Q.

The term independent of x in the expansion of ((x+1)(x2/3+1–x1/3)–(x–1)(x–x1/2))10, x>1, is:         [2025]

1 120  
2 240  
3 210  
4 150  

Ans.

(3)

We have, ((x+1)(x23+1–x13)–(x–1)(x–x12))10

=((x13+1)–(x+1x))10=(x13–1x)10

Tr+1=Cr10(x)10–r3(–1)r(x)–r2

Since, the term is independent of x, then 10–r3–r2=0

⇒  (20 – 2r) – 3r = 0 ⇒ r = 4.

Hence, the required term is C410(–1)4=210.