MATHS M.C.Q (1 Marks)

1. $\pm\ 3$

1. $\frac{10!}{(5!)^2}$

Solution:

Given: (1 + x)10

The greatest coefficient will always occur in the middle term.

Hence, the total number of terms in an expansion is 11. (i.e. 10 + 1 = 11)

Therefore, middle term $ =\Big[\big(\frac{10}{2}\big)+1\Big]=5+1=6\text{th }\text{term}.$

So, T6 = ¹⁰C₅ × x⁵

Therefore, the coefficient of the greatest term = ¹⁰C₅ $=\frac{10!}{(5!)^2}.$

4. 4AB

Solution:

If A and B denote respectively the sums of odd terms and even terms in the expansion $(\text{x}+\text{a})^{\text{n}}$

Then, $(\text{x}+\text{a})^{\text{n}}=\text{A}+\text{B}\ ...(\text{i})$

$(\text{x}-\text{a})^{\text{n}}=\text{A}-\text{B}\ ...(\text{ii})$

Squaring and subtraction equation (ii) from(i) we get,

$ (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=(\text{A}+\text{B})^{2}-(\text{A}-\text{B})^{2}$

$\Rightarrow (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=4\text{AB}$

3. 10

Solution:

$\text{T}_{5}=\text{T}_{4+1}$

$={^\text{n}}\text{C}_{\text{4}}\big(\text{a}^{\frac{2}{3}}\big)^{\text{n-4}}(\text{a}^{-1})^{4}$

$={^\text{n}}\text{C}_{\text{4}}\ \text{a}^{\big(\frac{2\text{n}-8}{3}-4\big)}$

For this term to be independent of a, we must have

$\frac{2\text{n}-8}{3}-4=0$

$\Rightarrow 2\text{n}-20=0$

$\Rightarrow \text{n}=10$

2. 301

Solution:

$\Big(27^{\frac{1}{6}}+32^{\frac{1}{10}}\text{x}\Big)^{600}$

$=\Big(3^{\frac{1}{2}}+2^{\frac{1}{2}}\Big)^{600}$

Total number of integral terms will be

$=\frac{600}{\text{L}.\text{C}.\text{M}(2,2)}+1$

$=\frac{600}{2}+1$

$=301$

2. 252

Solution:

Hence, n, i.e., 10, is an even number.

$\therefore$ Middle term $=\Big(\frac{10}{2}+1\Big)^{\text{th}}$ term = 6^th term

Thus, we have

$\text{T}_{6}=\text{T}_{5+1}$

$={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{10-5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$

$=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2}\times\frac{2^{5}}{3^{5}}\times\frac{3^{5}}{2^{5}}$

$=252$

3. r = 12

Solution:

Here,

$\text{T}^{\text{r}}={^\text{15}}\text{C}_{\text{r}-1}(\text{x}^{4})^{15-\text{r}+1}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}-1}$

$=(-1)^{\text{r}}\times{^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{64-4\text{r}-3\text{r}+3}$

For this term to contain x^-17, we must have

$67-7\text{r}=-17$

$\Rightarrow \text{r}=12 $

4. 540

Solution:

$\text{r}=\frac{6\times\frac{5}{2}-3}{\frac{5}{2}+\frac{3}{2}}=\frac{15-3}{4}=3$

$\therefore$ Coefficient of x³ is ⁶C₃​3³

^{$=\frac{6\times5\times4}{3\times2\times1}\cdot27$}

= 5 × 4 × 27 = 540

3. 9

Solution:

Coefficient of (r + 1)^th term = Coefficient of (n + 3)^th

Then, we have

${^\text{20}}\text{C}_{\text{n}}={^\text{20}}\text{C}_{\text{n}+2}$

$\Rightarrow 2\text{n}+2=20$

$\Rightarrow \text{n}=9$

4. 2^n-1

Solution:

(1 + x)ⁿ = C₀ ​+ C_1​x + C₂​x² + C₃​x³ + ... + C_n​xⁿ

Putting x = 1 and x = 1 and subtracting, we get.

2ⁿ = 2(C₁ ​+ C_3​ + C₅ ​+ ...)

$\therefore$ C_1​ + C_{3 ​}+ C₅ ​+ ... = 2^n-1

Or the sum of the coefficients of the odd power of x is 2^n-1.

1. 5

Solution:

Coefficients of $(2\text{r}+3)^{\text{th}}$ and $(\text{r}-1)^{\text{th}}$ terms in the given expansion are ${^\text{15}}\text{C}_{\text{2r}+2}$ and ${^\text{15}}\text{C}_{\text{2r}-2}.$

Then, we have

${^\text{15}}\text{C}_{\text{2r}+2}={^\text{15}}\text{C}_{\text{r}-2}$

$\Rightarrow 2\text{r}+2=\text{r}-2$ or $2\text{r}+2+\text{r}-2=15$

$\Rightarrow \text{r}=-4$ or $\text{r}=5$

Neglecting the negative value, We have

$\text{r}=5$

3. 10c³

^Solution:

 Given:$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^5$

$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}=\ ^{5}\text{C}_{\text{r}}\times(\text{y}^2)^{\text{r}}\times\big(\frac{\text{c}}{\text{y}}\big)^{5-\text{r}}$

$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}={5}\text{C}{\text{r}}\times\text{y}^{2{\text{r}}}\times\big(\frac{\text{c}^{5-\text{r}}}{\text{y}^{5-\text{r}}}\big)$

On solving this, we get r = 3.

Hence, the coefficient of y = ⁵C_{3 ×} c³ = 10c³.

1. 3

Solution:

The coefficient of x in the given expansion where x occurs at the (r + 1)^th term.

We have,

${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$

$={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$

For it to contain x, we must have

$10-3\text{r}=1$

$\Rightarrow \text{r}=3$

Coefficient of x in the given expansion,

$={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}$

Now, we have

$10\lambda^{3}=270$

$\Rightarrow \lambda^{3}=27$

$\Rightarrow \lambda=3$

2. 924

Solution:

(x + y)ⁿ, Sum of coefficient = 4096

When x = y = 1, if n = 12

⇒ (1 + 1)¹² = 2¹² = 4096

⇒ Hence, greatest coefficient

${^{\text{n}}}\text{C}_{\frac{\text{n}}{2}}={^{12}}\text{C}_{6}=\frac{12!}{6!6!}=924$

Hence, this is the answer.

1. 649

Solution:

Total number of integral terms are

$\frac{5832}{\text{L}.\text{C}.\text{M}(3,9)}+1$

$=\frac{5832}{9}+1$

$=648+1$

$=649$

1. $\frac{5}{4}$

1. 35

Solution:

Substituting x = 1, we get the sum of the coefficients as

(2)ⁿ = 128

$\therefore$ n = 7

Hence writing the general term, we get

$\text{T}_{\text{r}+1}={^{7}}\text{C}_{\text{r}}\text{x}\frac{63-11\text{r}}{6}$

Hence for the coefficient of x⁵

63 - 11r = 6(5)

63 - 11r = 30

33 - 11r = 0

$\therefore$ r = 3

Hence coefficient is ⁷C₃​ = 35.

1. 210c⁴

Solution:

$\Big(\text{y}+\frac{\text{c}}{\text{y}^{2}}\Big)^{10}$

$\text{T}_{\text{r}+1}={^{10}}\text{C}_{\text{r}}\text{y}^{10-3\text{r}}\text{c}^{\text{r}}$

Hence for y^-2

10 - 3r = -2

12 = 3r

r = 4

Coefficient will be

¹⁰C_4​c⁴

= 210c⁴

4. $\frac{9}{7}$

Solution:

Coefficients of x² = Coefficients of x³

${^\text{9}}\text{C}_{\text{2}}\times3^{9-2}\text{a}^{2}={^\text{9}}\text{C}_{\text{3}}\times3^{9-3}\ \text{a}^{3}$

$\Rightarrow \text{a}=\frac{{^\text{9}}\text{C}_{\text{2}}}{{^\text{9}}\text{C}_{\text{3}}}\times3$

$=\frac{9!\times3!\times6!\times3}{2!\times7!\times9!}$

$=\frac{9}{7}$

1. 924a⁶b⁶

Solution:

The middle term will be the 7th term.

Hence ^{$\text{T}_{6+1}=^{12}\text{C}_6\big(\frac{\text{a}}{\text{x}}\big)^6(\text{bx})^6=924\text{a}^6\text{b}^6$}

3. 50

Solution:

For the above question $\text{T}_{\text{r}+1}={^{55}}\text{C}_{\text{r2}}11-\frac{\text{r}}{5}3\frac{\text{r}}{10}$

Hence we will have rational terms at r = 0, 10, 20, 30, 40, 50 respectively.

Hence there will be 6 rational terms.

The total number of terms will be

55 + 1

= 56 terms.

Hence the number of irrational terms will be

56 - 6

= 50 terms.

2. 97

Solution

$\text{T}_{\text{r}+1}​=^{100}\text{C}_\text{r}​5^{\frac{(\text{r}-100)}{6}}2^{\frac{\text{r}}{8}}$

Hence we get rational terms when

r = 8k where k is an integer and $\frac{8\text{k}-100}{6}$ is an integer

r = 16, 40, 64, 88

Hence we get in total 4 rational terms.

However, total number of terms will be 101

Hence total number of irrational terms is 101 - 4

= 97 terms.

3. 9

Solution:

rth term in the given expansion is ${^\text{20}}\text{C}_{\text{r}-1}\Big(2\text{x}^{2}\Big)^{12-\text{r}+1}\Big(\frac{-1}{\text{x}}\Big)^{\text{r}-1}$

$=(-1)^{\text{r}-1}\ {^\text{20}}\text{C}_{\text{r}-1}\ 2^{13-\text{r}}\ \text{x}^{26-2\text{r}-\text{r}+1}$

For this term to be independent of x, we must have,

$27-3\text{r}=0$

$\Rightarrow \text{r}=9$

Hence, the term in the expansion is independent.

2. 32

Solution:

$\text{T}_\text{r}=^{124}\text{C}_{\text{r}-1}(\sqrt3)^{125-\text{r}}(\sqrt[4]{5})^{\text{r}-1}$

When both the terms are rational, T_r​ will be rational.

Hence, $\frac{125-\text{r}}{2}$ and $\frac{\text{r}-1}{4}$​ both must be integers.

Therefore, r must be of the form 4k + 1, where k is an integer. 

There are 125 terms in the expansion.

Hence, k can assume values from 0 to 31.

Hence, there are 32 values of k and 32 rational terms in the expansion.

2. 7, 14

Solution:

Coefficients of the 5^th, 6^th and 7^th terms in the given expansion are ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}.$

These coefficients are in AP.

Thus, we have

$2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$

On dividing both sides by ${^\text{n}}\text{C}_{\text{5}},$ we get

$2=\frac{{^\text{n}}\text{C}_{\text{4}}}{{^\text{n}}\text{C}_{\text{5}}}+\frac{{^\text{n}}\text{C}_{\text{6}}}{{^\text{n}}\text{C}_{\text{5}}}$

$\Rightarrow 2=\frac{5}{\text{n}-4}+\frac{\text{n}-5}{6}$

$\Rightarrow 12\text{n}-48=30+\text{n}^{2}-4\text{n}-5\text{n}+20$

$\Rightarrow \text{n}^{2}-21\text{n}+98=0$

$\Rightarrow (\text{n}-14)(\text{n}-7)=0$

$\Rightarrow \text{n}=7,14$

3. 41

Solution:

Total number of terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is 45 + 1 = 46

The general term in the expansion is $\text{T}{\text{r}+1​}={^{45}}\text{C}_{\text{r}}\times4\frac{45-\text{r}}{5}\times7^{\frac{\text{r}}{10}}\text{T}_{\text{r}+1}$ is rational if r = 0, 10, 20, 30, 40

$\therefore$ Number of rational terms = 5

$\therefore$ Number of irrational terms = 46 - 5 = 41