JEE Main 2025 — Binomial Theorem Question with Solution

To find the sum of $u$ and $v$, we first need to expand the expression:

${\left(x+\sqrt{x^3-1}\right)}^5+{\left(x-\sqrt{x^3-1}\right)}^5$

Using the Binomial Theorem, the expansion yields:

$=2\left({}^5{C}_0\cdot x^5+{}^5{C}_2\cdot x^3(x^3-1)+{}^5{C}_4\cdot x(x^3-1{)}^2\right)$

Simplifying this, we obtain:

$=2\left(5x^7+10x^6+x^5-10x^4-10x^3+5x\right)$

From this expansion, we can identify the coefficients:

The coefficient of $x^7$ is $\alpha =10$

The coefficient of $x^5$ is $\beta =2$

The coefficient of $x^3$ is $\gamma =-20$

The coefficient of $x$ is $\delta =10$

Given the equations:

$\alpha u+\beta v=18$

$\gamma u+\delta v=20$

Substituting in the coefficients:

$10u+2v=18$

$-20u+10v=20$

By solving these equations, we find:

From $10u+2v=18$, simplify to $5u+v=9$.

From $-20u+10v=20$, simplify to $-2u+v=2$.

Solving these linear equations simultaneously, we find:

Subtracting equation 2 from equation 1:

$5u+v=9$

$-(-2u+v=2)$

This yields:

$7u=7\Rightarrow u=1$

Substitute $u=1$ back into $5u+v=9$:

$5(1)+v=9\Rightarrow v=4$

Thus, the sum $u+v$ is:

$u+v=1+4=5$