NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem

NCERT Exemplar Class 11 Maths Solutions Chapter 8: Excercise-1.3

Question:1

Find the term independent of x, where x≠0, in the expansion of

Answer:

............... (Given)
……. (from standard formula of Tr+1)
…… (i)
Now, for x,
30 – 3r = 0
→ r = 10
Substituting the value of r in eq (i),

Question:2

if the term free from x is the expansion of is 405, then find the value of k.

Answer:

................... (Given)
……. (from standard formula of Tr+1)

…….. (i)
Now, for x,
(10 – 5r)/2 = 0
→ r = 2
Substituting the value of r in eq (i),

= 405


Thus k² = 9
Thus, k = ±3

Question:3

Find the coefficient of x in the expansion of

Answer:

(1-3x+7x²) (1-x)¹⁶ ….. (given)
= (1-3x+7x²) (¹⁰⁶C₀ - ¹⁶C₁x¹ + ¹⁶C₂x² + …… + ¹⁶C₁₆x¹⁶)
= (1-3x +7x²) (1-16x+120x²+…..)
Thus, coefficient of x = -19

Question:4

Find the term independent of x in the expansion of

Answer:

Given: (3x – 2/x²)15
Thus,
Now, for independent term x,
15 – 3r = 0
Thus, r = 5
Thus, the term x is
T₅₊₁ = ¹⁵C₅3^15-5(-2)⁵
= -15 x 14 x 13 x 12 x 11 x 10!/5x4x3x2x1x10! .3¹⁰.2⁵
= -3003 x 3¹⁰x 2⁵

Question:5

Find the middle term (terms) in the expansion of
i) ii)

Answer:

(i) Given: (x/a – a/x)¹⁰
Where, n = 10(even).
The middle term = (10/2 + 1)th term, viz., the sixth term
Thus, T₆ = T₅₊₁ ¹⁰C₅ (x/a)^10-5 (-a/x)⁵
= - ¹⁰C₅ (x/a)⁵(a/x)⁵
= - 10 x 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 (x/a)⁵(x/a)^-5
= -252
(ii) Given: (3x – x³/6)⁹
Where, n = 9 (odd)
Now, here are two middle terms-
(9+1/2)^th ,viz., the tenth term
& (9+1/2 + 1)^th, viz., the sixth term
Thus, T₅ = T₄₊₁
= ⁹C₄ (3x)^9-4 (-x³/6)⁴
= 9x8x7x6x5!/4x3x2x1x5! . 3⁵x⁵x¹²6^-4
= 189/8 x¹⁷
Now,
T₆ = T₅₊₁
= ⁹C₅(3x)^9-5 (-x³/6)⁵
= - 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 . 3⁴.x⁴.x¹⁵.6^-4
= - 21/16 x¹⁹

Question:6

Find the coefficient of x¹⁵ in the expansion of (x – x²)¹⁰

Answer:

Given: (x – x²)¹⁰
Thus, T_r+1 = ¹⁰C_r x^10-r(-x²)^r
= (-1)^{r 10}C_rx^10-rx^2r
= (-1)^{r 10}C_rx^10+r
Now, for coefficient of x¹⁵
We have,
10 + r = 15
Thus, r = 5
Thus, T₅₊₁ = (-1)⁵.¹⁰C₅x¹⁵
Therefore,
Coefficient of x¹⁵ = -10x9x8x7x6x5!/5x4x3x2x1x5!
= -252

Question:7

Find the coefficient of x^1/17 in the expansion of (x⁴ – 1/x³)¹⁵

Answer:

Given: (x⁴ – 1/x³)¹⁵
Thus,
T_r+1 = ¹⁵C_r(x⁴)^15-r(-1/x³)^r
= ¹⁵C_r x^60-4r(-1)^r x^-3r
= ¹⁵C_r x^60-7r(-1)^r
Now, for coefficient x^-17,
60 – 7r = -17
7r = 77
r = 11
Thus, T₁₁₊₁ = ¹⁵C₁₁ x^60-77(-1)¹¹
Therefore,
Coefficient of x^-17 = -15 x 14 x 13 x 12 x 11!/11! x 4 x 3 x 2 x 1
= -1365

Question:8

Find the sixth term of the expansion (y^1/2 + x^1/3)ⁿ, if the binomial coefficient of the third term from the end is 45.
Answer:

Given: (y^1/2 + x^1/3)ⁿ
From the end, binomial coefficient of 3^rd term = 45
Thus, ⁿC_n-2 = 45 , ⁿC₂ = 45
i.e., n(n-1)(n-2)!/2!(n-2)! = 45
Now, n(n-1) = 90,
n² – n – 90 = 0,
(n – 10)(n + 9) = 0
n = 10 …. (since n is not equal to -9)
Now, the 6^th term,
=¹⁰C₅ (y^1/2)^10-5 (x^1/3)⁵
= 252y^5/2.x^5/3

Question:9

Find the value of r, if the coefficients of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal.

Answer:

Given: (1+x)¹⁸
Now, T_(2r+3)+1 is the (2r + 4)^th term
Thus, T_(2r+3)+1 = ¹⁸C_2r+3 (x)^2r+3
Now, T_(r-3)+1 is the (r-2)^th term
Thus, T_(r-3)+1 = ¹⁸C_r-3 x^r-3
Now, it is given that the terms are equal in expansion,
Thus,
¹⁸C_2r+3 = ¹⁸C_r-3
i.e., 2r + 3 + r – 3 = 18 …… (ⁿC_x = ⁿC_y gives x + y = n)
Thus, 3r = 18
Therefore, r = 6

Question:10

If the coefficient of second, third and fourth terms in the expansion of (1 + x)^2” are in A.P., then show that 2n² – 9n + 7 = 0.
Answer:

Given: (1+x)²ⁿ
Now, ²ⁿC₁, ²ⁿC₂, ²ⁿC₃ are the coefficients of the first second and the third terms respectively.
It is also given that the coefficients are in A.P.
Thus, 2. ²ⁿC₂ =²ⁿC₁ + ²ⁿC₃
Thus, 2[2n(2n-1)(2n-2)!/2x1x(2n-2)!] = 2n + 1n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!
N(2n-1) = n = n(2n-1)(n-1)/3
3(2n-2) = 3+ (2n² – 3n + 1)
6n – 3 = 2n² – 3n + 4
Thus, 2n² – 9n + 7 = 0

Question:11

Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)^11.

Answer:

Given: (1+x+x²+x³)¹¹
That is equal to [(1+x) + x²⁽1 + x)]¹¹
= [(1+x) (1+x²)]¹¹
= (1+x)¹¹.(1+x²)¹¹
= (¹¹C₀ + ¹¹C₁x + ¹¹C₂x² + ¹¹C₃x³ + ¹¹C₄x⁴ + ….) (_¹¹C₀ + ¹¹C₁x² + ¹¹C₂x4 + ….)
Thus, the coefficient of x⁴ =¹¹C₀ x ¹¹C4 + ¹¹C1 x ¹¹C2+¹¹C2x ¹¹C₀
= 330 + 605 + 55
= 990

Question:12

If p is a real number and the middle term in the expansion (p/2 + 2)⁸ is 1120, then find the value of p.

Answer:

Given: (p/2 + 2)⁸
Now, here index is n = 8,
Since there is only one middle term, viz., (8/2 + 1)th i.e., the fifth term
T₅ = T₄₊₁
= ⁸C₄(p/2)^8-4. 2⁴
Now,
1120 = ⁸C₄p⁴
1120 = 8x7x6x5x4!/4!x4x3x2x1 . p⁴
1120 = 7x2x5xp⁴
Thus, p⁴ = 1120/70
p⁴= 16
p² = 4
Thus, p = ±2

Question:13

Show that the middle term in the expansion ofis
Answer:

Given: (x – 1/x)²ⁿ
Now, here the index = 2n(even)
There’s only one middle term, i.e., (2n/2 + 1)^th term, viz., (n + 1)^th term
T_n+1 = ²ⁿC_n(x)^2n-n(-1/x)ⁿ
= ²ⁿC_n(-1)ⁿ
= (-1)ⁿ(2n!)/n!.n!
= (-1)ⁿ 1.2.3.4.5. …. (2n-1).(2n)/n!.n!
= (-1)ⁿ [1.3.5…. (2n-1)].2ⁿ[1.2.3…..n]/(1.2.3…n).n!
= (-2)ⁿ[1.3.5…(2n-1)].2ⁿ/n!

Question:14

Find n in the binomial if the ratio of 7th term from the beginning to the 7th term from the end is

Answer:

Given
Now, from the beginning,
Seventh term is
...................(i)
& from the end,
The seventh term is the same ,i.e.,
.............(ii)
Now, it is given that

Thus

Thus

Thus

Thus n=9

Question:15

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
(i) O²– E²= (x²– a²)n (ii) 4OE = (x + a)²ⁿ- (x - a)²ⁿ

Answer:

(i) (x+a)ⁿ= ⁿC₀xⁿ + ⁿC₁x^n-1 a¹ + ⁿC₂x^n-2 a² + ⁿC₃x^n-3 a³ + ….. + ⁿC_n aⁿ
Now, sum of odd terms,
O = ⁿC₀xⁿ+ ⁿC₂x^n-2 a² + …….
& sum of even terms,
E = ⁿC₁x^n-1 a¹ + ⁿC₃x^n-3 a³ + …..
Now, (x+a)ⁿ = O+E & (x-a)ⁿ = O – E
Thus, (O + E)(O – E) = (x + a)ⁿ(x – a)ⁿ
Thus, O² – E² = (x² – a²)ⁿ
(ii)Here,
4OE = (O + E)² – (O – E)²
= [(x+a)ⁿ]² – [(x-a)ⁿ]²
= (x+a)²ⁿ – (x-a)²ⁿ

Question:16

If xⁿ occurs expantion of then prove that its coefficient is

Answer:

Given: (x² + 1/x)²ⁿ
Thus, T_r+1 =²ⁿC_r(x²)^2n-r(1/x)^r
= ²ⁿC_rx^4n-3r
Now, if xn will occur in the expansion, then,
Let us consider 4n – 3r = p
Thus,
3r = 4n – p
r = 4n – p/3
Thus, coefficient of xp = ²ⁿC_r
= (2n)!/r!(2n-r)!
=

Question:17

Find the term independent of x in the expansion of(1+x+2x³)(3x²/2 – 1/3x)⁹

Answer:

Given: (1+x+2x³)( 3x²/2 – 1/3x)⁹
Now, let us consider (3x²/2 – 1/3x)⁹
T_r+1 = ⁹C_r (3/2 x²)^9-r (-1/3x)^r
= ⁹C_r (3/2)^9-r (-1/3)^r x^18-3r
Thus, in the expansion of (1+x+2x³)(3/2x² – 1/3x)⁹ the general term is:
⁹C_r (3/2)^9-r (-1/3)^r x^18-3r + ⁹C_r (3/2)^9-r (-1/3)^r x^19-3r + 2.⁹C_r (3/2)^9-r (-1/3)^r x^21-3r
Now, for the independent term x,
Let us put 18 – 3r = 0,
19 – 3r = 0
& 21 – 3r = 0,
We get,
R = 6 & r = 7
Now, since the second term is not independent of x, it will be:
⁹C₆ (3/2)^9-6 (-1/3)⁶ + 2.⁹C_r (3/2)^9-7 (-1/3)⁷
= 9x8x7x6!/6!x3x2 . 1/2³.3³ – 2. 9x8x7!/7!x2x1. 3²/2².1/3⁷
= 84/8.1/3³ – 36/4.2/3⁵
= 17/54

Question:18

The total number of terms in the expansion of (x + a)¹⁰⁰ + (x – a)¹⁰⁰ after simplification is
(A) 50 (B) 202 (C) 51 (D) none of these

Answer:

(c) 51.
Now, (x+a)¹⁰⁰ + (x-a)¹⁰⁰ = (¹⁰⁰C₀x¹⁰⁰ + ¹⁰⁰C₂x⁹⁹a + ¹⁰⁰C₂x⁹⁸a² + ….) + (¹⁰⁰C₀x¹⁰⁰ - ¹⁰⁰C₂x⁹⁹a + ¹⁰⁰C₂x⁹⁸a²)
= 2(¹⁰⁰C₀x¹⁰⁰ + ¹⁰⁰C₂x⁹⁸a² +…..+¹⁰⁰C₁₀₀a¹⁰⁰)
Therefore, there are 51 terms.

Question:19

Given the integers r > 1, n > 2, and coefficients of (3r)^th and (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(A) n = 2r (B) n = 3r (C) n = 2r + 1 (D) none of these

Answer:

(a) n =2r
Given: (1+x)²ⁿ
Thus, T_3r = T_(3r-1)+1
= ²ⁿC_3r-1x^3r-1
& T_r+2 = T_(r+1)+1
= ²ⁿC_r+1x^r+1
Now, ²ⁿC_3r-1x^3r-1 = ²ⁿC_r+1x^r+1 ……… (given)
Thus, 3r – 1 + r + 1 = 2n
Thus, n = 2r

Question:20

The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1: 4 are
(A) 3^rd and 4^th (B) 4^th and 5^th (C) 5^th and 6^th (D) 6^th and 7^th

Answer:

(c) 5^th & 6^th
Let us consider (r+1)^th& (r+2)^th as the 2 successive terms in the expansion of (1+x)²⁴
Now, T_r+1 = ²⁴C_rx^r& T_r+2 = ²⁴C_r+1x^r+1
Now, ²⁴C_r/²⁴C_r+1 = ¼ ……. (given)
Thus,

Thus, (r+1)r! (23 – r)! / r!(24 – r)(23 – r)! = ¼
r + 1/24 – r = ¼
4r + 4 = 24 – r
t = 4
Thus, T₄₊₁ = T₅
& T₄₊₂ = T₆
Thus, opt (c).

Question:21

The coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^{2n ~1} are in the ratio
(a) 1 : 2
(b) 1 : 3
(c) 3 : 1
(d) 2:1

Answer:

The answer is the option (d) 2:1
²ⁿC_n is the coefficient of xⁿ in the expansion of (1+x)²ⁿ
&^2n-1C_n is the coefficient of xⁿ in the expansion of (1+x)^2n-1

= (2n)!n!(n-1)!/(2n-1)!n!n!
= 2n(2n-1)!n!(n-1)!/n!n(n-1)!(2n-1)!
= 2/1
= 2:1

Question:22

If the coefficients of 2^nd, 3^rd and the 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then the value of n is
(a) 2
(b) 7
(c) 11
(d) 14

Answer:

The answer is the option (b) 7
(1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ⁿC₃x³ + ….. + ⁿC_nxⁿ
Now, ⁿC₁, ⁿC₂, ⁿC₃ are the coefficients of the second, third and fourth terms respectively.
2 ⁿC₂ =ⁿC₁ + ⁿC₃ …… since they are in A.P.
2[n!/(n-2)!2!] = n + n!/ 3!(n-3)!
2[n(n-1)/2!] = n + n(n-1)(n-2)/3!
(n-1) = 1 + (n-1)(n-2)/6
6n – 6 = 6 + n² – 3n + 2
n² – 9n + 14 = 0
(n-7)(n-2) = 0
Thus, n = 2 or n = 7
Now, n = 2 is not possible,
Thus, n = 7

Question:23

If A and B are coefficients of xⁿ in the expansions of (1 + x)²ⁿ and (1 + x)^2n–1 respectively, then A/B equals to
(a) 1
(b) 2
(c)
(d)

Answer:

The answer is the option (b) 2
²ⁿC_{n i}s the coefficient of xn in the expansion of (1+x)²ⁿ
Thus, A = ²ⁿC_n
&^2n-1C_n is the coefficient of xⁿ in the expansion of (1+x)^2n-1
Thus, B = ^2n-1C_n
Thus, A/B = ²ⁿC_n/^2n-1C_n
= 2/1
=2:1

Question:24

If the middle term of is equal to then the value of x is
(a) (b) (c) (d)

Answer:

The answer is the option (c) nπ + (-1)ⁿπ/6
Now, n = 10 (even)
Thus, there is only one middle term i.e., the 6^th term
Thus, T₆ = T₅₊₁
= ¹⁰C₅(1/x)^10-5 (x sin x)⁵
63/8 = ¹⁰C₅ sin⁵x …… (given)
63/8 = 252 x sin⁵x
Thus, sin⁵x = 1/32
Sin x = ½
Sin x = π/6
Thus, x = nπ + (-1)ⁿπ/6

Question:25

The largest coefficient in the expansion of (1 + x)³⁰ is _________________ .
Answer:

³⁰C_30/2 = ³⁰C₁₅ is the largest coefficient in the expansion of (1+x)³⁰

Question:27

In the expansion of (x² – 1/x²)¹⁶, the value of constant term is_________________.

Answer:

Here, T_r+1 = ¹⁶C_r(x²)^16-r(-1/x²)^r
= ¹⁶C_rx^32-4r(-1)^r
Now, for constant term,
32 – 4r = 0
Thus, r = 8
Thus, T₈₊₁ = ¹⁶C₈

Question:28

If the seventh terms from the beginning and the end in the expansion of are equal, then n equals _________________

Answer:

Given:
Now, referring to que 14.

Thus

Thus
6^(n-12)/3 = 6⁰
(n-12)/3 = 0
Thus, n = 12

Question:29

The coefficient of in the expansion of is ______________

Answer:

Given: (1/a – 2b/3)¹⁰
Thus, T_r+1 = ¹⁰C_r(1/a)^10-r (-2b/3)^r
Now, for the coefficient of a^-6b⁴,
10 – r = 6
Thus, r = 4
Therefore the coefficient of a^-6b⁴ = ¹⁰C₄(-2/3)⁴
= 10.9.8.7.6!/6!.4.3.2.1 . 2⁴/3⁴
= 1120/27

Question:30

Middle term in the expansion of (a³ + ba)²⁸is _________

Answer:

Given: (a³ + ba)²⁸
Here, n = 28 (even)
Thus, there is only one middle term (28/2 + 1)th or 15^th term.
Thus, the middle term, T₁₅ = T₁₄₊₁
= ²⁸C₁₄(a³)^28-14(ba)¹⁴
= ²⁸C₁₄a⁴²b¹⁴a¹⁴
= ²⁸C₁₄a⁵⁶b¹⁴

Question:31

The ratio of the coefficients of x^p and x^q in the expansion of (1 +x)^p+q is _______

Answer:

Given: (1+x)^p+q
Thus, coefficient of x^p = ^p+qC_p
& Coefficient of x^q = ^p+qC_q
Therefore,
^p+qC_p/ ^p+qC_q = ^p+qC_p /^p+qC_p
= 1:1

Question:32

The position of the term independent of x in the expansion of is _____________.

Answer:

Given:
Thus,
Now, for constant term,
10 – 5r = 0
Thus, r = 2
Therefore, the third term is independent of x.

Question:33

If 25¹⁵ is divided by 13, the reminder is _________ .

Answer:

Now, 25¹⁵ = (26 – 1)¹⁵
= ¹⁵C₀26¹⁵ – ¹⁵C₁26¹⁴ + …. + ¹⁵C₁₄26 – ¹⁵C₁₅
= (¹⁵C₀26¹⁵ – ¹⁵C₁26¹⁴ + …. + ¹⁵C₁₄26 – 13) + 12
Therefore, it is clear that, when we divide 25¹⁵ by 13, we get 12 as the remainder.

Question:34

The sum of the series is

Answer:

The given statement is False.
Now, from what’s given, we can say that,
= ²⁰C₀ + ²⁰C₁ + ²⁰C₂ + ²⁰C₃ + …… + ²⁰C₁₀
= ²⁰C₀ + ²⁰C₁ + ….. + ²⁰C₁₀ + ²⁰C₁₁ + …… + ²⁰C₂₀ – (²⁰C₁₁ + ….. + ²⁰C₂₀)
= 2²⁰ – (²⁰C₁₁ + …… + ²⁰C₂₀)

Question:35

The expression 7⁹ + 9⁷ is divisible by 64.

Answer:

The given statement is True.
Given: 7⁹ + 9⁷ is divisible by 64.
Now,
7⁹ + 9⁷ = (1+8)⁷ – (1-8)⁹
= [⁷C₀ + ⁷C₁.8 + ⁷C₂.(8)² + ⁷C₃.(8)³ + ….. + ⁷C₇.(8)⁷] – [⁹C₀ + ⁹C₁.8 + ⁹C₂.(8)² - ⁹C₃.(8)³ + ….. ⁹C₉.(8)⁰]
= (7x8 + 9x8) + (21 x 8² – 36 x 8²) + …..
= 128 + 64(21 – 36)
= 64[2 + (21 – 36) + …], viz. divisible by 64.

Question:36

The number of terms in the expansion of [(2x + y³)⁴]⁷ is 8.

Answer:

The given statement is False.
Given: [(2x + y³)⁴]⁷ = (2x + y³)²⁸
Thus, no. of terms = 28 + 1
= 29

Question:37

The sum of coefficients of the two middle terms in the expansion of (1+x)^2n-1 is equal to ^{2n – 1}C_n.

Answer:

The given statement is False.
Given: (1 + x)^2n-1
Now, no. of terms = 2n – 1 + 1
= 2n (even)
Thus, the middle term = 2n/2^th terms & (2n/2 + 1)^th terms
= n^th terms & (n+1)^th terms
Now, the coefficient of-
n^th term = ^{2n – 1}C_n-1
& (n + 1)^th term = ^2n-1C_n
Now, sum of the coefficients = ^{2n – 1}C_n-1 + ^2n-1C_n
= ^2n-1C_n-1 + ^2n-1C_n
= ²ⁿC_n

Question:38

The last two digits of the numbers 3⁴⁰⁰ are 01.

Answer:

The given statement is true.
Given: 3⁴⁰⁰ = (9)²⁰⁰
= (10 – 1)²⁰⁰
Thus, (10 – 1)²⁰⁰ = ²⁰⁰C₀(10)²⁰⁰ – ²⁰⁰C₁(10)¹⁹⁹ + ….. – ²⁰⁰C₁₉₉(10)¹ + ²⁰⁰C₂₀₀(1)²⁰⁰
= 10²⁰⁰ – 200 x 10¹⁹⁹ + …. – 10 x 200 + 1
Therefore, the last two digits will be 01.

Question:39

If the expansion of (x-1/x²)²ⁿ contains a term independent of x, then n is a multiple of 2.

Answer:

The given statement is False.
Given:
Now,
T_r+1 = ²ⁿC_r(x)^2n-r (-1/x²)^r
= ²ⁿC_r(x)^2n-r-2r(-1)^r
= ²ⁿC_r(x)^2n-3r (-1)^r(-1)^r
Now, for the independent term of x,
2n – 3r = 0
Thus, r = 2n/3, viz. not an integer
Thus, this expression is not possible to be true.

Question:40

Number of terms in the expansion of (a + b)ⁿ where n ∈ N is one less than the power n.

Answer:

The given statement is False.
In the given expression (a+b)ⁿ, the no. of terms is just 1 more than n, i.e., n + 1.