{a}_{0}+\text{a}_{2}+...+\text{a}_{2\text{n}})$
Hence, the value of $\text{a}_{0}+\text{a}_{2}+\text{a}_{4}+..\text{a}_{2\text{n}}$ is $\frac{3^{\text{n}}+1}{2}.$

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Question 71 Mark

Write the sum of the coefficients in the expansion of $(1-3\text{x}+\text{x}^{2})^{111}.$

Answer

$(1-3\text{x}+\text{x}^{2})^{111}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{111}+{^\text{111}}\text{C}_{\text{1}}(1)^{100}(-3\text{x}+\text{x}^{2})+{^\text{111}}\text{C}_{\text{2}}(1)^{109}\$-3\text{x}+\text{x}^{2})^{2+....... +{^\text{111}}\text{C}_{\text{111}}}(-3\text{x}+\text{x}^{2})^{111}$
Let x = 1 on both sides,
$(1-3+1)^{111}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{111}+{^\text{111}}\text{C}_{\text{1}}(1)^{100}(-3+1)+{^\text{111}}\text{C}_{\text{2}}(1)^{109}\$-3+1)^{2+....... +{^\text{111}}\text{C}_{\text{111}}}(-3+1)^{111}$
The sum of the coefficients is $(-1)^{111}= -1.$

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Question 81 Mark

Find the number of terms in the expansion of $(\text{a}+\text{b}+\text{c})^{\text{n}}.$

Answer

We have,
$(\text{a}+\text{b}+\text{c})^{\text{n}}=\big[\text{a}+(\text{b}+\text{c})\big]^{\text{n}}$
$=\text{a}^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}(\text{b}+\text{c})^{1}+{^\text{n}}\text{C}_{\text{2}}\text{a}^{\text{n}-2}(\text{b}+\text{c})^{2}....+{^\text{n}}\text{C}_{\text{n}}(\text{b}+\text{c})^{\text{n}}$
Further, expanding each term of R.H.S., we note that,
First term consists of 1 term.
Second term on simplification gives 2 terms.
Third term on expansion gives 3 terms.
Similarly, fourth term on expansion gives 4 terms and so on.
$\therefore$ The total number of term $=1+2+3+...+(\text{n}+1)=\frac{(\text{n}+1)(\text{n}+2)}{2}.$

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Question 91 Mark

If a and b are coefficients of $x^n$ in the expansions of $(1+\text{x})^{2\text{n}}$ and $(1+\text{x})^{2\text{n}-1}$ respectively, then write the relation between a and b.

Answer

The coefficients of $x^n$ in $(1+\text{x})^{2\text{n}}$ is twice the coefficient of $x^n$ in $(1+\text{x})^{2\text{n}-1}.$
$\therefore$ a = 2b

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Question 101 Mark

Write the middle term in the expansion of $\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}.$

Answer

$\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}$
Hence n = 10, which is even number.
So, $\Big(\frac{10}{2}+1\Big)\text{th}$ term i.e., 6th term is the middle term.
$\text{T}_{6}=\text{T}_{5+1}={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{10-5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$
$={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$
$=252$

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Question 111 Mark

Which term is independent of x, in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^{9}?$

Answer

$\Big(\text{x}+\frac{1}{\text{x}}\Big)^{9}$
Let (r + 1)th term be the independent of x in the given expression.
$\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}(\text{x})^{9-\text{r}}\Big(-\frac{1}{3\text{x}^{2}}\Big)^{\text{r}}$
$={^\text{9}}\text{C}_{\text{r}}\Big(-\frac{1}{3\text{}}\Big)^{\text{r}}(\text{x})^{9-\text{r}-2\text{r}}$
$={^\text{9}}\text{C}_{\text{r}}\Big(-\frac{1}{3\text{}}\Big)^{\text{r}}(\text{x})^{9-3\text{r}}$
This term is independent of x, if
9 - 3r = 0
⇒ r = 3
So, (3 + 1)th i.e., 4th term is independent of x.

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Question 121 Mark

Find the sum of the coefficients of two middle terms in the binomial expansion of $(1+\text{x})^{2\text{n}-1}.$

Answer

$(1+\text{x})^{2\text{n}-1}$
Here,$n$ is an odd number.
Therefore, the middle term are $\Big(\frac{2\text{n}-1+1}{2}\Big)^{\text{th}}$ and $\Big(\frac{2\text{n}-1+1}{2}+1\Big)^{\text{th}},$ $n^{th}$ and $(\text{n}+1)^{\text{th}}$ terms.
Now, we have
$\text{T}_{\text{n}}=\text{T}_{\text{n}-1+1}$
$={^\text{2n-1}}\text{C}_{\text{n}-1}(\text{x})^{\text{n}-1}$
And,
$\text{T}_{\text{n}}=\text{T}_{\text{n}+1}$
$={^\text{2n-1}}\text{C}_{\text{n}}(\text{x})^{\text{n}}$
$\therefore$ The coefficient of two middle terms are ${^\text{2n-1}}\text{C}_{\text{n}-1}$ and ${^\text{2n-1}}\text{C}_{\text{n}}.$
Now,
$={^\text{2n-1}}\text{C}_{\text{n}-1}+{^\text{2n-1}}\text{C}_{\text{n}}={^\text{2n}}\text{C}_{\text{n}}$
Hence, the sum of the coefficients of two middle terms in the binomial expansion of $(1+\text{x})^{2\text{n}-1}$ is ${^\text{2n}}\text{C}_{\text{n}}.$

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Question 131 Mark

If a and b denote the sum of the coefficients in the expansions of $(1-3\text{x}+10\text{x}^{2})^\text{n}$and $(1+\text{x}^{2})^{\text{n}}$ respectively, then write the relation between a and b.

Answer

$(1-3\text{x}+10\text{x}^{2})^\text{n}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(-3\text{x}+10\text{x}^{2})+....+{^\text{n}}\text{C}_{\text{n}}(-3\text{x}+10\text{x}^{2})^{\text{n}}$
Let x = 1 on both sides.
$(1-3+10)^{\text{n}}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}}(-3+10)+...+{^\text{n}}\text{C}_{\text{n}}(-3+10)^{\text{n}}$
$\therefore 8^{\text{n}}=\text{a}$
$(1+\text{x}^{2})^{\text{n}}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(\text{x}^{2})+.....{^\text{n}}\text{C}_{\text{n}}(\text{x}^{2})^{\text{n}}$
Let x = 1 on both sides.
$(1+1)^{\text{n}}$
$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(\text{x}^{2})+.....{^\text{n}}\text{C}_{\text{n}}(\text{x}^{2})^{\text{n}}$
$\therefore 2^{\text{n}}=\text{b}$
$\therefore \text{a}=\text{b}^{3}$

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Question 141 Mark

Write the coefficient of the middle term in the expansion of $(1+\text{x})^{2\text{n}}.$

Answer

$(1+\text{x})^{2\text{n}}$
Here k = 2n, which is number.
So, $\Big(\frac{2\text{n}}{2}+1\Big)\text{th}$ term i.e. (n + 1)th term is the middle term.
Coefficient of (r + 1)th term in the binomial expansion of $(1+\text{x})^{\text{k}}$ is ${^\text{k}}\text{C}_{\text{r}}.$
Coefficient of (n + 1)th term in the binomial expansion of $(1+\text{x})^{\text{2}\text{n}}$ is ${^\text{2n}}\text{C}_{\text{n}}.$

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Question 151 Mark

If $a$ and $b$ are the coefficients of $x^n$ in the expansion of $(1+\text{x})^{2\text{n}}$ and $(1+\text{x})^{2\text{n}-1}$ respectively, find $\frac{\text{a}}{\text{b}}.$

Answer

Coefficients of $x^n$ in the expansion of $(1+\text{x})^{2\text{n}}$ is ${^\text{2n}}\text{C}_{\text{n}}=\text{a}.$
Coefficients of $x^n$ in the expansion of $(1+\text{x})^{2\text{n}-1}$ is ${^\text{2n-1}}\text{C}_{\text{n}}=\text{b}.$
Now,
$\frac{\text{a}}{\text{b}}=\frac{{^\text{2n}}\text{C}_{\text{n}}}{{^\text{2n-1}}\text{C}_{\text{n}}}$
$=\frac{\frac{(2\text{n})!}{\text{n}!\text{n}!}}{\frac{(2\text{n}-1)!}{\text{n}!(\text{n}-1)!}}$
$=\frac{2\text{n}}{\text{n}}$
$=2$
Hence, $\frac{\text{a}}{\text{b}}=2.$

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Question 161 Mark

If a and b denote respectively the coefficients of $x^m$ and $x^n$ in the expansion of $(1+\text{x})^{\text{m}+\text{n}},$ then write the relation between a and b.

Answer

In the binomial expansion of $(1+\text{x})^{\text{m}+\text{n}}$ coefficients of $x^m$ and $x^n$ are equal.
$\therefore a = b$

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Question 171 Mark

Write the number of terms in the expansion of $\Big[(2\text{x}+\text{y}^{3})^{4}\Big]^{7}.$

Answer

In the binomial expansion of $(\text{a}+\text{b})^{\text{n}},$ total number of term will be (n + 1).
Now, $\Big[(2\text{x}+\text{y}^{3})^{4}\Big]^{7}=(2\text{x}+\text{y})^{28}$
Therefore, in the expansion of $\Big[(2\text{x}+\text{y}^{3})^{4}\Big]^{7},$ total number of term will be 28 + 1 = 29.

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Question 181 Mark

Write the middle term in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^{10}.$

Answer

$\Big(\text{x}+\frac{1}{\text{x}}\Big)^{10}$
Hence n = 10, which is even number.
So, $\Big(\frac{10}{2}+1\big)\text{th}$ term i.e., 6th term is the middle term.
$\text{T}_{6}=\text{T}_{5+1}={^\text{10}}\text{C}_{\text{5}}(\text{x})^{10-5}\Big(\frac{1}{5}\Big)^{5}$
$={^\text{10}}\text{C}_{\text{5}}$

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