JEE Main 2025 Jan 22 Shift 2, Mathematics Q1: Introduction to Binomial Expansion

JEE Main2025Jan 22, Shift 2Mathematics

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 :

Solution

Using the binomial expansion of $(a + b)^{5} + (a - b)^{5}$ , only even powers of $b$ survive: $(a + b)^{5} + (a - b)^{5} = 2 [(0 5) a^{5} + (2 5) a^{3} b^{2} + (4 5) a b^{4}]$ .

With $a = x$ and $b = x^{3} - 1$ , so $b^{2} = x^{3} - 1$ and $b^{4} = (x^{3} - 1)^{2}$ :

$= 2 [x^{5} + 10 x^{3} (x^{3} - 1) + 5 x (x^{3} - 1)^{2}]$

$= 2 [x^{5} + 10 x^{6} - 10 x^{3} + 5 x (x^{6} - 2 x^{3} + 1)]$

$= 2 [5 x^{7} + 10 x^{6} + x^{5} - 10 x^{4} - 10 x^{3} + 5 x]$ .

Reading off coefficients: $α = 10$ (for $x^{7}$ ), $β = 2$ (for $x^{5}$ ), $γ = - 20$ (for $x^{3}$ ), $δ = 10$ (for $x$ ).

The system becomes:

$10 u + 2 v = 18 \Rightarrow 5 u + v = 9$ ,

$- 20 u + 10 v = 20 \Rightarrow - 2 u + v = 2$ .

Subtracting: $7 u = 7 \Rightarrow u = 1$ , and then $v = 4$ .

$∴ u + v = 5$ .

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JEE Main 2025 Jan 22 Shift 2, Mathematics Q1: Introduction to Binomial Expansion

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