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Question

# In the expansion of (x+1x2/3âˆ’x1/3+1âˆ’xâˆ’1xâˆ’x1/2)10, the term which does not contain x, is equal to

A
10C0
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B
10C7
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C
10C4
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D
10C6
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Solution

## The correct option is D 10C6=(x+1x2/3−x1/3+1−x−1x−x1/2)10=((x+1)(x1/3+1)(x2/3−x1/2+1)(x1/3+1)−(√x+1)(√x−1)√x(√x−1))10=((x+1)(x1/3+1)x+1−√x+1√x)10 [a3+b3=(a+b)(a3+b2−ab)(x1/3)3+13=(x1/3+1)(x2/3+1−x1/3)x+1=]=(x1/3+1−1+1√x)10=(x1/3+1x1/2)10Thus Tr+1=10Cr(x1/3)r(1x1/2)10−rFor term without x r3−10−r2=05=5r6r=6Thus T7=10C6x0T7=10C6⇒ term without x

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