NCERT Class XI Mathematics - Binomial Theorem - Solutions
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Question : 32
Total: 36
Find the value of ( a 2 + √ a 2 − 1 ) 4 + ( a 2 − √ a 2 − 1 ) 4
Solution:
We have, ( a 2 + √ a 2 − 1 ) 4
C 0 ( a 2 ) 4 +
C 1 ( a 2 ) 3 ( √ a 2 − 1 ) +
C 2 ( a 2 ) 2 ( √ a 2 − 1 ) 2 +
C 3 ( a 2 ) ( √ a 2 − 1 3 ) +
C 4 ( √ a 2 − 1 ) 4
and( a 2 − √ a 2 − 1 ) 4
C 0 ( a 2 ) 4 −
C 1 ( a 2 ) 3 ( √ a 2 − 1 ) +
C 2 ( a 2 ) 2 ( √ a 2 − 1 ) 2 −
C 3 ( a 2 ) ( √ a 2 − 1 ) 3 +
C 4 ( √ a 2 − 1 ) 4
Adding (i) and (ii), we get
( a 2 + √ a 2 − 1 ) 4 + ( a 2 − √ a 2 − 1 ) 4 = 2 [
C 0 ( a 2 ) 4 +
C 2 ( a 2 ) 2 ( √ a 2 − 1 ) 2 +
C 4 ( √ a 2 − 1 ) 4 ]
=2 [ a 8 + 6 a 4 ( a 2 − 1 ) + ( a 2 − 1 ) 2 ] 2 [ a 8 + 6 a 6 − 6 a 4 + a 4 − 2 a 2 + 1 ]
=2 a 8 + 12 a 6 − 10 a 4 − 4 a 2 + 2
and
Adding (i) and (ii), we get
=
=
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