Find the term containing x¹⁰ in the expansion of (2x² − 3/x)¹¹.

Suppose (r + 1)^th term contains x¹⁰ in the binomial expansion of (2x² − 3/x)¹¹ when x ≠ 0.

Now, T_r+1 = ¹¹C_r(2x²)^11−r(−3/x)^r = (−1)^r11C_r(2)^11−r3^rx^22−3r

This term will contain x¹⁰ if 22 − 3r = 10

⇒ r = 4

So, (4 + 1)^th term, i.e., 5^th term will contain x¹⁰.