NIMCET 2026 — Mathematics PYQ

To find the coefficient of $x^{10}$, we examine the general term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$ for both binomial expressions.

Part 1: Expansion of ${\left(x^2+\frac{1}{x}\right)}^{12}$

The general term is:

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{12}{r}\right)(x^2{)}^{12-r}{\left(\frac{1}{x}\right)}^r=(\genfrac{}{}{0ex}{}{12}{r})x^{24-2r}\cdot x^{-r}=(\genfrac{}{}{0ex}{}{12}{r})x^{24-3r}$$

Set $24-3r=10\Rightarrow 3r=14$. Since $r$ must be an integer, there is no term containing $x^{10}$ in this part. The coefficient is $0$.

Part 2: Expansion of ${\left(x+\frac{1}{x^2}\right)}^{12}$

The general term is:

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{12}{r}\right)(x{)}^{12-r}{\left(\frac{1}{x^2}\right)}^r=(\genfrac{}{}{0ex}{}{12}{r})x^{12-r}\cdot x^{-2r}=(\genfrac{}{}{0ex}{}{12}{r})x^{12-3r}$$

Set $12-3r=10\Rightarrow 3r=2$. Again, $r$ is not an integer. There is no term containing $x^{10}$ in this part. The coefficient is $0$.

Conclusion:

Since neither expansion contains a term with $x^{10}$, the sum of the coefficients is $0+0=0$.

Correct Option: (d) $0$