JEE Main 2022 — Mathematics Algebra

Given,

$$\Delta =|\begin{matrix}P! & (P+1)! & (P+2)! \\ (P+1)! & (P+2)! & (P+3)! \\ (P+2)! & (P+3)! & (P+4)!\end{matrix}|$$Δ=∣P!(P+1)!(P+2)!​(P+1)!(P+2)!(P+3)!​(P+2)!(P+3)!(P+4)!​∣

Now using the factorial concept and taking common terms we get,

$$\Rightarrow \Delta =P!(P+1)!(P+2)!|\begin{matrix}1 & 1 & 1 \\ P+1 & P+2 & P+3 \\ (P+2)(P+1) & (P+3)(P+2) & (P+4)(P+3)\end{matrix}|$$⇒Δ=P!(P+1)!(P+2)!∣1P+1(P+2)(P+1)​1P+2(P+3)(P+2)​1P+3(P+4)(P+3)​∣

Now on solving determinant we get, $|\begin{matrix}1 & 1 & 1 \\ P+1 & P+2 & P+3 \\ (P+2)(P+1) & (P+3)(P+2) & (P+4)(P+3)\end{matrix}|$∣1P+1(P+2)(P+1)​1P+2(P+3)(P+2)​1P+3(P+4)(P+3)​∣

Using operation {C}_{1}\rightarrow {C}_{1}-{C}_{2}&{C}_{2}\rightarrow {C}_{2}-{C}_{3}

$$|\begin{matrix}0 & 0 & 1 \\ -1 & -1 & P+3 \\ -2P-4 & -2P-6 & (P+4)(P+3)\end{matrix}|$$∣0−1−2P−4​0−1−2P−6​1P+3(P+4)(P+3)​∣

$$=2P+6-(2P+4)=2$$=2P+6−(2P+4)=2

Now putting the value of determinant we get,

$$\Rightarrow \Delta =2P!(P+1)!(P+2)!$$⇒Δ=2P!(P+1)!(P+2)!

$$\Rightarrow \Delta =2{P}^{3}(P-1)!(P+1)(P-1)!(P+2)(P+1)(P-1)!$$⇒Δ=2P3(P−1)!(P+1)(P−1)!(P+2)(P+1)(P−1)!

Which is divisible by ${P}^{\alpha }$Pα and ${(P+2)}^{\beta }$(P+2)β

So, $\alpha =3,\beta =1$α=3,β=1