NCERT Solutions For Class 11 Maths chapter-8 Binomial Theorem Miscellaneous Exercise
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NCERT Solutions For Class 11 Maths chapter-8 Binomial Theorem Miscellaneous Exercise
NCERT Solutions for Class-11 Maths Chapter-8 Binomial Theorem
NCERT Solutions for Class-11 Maths Chapter-8 Binomial Theorem Miscellaneous Exercise prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class-11 maths solutions. You can download NCERT solutions of all chapters from Physics Wallah NCERT solutions of class 11.
NCERT Solutions for Class-11 Maths Miscellaneous Exercise
Question 1. Find a, b and n in the expansion of (a + b) n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Solution :
Given:
(a + b)
n
T
r+1
=
n
C
r
a
n-t
b
r
The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,
T1 =
n
C
0
a
n-0
b
0
= an = 729….. 1
T2 =
n
C
1
a
n-1
b
1
=
n
a
n-1
b = 7290…. 2
T3 =
n
C
2
a
n-2
b
2
= {n (n -1)/2 }a
n-2
b
2
= 30375……3
Dividing 2 by 1 we get
Question 2. Find a if the coefficients of x 2 and x 3 in the expansion of (3 + ax) 9 are equal.
Solution :
Question 3. Find the coefficient of x 5 in the product (1 + 2x) 6 (1 – x) 7 using binomial theorem.
Solution :
Using Binomial Theorem,
Question 4. If a and b are distinct integers, prove that a – b is a factor of a n – b n , whenever n is a positive integer.
[Hint: write an = (a – b + b) n and expand]
Solution :
n order to prove that (a – b) is a factor of (a
n
– b
n
), it has to be proved that
a
n
– b
n
= k (a – b) where k is some natural number.
a can be written as a = a – b + b
an = (a – b + b)
n
= [(a – b) + b]
n
=
n
C
0
(a – b)
n
+
n
C
1
(a – b)
n-1
b + …… +
n
C
n
b
n
a
n
– b
n
= (a – b) [(a –b)
n-1
+
n
C
1
(a – b)
n-1
b + …… +
n
C
n
b
n
]
a
n
– b
n
= (a – b)
k
Where k = [(a –b)
n-1
+
n
C
1
(a – b)
n-1
b + …… +
n
C
n
b
n]
is a natural number
Question 5. Evaluate: (√3 + √2) 6 - (√3 - √2) 6
Solution :
Question
6. Find the value of
Solution :
Question 7. Find an approximation of (0.99) 5 using the first three terms of its expansion.
Solution :
Here .99 can be written as
0.99 = 1 – 0.01
Now by applying binomial theorem we get
(o. 99)5 = (1 – 0.01)
5
= 5C0 (1)
5
–
5
C
1
(1)
4
(0.01) +
5
C
2
(1)
3
(0.01)
2
= 1 – 5 (0.01) + 10 (0.01)
2
= 1 – 0.05 + 0.001
= 0.951
Question
8. Find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
is
Solution :
Question
9. Expand using binomial theorem
Solution :
We have
Question 10. Find the expansion of (3x 2 – 2ax + 3a 2 ) 3 using binomial theorem.
Solution :
Here
= We know that (a + b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
Putting a = 3x
2
& b = -a (2x-3a), we get
[3x
2
+ (-a (2x-3a))]
3
= (3x2)
3
+3(3x
2
)
2
(-a (2x-3a)) + 3(3x
2
) (-a (2x-3a))
2
+ (-a (2x-3a))
3
= 27x
6
– 27ax
4
(2x-3a) + 9a2x
2
(2x-3a)
2
– a
3
(2x-3a)
3
= 27x
6
– 54ax
5
+ 81a2x
4
+ 9a2x
2
(4x
2
-12ax+9a
2
) – a3 [(2x)3 – (3a)3 – 3(2x)
2
(3a) + 3(2x)(3a)
2
]
= 27x
6
– 54ax
5
+ 81a
2
x
4
+ 36a
2
x
4
– 108a
3
x
3
+ 81a
4
x
2
– 8a3x3 + 27a
6
+ 36a
4
x2 – 54a
5
x
= 2
7x6
– 54ax
5
+ 117a
2
x
4
– 116a
3
x
3
+ 117a
4
x
2
– 54a
5
x + 27a
6
Thus, (3x
2
– 2ax + 3a
2
)
3
= 27x
6
– 54ax
5
+ 117a2x
4
– 116a
3
x
3
+ 117a
4
x
2
– 54a
5
x + 27a
6
