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## Sunday, 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 8 - Binomial Theorem

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 8 – Binomial Theorem

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#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 8 – Binomial Theorem

Question 1.
In the expansion of (a + b)n, if n is even then the middle term is
(a) (n/2 + 1)th term
(b) (n/2)th term
(c) nth term
(d) (n/2 – 1)th term

Answer: (a) (n/2 + 1)th term
Hint:
In the expansion of (a + b)n,
if n is even then the middle term is (n/2 + 1)th term

Question 2.
In the expansion of (a + b)n, if n is odd then the number of middle term is/are
(a) 0
(b) 1
(c) 2
(d) More than 2

Hint:
In the expansion of (a + b)n,
if n is odd then there are two middle terms which are
{(n + 1)/2}th term and {(n+1)/2 + 1}th term

Question 3.
The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
(a) 4815
(b) 4851
(c) 8451
(d) 8415

Hint:
Given, x + y + z = 100;
where x = 1, y = 1, z = 1
Let u = x – 1, v = y – 1, w = z – 1
where u = 0, v = 0, w = 0
Now, equation becomes
u + v + w = 97
So, the total number of solution = 97+3-1C3-1
= 99C2
= (99 з 98)/2
= 4851

Question 4.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)В
(c) 10!/(5! з 4!)В
(d) 10!/(5! з 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/2 = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)В

Question 5.
The fourth term in the expansion (x – 2y)12 is
(a) -1670 x9 з yГ
(b) -7160 x9 з yГ
(c) -1760 x9 з yГ
(d) -1607 x9 з yГ

Answer: (c) -1760 x9 з yГ
Hint:
4th term in (x – 2y)12 = T4
= T3+1
= 12C3 (x)12-3 з(-2y)Г
= 12C3 x9 з(-8yГ)
= {(12з11з10)/(3з2з1)} з x9 з(-8yГ)
= -(2з11з10з8) з x9 з yГ
= -1760 x9 з yГ

Question 6.
If n is a positive integer, then (v3+1)2n+1 + (v3-1)2n+1 is
(a) an even positive integer
(b) a rational number
(c) an odd positive integer
(d) an irrational number

Hint:
Since n is a positive integer, assume n = 1
(v3+1)Г + (v3-1)Г
= {3v3 + 1 + 3v3(v3 + 1)} + {3v3 – 1 – 3v3(v3 – 1)}
= 3v3 + 1 + 9 + 3v3 + 3v3 – 1 – 9 + 3v3
= 12v3, which is an irrational number.

Question 7.
If the third term in the binomial expansion of (1 + x)m is (-1/8)xВ then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}xВ + ……..
Now, {m(m – 1)/2}xВ = (-1/8)xВ
? m(m – 1)/2 = -1/8
? 4mВ – 4m = -1
? 4mВ – 4m + 1 = 0
? (2m – 1)В = 0
? 2m – 1 = 0
? m = 1/2

Question 8.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)В
(c) 10!/(5! з 4!)В
(d) 10!/(5! з 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/ = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)В

Question 9.
The coefficient of xn in the expansion of (1 – 2x + 3xВ – 4xГ + ……..)-n is
(a) (2n)!/n!
(b) (2n)!/(n!)В
(c) (2n)!/{2з(n!)В}
(d) None of these

Hint:
We have,
(1 – 2x + 3xВ – 4xГ + ……..)-n = {(1 + x)-2}-n
= (1 + x)2n
So, the coefficient of xnC3 = 2nCn = (2n)!/(n!)В

Question 10.
The value of n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T1 = nC0 з an-0 з b0 = 729
? an = 729 ……………. 1
T2 = nC1 з an-1 з b1 = 7290
? n
an-1 з b = 7290 ……. 2
T3 = nC2 з an-2 з bВ = 30375
? {n(n-1)/2}
an-2 з bВ = 30375 ……. 3
Now equation 2/equation 1
n
an-1 з b/an = 7290/729
? nзb/n = 10 ……. 4
Now equation 3/equation 2
{n(n-1)/2}
an-2 з bВ /n
an-1 з b = 30375/7290
? b(n-1)/2a = 30375/7290
? b(n-1)/a = (30375з2)/7290
? bn/a – b/a = 60750/7290
? 10 – b/a = 6075/729 (60750 and 7290 is divided by 10)
? 10 – b/a = 25/3 (6075 and 729 is divided by 243)
? 10 – 25/3 = b/a
? (30-25)/3 = b/a
? 5/3 = b/a
? b/a = 5/3 …………….. 5
Put this value in equation 4, we get
n з 5/3 = 10
? 5n = 30
? n = 30/5
? n = 6
So, the value of n is 6

Question 11.
The coefficient of y in the expansion of (yВ + c/y)5 is
(a) 10c
(b) 10cВ
(c) 10cГ
(d) None of these

Hint:
Given, binomial expression is (yВ + c/y)5
Now, Tr+1 = 5Cr з (yВ)5-r з (c/y)r
= 5Cr з y10-3r з Cr
Now, 10 – 3r = 1
? 3r = 9
? r = 3
So, the coefficient of y = 5C3 з cГ = 10cГ

Question 12:
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

Hint:
Given, (1.1)10000 = (1 + 0.1)10000
10000C0 + 10000C1 з (0.1) + 10000C2 з(0.1)В + other +ve terms
= 1 + 10000з(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)10000 is greater than 1000

Question 13.
If n is a positive integer, then (v5+1)2n + 1 – (v5-1)2n + 1 is
(a) an odd positive integer
(b) not an integer
(c) none of these
(d) an even positive integer

Hint:
Since n is a positive integer, assume n = 1
(v5+1)В + 1 – (v5-1)В + 1
= (5 + 2v5 + 1) + 1 – (5 – 2v5 + 1) + 1 {since (x+y)В = xВ + 2xy + yВ}
= 4v5 + 2, which is not an integer

Question 14.
if n is a positive ineger then 23nn – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Hint:
Given, 23n – 7n – 1 = 23зn – 7n – 1
= 8n – 7n – 1
= (1 + 7)n – 7n – 1
= {nC0 + nC1 7 + nC2 7В + …….. + nCn 7n} – 7n – 1
= {1 + 7n + nC2 7В + …….. + nCn 7n} – 7n – 1
= nC2 7В + …….. + nCn 7n
= 49(nC2 + …….. + nCn 7n-2)
which is divisible by 49
So, 23n – 7n – 1 is divisible by 49

Question 15.
If the third term in the binomial expansion of (1 + x)m is (-1/8)xВ then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}xВ + ……..
Now, {m(m – 1)/2}xВ = (-1/8)xВ
? m(m – 1)/2 = -1/8
? 4mВ – 4m = -1
? 4mВ – 4m + 1 = 0
? (2m – 1)В = 0
? 2m – 1 = 0
? m = 1/2

Question 16.
In the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other, then the value of n is
(a) 10
(b) 15
(c) 20
(d) 25

Hint:
Given, in the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other
? nC3 = nC12
This is possible when n = 15
Because 15C13 = 15C12

Question 17.
In the binomial expansion of (71/2 + 51/3)37, the number of integers are
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Given, (71/2 + 51/3)37
Now, general term of this binomial Tr+1 = 37Cr з (71/2)37-r з (51/3)r
? Tr+1 = 37Cr з 7(37-r)/2 з (5)r/3
This General term will be an integer if 37Cr is an integer, 7(37-r)/2 is an integer and (5)r/3 is an integer.
Now, 37Cr will always be a positive integer.
Since 37Cr denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So, 37Cr is an integer.
Again, 7(37-r)/2Cr will be an integer if (37 – r)/2 is an integer.
So, r = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37 …………. 1
And if (5)r/3 is an integer, then r/3 should be an integer.
So, r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 ………….2
Now, take intersection of 1 and 2, we get
r = 3, 9, 15, 21, 27, 33
So, total possible value of r is 6
Hence, there are 6 integers are in the binomial expansion of (71/2 + 51/3)37

Question 18.
If a and п are the roots of the equation xВ – x + 1 = 0 then the value of a2009 + п2009 is
(a) 0
(b) 1
(c) -1
(d) 10

Hint:
Given, xВ – x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 Б v(1 – 4з1з1) }/2
? x = {1 Б v(1 – 4) }/2
? x = {1 Б v(-3)}/2
? x = {1 Б v(3 з -1)}/2
? x = {1 Б v3 з v-1}/2
? x = {1 Б iv3}/2 {since i = v-1}
? x = {1 + iv3}/2, {–1 – iv3}/2
? x = -{-1 – iv3}/2, -{-1 + iv3}/2
? x = w, wВ {since w = {-1 + iv3}/2 and wВ = {-1 – iv3}/2 }
Hence, a = -w, п = wВ
Again we know that wГ = 1 and 1 + w + wВ = 0
Now, a2009 + п2009 = a2007 з aВ + п2007 з пВ
= (-w)2007 з (-w)В + (-wВ)2007 з (-wВ)В {since 2007 is multiple of 3}
= -(w)2007 з (w)В – (wВ)2007 з (w4)
= -1 з wВ – 1 з wГ з w
= -1 з wВ – 1 з 1 з w
= -wВ – w
= 1 {since 1 + w + wВ = 0}
So, a2009 + п2009 = 1

Question 19.
The general term of the expansion (a + b)n is
(a) Tr+1 = nCr з ar з br
(b) Tr+1 = nCr з ar з bn-r
(c) Tr+1 = nCr з an-r з bn-r
(d) Tr+1 = nCr з an-r з br

Answer: (d) Tr+1 = nCr з an-r з br
Hint:
The general term of the expansion (a + b)n is
Tr+1 = nCr з an-r з br

Question 20.
The coefficient of xn in the expansion (1 + x + xВ + …..)-n is
(a) 1
(b) (-1)n
(c) n
(d) n+1

Hint:
We know that
(1 + x + xВ + …..)-n = (1 – x)-n
Now, the coefficient of x = (-1)n з nCn
= (-1)n

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