Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $\mathrm{x}^{7}, \mathrm{x}^{5}, \mathrm{x}^{3}$ and $x$ respectively in the expansion of $\left(\mathrm{x}+\sqrt{\mathrm{x}^{3}-1}\right)^{5}+\left(\mathrm{x}-\sqrt{\mathrm{x}^{3}-1}\right)^{5}, \mathrm{x}>1$. If u and v satisfy the equations $\alpha \mathrm{u}+\beta \mathrm{v}=18$, $\gamma \mathrm{u}+\delta \mathrm{v}=20$, then $\mathrm{u}+\mathrm{v}$ equals :

Binomial Theorem — JEE Main · 2025 · 22 Jan 2025 · Shift 2 Question with Solution

Let

α, β, γ

and

δ

be the coefficients of

x^{7}, x^{5}, x^{3}

and

x

respectively in the expansion of

(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, x > 1

. If u and v satisfy the equations

α u + β v = 18

γ u + δ v = 20

,
then

u + v

equals :

Answer: (A)

The two expansions are conjugate in

x^{3} - 1

, so on addition the odd powers of

x^{3} - 1

cancel:

(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5} = 2 [^{5} C_{0} x^{5} +^{5} C_{2} x^{3} (x^{3} - 1) +^{5} C_{4} x (x^{3} - 1)^{2}] .

Expanding:

x^{3} (x^{3} - 1) = x^{6} - x^{3}, x (x^{3} - 1)^{2} = x (x^{6} - 2 x^{3} + 1) = x^{7} - 2 x^{4} + x .

The bracket equals

x^{5} + 10 (x^{6} - x^{3}) + 5 (x^{7} - 2 x^{4} + x) = 5 x^{7} + 10 x^{6} + x^{5} - 10 x^{4} - 10 x^{3} + 5 x,

and doubling:

10 x^{7} + 20 x^{6} + 2 x^{5} - 20 x^{4} - 20 x^{3} + 10 x .

Reading off coefficients:

α = 10

(of

x^{7}

β = 2

(of

x^{5}

γ = - 20

(of

x^{3}

δ = 10

(of

x

).

Substituting into the given equations:

10 u + 2 v = 18 \Rightarrow 5 u + v = 9, - 20 u + 10 v = 20 \Rightarrow - 2 u + v = 2.

Subtracting:

7 u = 7 \Rightarrow u = 1

, and then

v = 9 - 5 = 4

.

So

u + v = 5.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option