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Question

# The term independent of x expansion of (x+1x23−x13+1−x−1x−x12)10 is

A
4
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B
120
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C
210
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D
310
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Solution

## The correct option is C 210[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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