CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter 8 – Binomial Theorem Maths
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Question 6:
4 4
2 2 2 2
Find the value of
a a 1 a a 1
Solution 6:
Firstly, the expression x y x y
This can be done as
4 4
is simplified by using Binomial Theorem.
4 4 4 4 3 4 2 2 4 3 4 4
x y C x C x y C x y C xy C y
0 1 2 3 4
4 3 2 2 3 4
x 4x y 6x y 4xy y
4 4 4 4 3 4 2 2 4 3 4 4
x y C x C x y C x y C xy C y
0 1 2 3 4
4 3 2 2 3 4
x 4x y 6x y 4xy y
4 4 4 2 2 4
x y x y 2x 6x y y
Putting
x a
2
and
2
y a 1
, we obtain
2 2 2 2 2
4
2
2
2 2
a a 1 a a 1 2
a 6a a 1 a 1
4 4 2 4
2
8 4 2 2
2 a 6a a 1 a 1
8 6 4 4 2
2
a 6a 6a a 2a 1
8 6 4 2
2
a 6a 5a 2a 1
8 6 4 2
2a 12a 10a 4a 2
Question 7:
Find an approximation of 5
Solution 7:
0.99 1
0.01
0.99 1 0.01
5 5
0.99
5 5 5 4 5 3 2
C0 1 C1 1 0.01 C2
1 0.01
using the first three terms of expansion.
[Approximately]
150.01 100.01 2
10.05 0.001
1.001 0.05
0.951
Thus, the value of 5
0.99 is approximately 0.951.
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