MATHS M.C.Q (1 Marks)

4. $\text{None of these}$

2. The number of irrational terms = 19

Solution:

$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}4\frac{20-\text{r}}{3}6\frac{-\text{r}}{4}$

$={^\text{n}}\text{C}_{\text{r}}2\frac{40-2\text{r}}{3}-\frac{\text{r}}{4}3-\frac{\text{r}}{4}$

$={^\text{n}}\text{C}_{\text{r}}2\frac{100-11\text{r}}{12}3-\frac{\text{r}}{4}$

There are total of 21 terms.

Hence, we get rational terms for r = 20, 8

Hence there are in total 21 - 2 = 19 irrational terms.

The middle term is at r = 10 which is irrational.

1. 6th term

Solution:

The greatest term in the expansion of (x + y)n is the kth term. Where $\text{k}=\frac{[(\text{n}+1)\text{y}]}{[\text{x}+\text{y}] }...(1)$

On comparing the given expression with the general form, x = 3, y = 2x, n = 50

Now, substitute the values in the given expression, we get

Hence, k^th term $= \frac{\big[(50+1)(2\text{x})\big]}{[3+2\text{x}]}$

When $\text{x}=\frac{1}{5}$

K^th term $ =\frac{\Big[(51)\big(2\big(\frac{1}{5}\big)\big)\Big]}{\Big[3+2\big(\frac{1}{5}\big)\Big]}=6$

Hence, the 6th term is the largest term in the expansion of (3 + 2x)50, when $\text{x}=\frac{1}{5}.$

3. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$

Solution:

Given expression is $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$

Since, n = 10 (even), so there is only one middle term which is, 6^th term.

$\therefore\text{T}_6=\text{T}_{5+1}=\ ^{10}\text{C}_5\Big(\frac{1}{\text{x}}\Big)^{10-5}(\text{x}\sin\text{x})^5$

$\Rightarrow\frac{63}{8}=\ ^{10}\text{C}_5\sin^5\text{x}$ (given)

$\Rightarrow\frac{63}{8 }=252\times\sin^5\text{x}\Rightarrow\sin^5\text{x}=\frac{1}{32}$ $\Rightarrow\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$

$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$

1. $\frac{595}{48}$

Solution

Total number of terms are 8.

So, middle term will be the 4^th and 5^th term.

$\therefore\text{t}_{3+1}=^7\text{C}_3(-1)^{3}(3\text{x})^{7-3}\big(\frac{\text{x}^3}{6}\big)^3=-\frac{105\text{x}^{13}}{8}$

$\therefore\text{t}_{4+1}=^7\text{C}_4(-1)^{4}(3\text{x})^{7-4}\big(\frac{\text{x}^3}{6}\big)^4=\frac{35\text{x}^{15}}{48}$

So, $-\frac{105}{8}+\frac{35}{48}=-\frac{595}{48}$

4. All option are right

Solution:

We know a prime x > 6 can be always written as 6k + 1 or 6k - 1, when k is an integer.

Option A ⟶ 889 = 6 × 148 + 1

Option B ⟶ 997 = 6 × 166 + 1

Option C ⟶ 899 = 6 × 133 + 1

Option D ⟶ 1147 = 6 × 191 + 1

$\therefore$ All the options are correct.

3. 5^th and 6^th.

Solution:

Let the two successive terms in the expansion of (1 + x)²⁴ be (r + 1)(r + 2)^th terms.

Now, $\text{T}_{\text{r}+1}=\ ^{24}\text{C}_\text{r}\text{x}^\text{r}\ \text{and}\ \text{T}_{\text{r}+2}=\ ^{24}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$

Given that, $\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}$

$\Rightarrow\frac{\frac{(24)!}{\text{r}!(24-\text{r})!}}{\frac{(24)!}{(\text{r}+1)!(24-\text{r}-1)!}}=\frac{1}{4}$ $\Rightarrow\frac{(\text{r}+1)\text{r}!(23-\text{r})!}{\text{r}!(24-\text{r})(23-\text{r})!}=\frac{1}{4}\Rightarrow\frac{\text{r}+1}{24-\text{r}}=\frac{1}{4}$

$\Rightarrow4\text{r}+4=24-\text{r}\Rightarrow\text{r}=4$

$\therefore\text{T}_{4+1}=\text{T}_5\ \text{and}\ \text{T}_{4+2}=\text{T}_6$

Hence. 5^th and 6^th terms.

2. $2.$

Solution:

Given expression is (1 + x)²ⁿ

$\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$

$\therefore$ Coefficient of xn = ²ⁿC_n = A (Given)

In the expression of (1 + x)^{2n - 1}

$\text{T}_{\text{r}+1}=2^{\text{n}-1}\text{C}_\text{x}=\text{B}$ Given

So, $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{2}{1}$

Hence, the correct option is (b).

4. 2 : 1.

Solution:

General Term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$

In the expansion of (1 + x)²ⁿ, we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$

To get the coefficient of xⁿ, put r = n

$\therefore$ Coefficient of $\text{x}^\text{n}=\ ^{\text{2n}}\text{C}_\text{n}$

In the expansion of $(1+\text{x})^{2\text{n}-1},$ we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}-1}\text{C}_\text{r}\text{x}^\text{r}$

$\therefore$ Coefficient of $\text{x}^\text{n}\ \ \text{is}=\ ^{\text{2n}-1}\text{C}_{\text{n}-1}$

The required ratio is $\frac{^{\text{2n}}\text{C}_{\text{n}-1}}{^{\text{2n}-1}\text{C}_{\text{n}-1}}$

$=\frac{\frac{2\text{n}!}{\text{n}!(\text{n}!)}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(2\text{n}-1-\text{n}+1)!}}=\frac{\frac{2\text{n}!}{\text{n}!.\text{n}!}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(\text{n}!)}}$

$=\frac{2\text{n}!}{\text{n}!\text{n}!}\times\frac{(\text{n}-\text{n})!\cdot\text{n}!}{(2\text{n}-1)!}=\frac{2\text{n}(2\text{n}-1)!}{\text{n}!\text{n}(\text{n}-1)!}\times\frac{(\text{n}-1)!\cdot\text{n}!}{(2\text{n}-1)!}$

$=\frac{2}{1}=2:1$

Hence, the correct option is (d).

2. 7.

Solution:

$(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$

So, coefficients of 2^nd, 3^rd and 4th terms are ⁿC₁, ⁿC₂ and ⁿC₃, respectively.

Given that, ⁿC₁, ⁿC₂ and ⁿC₃, are in A.P.

$\therefore2\ ^\text{n}\text{C}_2=\ ^\text{n}\text{C}_1+\ ^\text{n}\text{C}_3$

$\Rightarrow2\Big[\frac{\text{n}!}{(\text{n}-2)!2!}\Big]=\text{n}+\frac{\text{n}!}{3!(\text{n}-3)!}$

$\Rightarrow2\Big[\frac{\text{n}(\text{n}-1)}{2!}\Big]=\text{n}+\frac{\text{n}(\text{n}-1)(\text{n}-2)}{3!}\Rightarrow(\text{n}-1)=1+\frac{(\text{n}-1)(\text{n}-2)}{6}$

$\Rightarrow6\text{n}-6=6+\text{n}^2-3\text{n}+2\Rightarrow\text{n}^2-9\text{n}+14=0$ $\Rightarrow(\text{n}-7)(\text{n}-2)=0$

$\therefore\text{n}=2\ \text{or}\ \text{n}=7$

Since n = 2 is not possible, so n = 7.

1. n = 2r.

Solution:

The given expression is $(1 + \text{x})^{2\text{n}}$

$\therefore\text{T}_{3\text{r}}=\text{T}_{(3\text{r}-1)+1}=\ ^{2\text{n}}\text{C}_{3\text{r}-1}\ \text{x}^{3\text{r}-1}$

and $\text{T}_{\text{r}+2}=\text{T}_{(\text{r}+1)+1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$

Given, $^{2\text{n}}\text{C}_{3\text{r}-1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}$

$\Rightarrow3\text{r}-1+\text{r}+1=2\text{n}\ \ [\because\ ^\text{n}\text{C}_\text{x}=\ ^\text{n}\text{C}_\text{y}\Rightarrow\text{x}+\text{y}=\text{n}]$

$\therefore\text{n}=2\text{r}$

1. $2$

Solution:

$\sum\frac{1}{3^{\text{k}}}{^\text{k}}\text{C}_{\text{r}}$

$=\frac{1}{3^{\text{k}}}\sum{^\text{k}}\text{C}_{\text{r}}$

$=\frac{2^{\text{k}}}{3^{\text{k}}}$

This is a G.P

Therefore, the sum of the series will be

$\text{S}=\frac{\frac{2}{3}}{1-\frac{20}{3}}=2$

2. $220\text{x}^{\frac{3}{2}}$

Solution:

Expansion is $\Big(\sqrt{\text{x}}+\frac{1}{\text{x}}\Big)^{12}$

$\text{T}_{\text{r}+1}=12_{\text{C}\text{r}}\big(\frac{1}{\text{x}}\big)^{\text{r}}\cdot(\sqrt{\text{x}})^{12-\text{r}}=12_{\text{C}\text{r}}\cdot\text{x}^{6-1.5\text{r}}$

4^th term is $\text{T}_4=12_{\text{C}3}\cdot\text{x}^{6-1.5\times3}=220.\text{x}^{\frac{3}{2}}$

2. 51

Solution:

In the above binomial expansion, the terms at the even places will get eliminated, and we would be left with twice the sum of the terms at odd places.

Hence there will be

$\frac{\text{n}}{2}+1$

$=\frac{100}{2}+1$

$=51\text{terms}$

4. 80

Solution:

General term of (1 - 2x)⁵ is given by

T_r+1 ​= ⁵C_r​ (-2x)^r

= ⁵C_r ​(-2)x^r

For coefficient of x⁴, power of x = 4

$\therefore$ r = 4

$\therefore$ Coefficient pf x⁴ = ⁵C₄​ (-2)⁴

= 5 × 16 = 80

3. 41

Solution:

The general term T_r+1 in the given expansion is given by ${^\text{45}}\text{C}_{\text{r}}\Big(4^{\frac{1}{5}}\Big)^{45-\text{r}}\Big(7^{\frac{1}{10}}\Big)^{\text{r}}$

For T_r+1 to be an integer, we must have $\frac{\text{r}}{5}$ and $\frac{\text{r}}{10}$ as integers i.e. $0\leq\text{r}\leq45$

$\therefore \text{r}=0,10,20,30,40$

Hence, there are 5 rational and 41, i.e. 46 - 5, irrational terms.

4. 101

Solution:

The general term T_r+1 in the given expansion is given by ${^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{600-\text{r}}\Big(35^{\frac{1}{2}}\text{x}\Big)^{\text{r}}$

$={^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{200-\frac{\text{r}}{3}}\times\Big(35^{\frac{\text{r}}{2}}\text{x}\Big)^{\text{r}}$

Now, T_r+1 is an integer if $\frac{\text{r}}{2}$ and $\frac{\text{r}}{2}$ are integers for all $0\leq\text{r}\leq600$

Thus, we have

r = 0, 6, 12, ....600

Since, It is an A.P

So, $600 = 0 + (\text{n} - 1)6$

$\Rightarrow \text{n}=101$

Hence, there are 101 terms with integral coefficients.

3. (m + 1)2ⁿ - 1

Solution:

No. of elements in b.c.d... = n

Choose a^k, where $\text{k}\in0,$ 0, 1, 2, .. m at a time = (m + 1) for every k, no. possible factors from n elements 

 = ⁿC_0​ + ⁿC_1​ + ⁿC₂​ +..... + ⁿC_n​ 

= 2ⁿ

Total factors = No. of possible powers of a × every k, no. possible factors from n elements

= (m + 1)2ⁿ

If factor 1 is excluded then

(m + 1)2n - 1

3. 2¹⁶ - 17

Solution:

Consider given the binomial expression,

$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}={^{16}}\text{C}_{2}+{^{16}}\text{C}_{3}+{^{16}}\text{C}_{3}+\ .....\ {^{16}}\text{C}_{16}$

$=2^{16}-17$

Hence, this is the answer.

2. ⁿC_r​3^r

Solution:

$(1-\text{x})^{\text{n}}(1+\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}}(1-\text{x})^{\text{n}-\text{r}}$

$\Rightarrow(1-\text{x}+2\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$

$\Rightarrow\sum_{\text{r}=0}^{\text{n}}{^\text{n}}\text{C}_{\text{r}}(1-\text{x})^{\text{n}-\text{r}}(2\text{x})^{\text{r}}$

$=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$

Comparing general term, we get $\text{a}^{\text{r}}={^\text{n}}\text{C}_{\text{r}}2^{\text{r}}$

1. 94

Solution:

As n = 100 hence there are 101 terms

The genral term for expansion is given as $\text{T}_{\text{r}+1}={^{100}}\text{C}_{\text{r}}\cdot3\frac{100-\text{r}}{5}\cdot7\frac{\text{r}}{3}$

So, we need to find such values of r for which $\frac{100-\text{r}}{5}$ and $\frac{\text{r}}{3}$ is a natural number.

For such value we find value of r i.e r = 0, 15, 30, 45, 60, 75, 90

hence there are 7 values for which it is rational.

So there are 101 - 7 = 94 irrational terms.

1. 1120

Solution:

Suppose (r + 1)^th term in the given expansion is independent of x.

Then, we have

$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}(2\text{x})^{\text{n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$

$={^\text{n}}\text{C}_{\text{r}}(2)^{\text{n}-\text{r}}\text{x}^{\text{n}-2\text{r}}$

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

$\text{n}-2\text{r}=0$

$\Rightarrow \text{r}=\frac{\text{n}}{2}$

$\therefore$ Required term $={^\text{n}}\text{C}_{\frac{\text{n}}{2}}\ 2^{\text{n}-\frac{\text{n}}{2}}=\frac{\text{n!}}{\big[(\frac{\text{n}}{2})\big]}\ 2^{\frac{\text{n}}{2}}$

We know,

Sum of the given expansion = 256

Thus, we have

$2^{\text{n}}.1^{\text{n}}=256$

$\Rightarrow \text{n}=8$

$\therefore$ Required term $=\frac{8!}{(4)!(4)!}2^{4}=1120$

1. A² - B²

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,

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

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

Multplying both the equations we get,

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

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

3. ${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$

Solution:

we have,

$(1+\text{x})^{21}+(1+\text{x})^{22}+...+(1+\text{x})^{30}$

$=(1+\text{x})^{21}\Big[\frac{(1+\text{x})^{10}-1}{(1+\text{x})+1}\Big]$

$=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$

Coefficient of x⁵ in the given expansion = Coefficient of x⁵ in $=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$

= Coefficient of x⁶ in $\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$

$={^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$

4. -330m⁷

Solution:

Let x^-3 occur at (r + 1)^th term in the given expansion.

Then, we have

$\text{T}_{r+1}={^\text{11}}\text{C}_{\text{r}}\ \text{x}^{11-\text{r}}\ \Big(\frac{-\text{m}}{\text{x}}\Big)^{\text{r}}$

$=(-1)^{\text{r}}\times{^\text{11}}\text{C}_{\text{r}}\ \text{m}^{\text{r}}\ \text{x}^{11-\text{r}-\text{r}}$

For this term to contain x^-3, we must have

$=11-2\text{r}=-3$

$\Rightarrow \text{r}=7$

Required coefficient $=(-1)^{7}\ {^\text{11}}\text{C}_{\text{7}}\ \text{m}^{7}$

$=-\frac{11\times10\times9\times8}{4\times3\times2}\ \text{m}^{7}$

$=-330\text{m}^{7}$

3. 5

Solution:

In the expansion of $\big(1+3\sqrt{2​}\text{x}\big)^{9}+\big(1-3\sqrt{2}​\text{x}\big)^{9}$ 2nd, 4th, 6th, 8th and 10th terms get cancelled.

$\therefore$ Number of non - zero terms in $2\Big[{^9}\text{C}_{0}+{^9}\text{C}_{2}(3\sqrt{2}\text{x})^{2}+\ ...\ +{^9}\text{C}_{8}(3\sqrt{2}\text{x})^{8}\Big]$ is 5.

4. $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$

Solution:

In general, if “n” is the even in the expansion of (a + b)ⁿ, then the number of terms will be odd. (i.e) n + 1.

Hence, the middle term of the expansion (a + b)ⁿ is $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}.$

1. 120

Solution:

(1 + x)¹⁰ = ¹⁰C₀​ +¹⁰C₁​x + ¹⁰C₂​x² + ........ + ¹⁰C₇​x⁷ + ¹⁰C₈​x⁸ + ¹⁰C₉​x⁹ + ¹⁰C₁₀​x¹⁰

So here, first term is ¹⁰C₀​ then 8^th term will be ¹⁰C₇​x⁷.

⇒  Coefficient of the 8^th term =¹⁰C₇​

$=\frac{10!}{7!3!}$

$=\frac{10\times9\times8\times7!}{7!\times3\times2\times1}$

$=120$

4. 51

Solution:

The number of rational terms will be

$1+\frac{300}{\text{L}.\text{C}.\text{M}(3,2)}$

$=1+\frac{300}{6}$

$=1+50=51$rational terms.

4. 3

Solution:

$(1+0.0001)^{10000}$

$=\big(1+\frac{1}{10000}\big)^{10000}$

$=\big(1+\frac{1}{\text{n}}\big)^{\text{n}}=\text{n}\text{c}_{0}(1)^{\text{n}}+\text{n}\text{c}_{1}(1)^{\text{n}-1}\cdot\frac{1}{\text{n}}+\text{n}\text{c}_{2}(1)^{\text{n}-2}\frac{1}{\text{n}^{2}}+\ ....$

$=1+\text{n}\cdot\frac{1}{\text{n}}+\frac{\text{n}(\text{n}-1)}{2}\cdot\frac{1}{\text{n}^2}+\ ....$

$=2+\frac{\text{n}(\text{n}-1)}{2\text{n}^{2}}+\ ....$ > 2 Integer just greater than 2 is 3.

2. 11!10!

Solution:

In pascals triangle middle terms has the highest value.

Therefore consider ²¹C_n​

_{$=\frac{21!}{(21-\text{n})!\text{n}!}$}

For (21 - n)!n!

to be least ²¹C_n​ has to be maximum.

Therefore, since 21 is odd we have two middle terms T₁₁​ and T₁₂​.

Hence for n = 11,

(21 - n)!n! = 11!(10)!

which is less than 12!(9!) for n = 12.

3. 5.013

Solution:

$ (126)^{\frac{1}{3}}$ can also be written as the cube root of 126.

Hence, $ (126)^{\frac{1}{3}}$is approximately equal to 5.013.

1. $\frac{405}{256}$

Solution:

Suppose x⁴ occurs at the (r + 1)^th term in the given expansion.

Then, we have

$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\Big(\frac{\text{x}}{2}\Big)^{10-\text{r}}\Big(\frac{-3}{2\text{x}^{2}}\Big)$

$=(-1)^{\text{r}}\ {^\text{10}}\text{C}_{\text{r}}\frac{3^{\text{r}}}{2^{10-\text{r}}}\text{x}^{10-\text{r}-2\text{r}}$

For this term to contain x⁴, we must have

$10-3\text{r}=4$

$\Rightarrow \text{r}=2$

$\therefore$ Required coefficient $={^\text{10}}\text{C}_{\text{2}}\frac{3^{2}}{2^{8}}=\frac{10\times9\times9}{2\times2^{8}}=\frac{405}{256}$