NIMCET Binomial Theorem Previous Year Questions (PYQs) – Page 2 of 2

First expression is:

$\left(x^2+\frac{1}{x}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x^2)^{12-k}\left(\frac{1}{x}\right)^k$

$T_{k+1}={}^{12}C_k x^{24-2k-k}$

$T_{k+1}={}^{12}C_k x^{24-3k}$

For coefficient of $x^{10}$,

$24-3k=10$

$3k=14$

$k=\frac{14}{3}$

This is not an integer, so $x^{10}$ term is not present in the first expression.

Now second expression is:

$\left(x+\frac{1}{x^2}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x)^{12-k}\left(\frac{1}{x^2}\right)^k$

$T_{k+1}={}^{12}C_k x^{12-k-2k}$

$T_{k+1}={}^{12}C_k x^{12-3k}$

For coefficient of $x^{10}$,

$12-3k=10$

$3k=2$

$k=\frac{2}{3}$

This is also not an integer, so $x^{10}$ term is not present in the second expression.

Therefore, the coefficient of $x^{10}$ is $0$.