Question
The term independent of x expansion of (x+1x23−x13+1−x−1x−x12)10 is
Answer: Option C
:
C
[x+1x23−x13+1−x−1x−x12]10
=⎡⎢
⎢
⎢⎣(x13)3+13x23−x13+1−{(√x)2−1}√x(√x−1)⎤⎥
⎥
⎥⎦10
=⎡⎢
⎢
⎢⎣(x13+1)(x23+1−x13)x23−x13+1−{(√x)2−1}√x(√x−1)⎤⎥
⎥
⎥⎦10
[(x13+1)−(√x+1)√x]10=(x13−x−12)10
∴ The general term is
Tr+1=10Cr(x13)10−r(−x−12)r=10Cr(−1)rx10−r3−r2
For independent of x, put
10−r3−r2=0
⇒ 20 - 2r - 3r = 0
⇒ 20 = 5r ⇒ r = 4
∴T5=10C4=10×9×8×74×3×2×1=210
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:
C
[x+1x23−x13+1−x−1x−x12]10
=⎡⎢
⎢
⎢⎣(x13)3+13x23−x13+1−{(√x)2−1}√x(√x−1)⎤⎥
⎥
⎥⎦10
=⎡⎢
⎢
⎢⎣(x13+1)(x23+1−x13)x23−x13+1−{(√x)2−1}√x(√x−1)⎤⎥
⎥
⎥⎦10
[(x13+1)−(√x+1)√x]10=(x13−x−12)10
∴ The general term is
Tr+1=10Cr(x13)10−r(−x−12)r=10Cr(−1)rx10−r3−r2
For independent of x, put
10−r3−r2=0
⇒ 20 - 2r - 3r = 0
⇒ 20 = 5r ⇒ r = 4
∴T5=10C4=10×9×8×74×3×2×1=210
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