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Class 11 Maths
Binomial Theorem
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
The coefficient of y in the expansion of (y² + c/y)
5
is
0%
10c
0%
10c²
0%
10c³
0%
None of these
Explanation
10c³
Hint:
Given, binomial expression is (y² + c/y)
5
Now, T
r+1
=
5
C
r
× (y²)
5-r
× (c/y)
r
=
5
C
r
× y
10-3r
× C
r
Now, 10 - 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y =
5
C
3
× c³ = 10c³
(1.1)
10000
is _____ 1000
0%
greater than
0%
less than
0%
equal to
0%
None of these
Explanation
greater than
Hint:
Given, (1.1)
10000
= (1 + 0.1)
10000
10000
C
0
+
10000
C
1
× (0.1) +
10000
C
2
×(0.1)² + other +ve terms
= 1 + 10000×(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)
10000
is greater than 1000
The fourth term in the expansion (x - 2y)
12
is
0%
-1670 x
9
× y³
0%
-7160 x
9
× y³
0%
-1760 x
9
× y³
0%
-1607 x
9
× y³
Explanation
-1760 x
9
× y³
Hint:
4th term in (x - 2y)
12
= T
4
= T
3+1
=
12
C
3
(x)
12-3
×(-2y)³
=
12
C
3
x
9
×(-8y³)
= {(12×11×10)/(3×2×1)} × x
9
×(-8y³)
= -(2×11×10×8) × x
9
× y³
= -1760 x
9
× y³
If n is a positive integer, then (√3+1)
2n+1
+ (√3−1)
2n+1
is
0%
an even positive integer
0%
a rational number
0%
an odd positive integer
0%
an irrational number
Explanation
an irrational number
Hint:
Since n is a positive integer, assume n = 1
(√3+1)³ + (√3−1)³
= {3√3 + 1 + 3√3(√3 + 1)} + {3√3 - 1 - 3√3(√3 - 1)}
= 3√3 + 1 + 9 + 3√3 + 3√3 - 1 - 9 + 3√3
= 12√3, which is an irrational number.
If the third term in the binomial expansion of (1 + x)
m
is (-1/8)x² then the rational value of m is
0%
2
0%
1/2
0%
3
0%
4
Explanation
1/2
Hint:
(1 + x)
m
= 1 + mx + {m(m - 1)/2}x² + ........
Now, {m(m - 1)/2}x² = (-1/8)x²
⇒ m(m - 1)/2 = -1/8
⇒ 4m² - 4m = -1
⇒ 4m² - 4m + 1 = 0
⇒ (2m - 1)² = 0
⇒ 2m - 1 = 0
⇒ m = 1/2
The greatest coefficient in the expansion of (1 + x)
10
is
0%
10!/(5!)
0%
10!/(5!)²
0%
10!/(5! × 4!)²
0%
10!/(5! × 4!)
Explanation
10!/(5!)²
Hint:
The coefficient of x
r
in the expansion of (1 + x)
10
is
10
C
r
and
10
C
r
is maximum for
r = 10/ = 5
Hence, the greatest coefficient =
10
C
5
= 10!/(5!)²
The coefficient of x
n
in the expansion of (1 - 2x + 3x² - 4x³ + ........)
-n
is
0%
(2n)!/n!
0%
(2n)!/(n!)²
0%
(2n)!/{2×(n!)²}
0%
None of these
Explanation
(2n)!/(n!)²
Hint:
We have,
(1 - 2x + 3x² - 4x³ + ........)
-n
= {(1 + x)
-2
}
-n
= (1 + x)
2n
So, the coefficient of x
n
C
3
=
2n
C
n
= (2n)!/(n!)²
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
0%
2
0%
4
0%
6
0%
8
Explanation
6
Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T
1
=
n
C
0
× a
n-0
× b
0
= 729
⇒ a
n
= 729 ................ 1
T
2
=
n
C
1
× a
n-1
× b
1
= 7290
⇒ n
a
n-1
× b = 7290 ....... 2
T
3
=
n
C
2
× a
n-2
× b² = 30375
⇒ {n(n-1)/2}
a
n-2
× b² = 30375 ....... 3
Now equation 2/equation 1
n
a
n-1
× b/a
n
= 7290/729
⇒ n×b/n = 10 ....... 4
Now equation 3/equation 2
{n(n-1)/2}
a
n-2
× b² /n
a
n-1
× b = 30375/7290
⇒ b(n-1)/2a = 30375/7290
⇒ b(n-1)/a = (30375×2)/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 × 5/3 = 10
⇒ 5n = 30
⇒ n = 30/5
⇒ n = 6
So, the value of n is 6
If α and β are the roots of the equation x² - x + 1 = 0 then the value of α
2009
+ β
2009
is
0%
0
0%
1
0%
-1
0%
10
Explanation
1
Hint:
Given, x² - x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 ± √(1 - 4×1×1) }/2
⇒ x = {1 ± √(1 - 4) }/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(3 × -1)}/2
⇒ x = {1 ± √3 × √-1}/2
⇒ x = {1 ± i√3}/2 {since i = √-1}
⇒ x = {1 + i√3}/2, {--1 - i√3}/2
⇒ x = -{-1 - i√3}/2, -{-1 + i√3}/2
⇒ x = w, w² {since w = {-1 + i√3}/2 and w² = {-1 - i√3}/2 }
Hence, α = -w, β = w²
Again we know that w³ = 1 and 1 + w + w² = 0
Now, α
2009
+ β
2009
= α
2007
× α² +
β
2007
× β²
= (-w)
2007
× (-w)² + (-w²)
2007
× (-w²)² {since 2007 is multiple of 3}
= -(w)
2007
× (w)² - (w²)
2007
× (w
4
)
= -1 × w² - 1 × w³ × w
= -1 × w² - 1 × 1 × w
= -w² - w
= 1 {since 1 + w + w² = 0}
So, α
2009
+ β
2009
= 1
The general term of the expansion (a + b)
n
is
0%
T
r+1
=
n
C
r
× a
r
× b
r
0%
T
r+1
=
n
C
r
× a
r
× b
n-r
0%
T
r+1
=
n
C
r
× a
n-r
× b
n-r
0%
T
r+1
=
n
C
r
× a
n-r
× b
r
Explanation
T
r+1
=
n
C
r
× a
n-r
× b
r
Hint:
The general term of the expansion (a + b)
n
is
T
r+1
=
n
C
r
× a
n-r
× b
r
The coefficient of x
n
in the expansion (1 + x + x² + .....)
-n
is
0%
1
0%
(-1)
n
0%
n
0%
n+1
Explanation
(-1)
n
Hint:
We know that
(1 + x + x² + .....)
-n
= (1 - x)
-n
Now, the coefficient of x = (-1)
n
×
n
C
n
= (-1)
n
If n is a positive integer, then (√5+1)
2n
+ 1 − (√5−1)
2n
+ 1 is
0%
an odd positive integer
0%
not an integer
0%
none of these
0%
an even positive integer
Explanation
not an integer
Hint:
Since n is a positive integer, assume n = 1
(√5+1)² + 1 − (√5−1)² + 1
= (5 + 2√5 + 1) + 1 - (5 - 2√5 + 1) + 1 {since (x+y)² = x² + 2xy + y²}
= 4√5 + 2, which is not an integer
In the expansion of (a + b)
n
, if n is even then the middle term is
0%
(n/2 + 1)
th
term
0%
(n/2)
th
term
0%
n
th
term
0%
(n/2 - 1)
th
term
Explanation
(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
In the expansion of (a + b)
n
, if n is odd then the number of middle term is/are
0%
0
0%
1
0%
2
0%
More than 2
Explanation
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
if n is a positive ineger then 2
3n
n - 7n - 1 is divisible by
0%
7
0%
9
0%
49
0%
81
Explanation
49
Hint:
Given, 2
3n
- 7n - 1 = 2
3×n
- 7n - 1
= 8
n
- 7n - 1
= (1 + 7)
n
- 7n - 1
= {
n
C
0
+
n
C
1
7 +
n
C
2
7² + ........ +
n
C
n
7
n
} - 7n - 1
= {1 + 7n +
n
C
2
7² + ........ +
n
C
n
7
n
} - 7n - 1
=
n
C
2
7² + ........ +
n
C
n
7
n
= 49(
n
C
2
+ ........ +
n
C
n
7
n-2
)
which is divisible by 49
So, 2
3n
- 7n - 1 is divisible by 49
In the binomial expansion of (7
1/2
+ 5
1/3
)
37
, the number of integers are
0%
2
0%
4
0%
6
0%
8
Explanation
6
Hint:
Given, (7
1/2
+ 5
1/3
)
37
Now, general term of this binomial T
r+1
=
37
C
r
× (7
1/2
)
37-r
× (5
1/3
)
r
⇒ T
r+1
=
37
C
r
× 7(
37-r
)
/2
× (5)
r/3
This General term will be an integer if
37
C
r
is an integer, 7(
37-r
)
/2
is an integer and (5)
r/3
is an integer.
Now,
37
C
r
will always be a positive integer.
Since
37
C
r
denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So,
37
C
r
is an integer.
Again, 7(
37-r
)
/2
C
r
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 (7
1/2
+ 5
1/3
)
37
The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
0%
4815
0%
4851
0%
8451
0%
8415
Explanation
4851
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+3-1
C
3-1
=
99
C
2
= (99 × 98)/2
= 4851
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
0%
10
0%
15
0%
20
0%
25
Explanation
15
Hint:
Given, in the binomial expansion of (a + b)
n
, the coefficient of fourth and thirteenth terms are equal to each other
⇒
n
C
3
=
n
C
12
This is possible when n = 15
Because
15
C
13
=
15
C
12
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