If $\alpha, \beta \neq 0$, and $f(n)=\alpha^{n}+\beta^{n}$ and $\left|\begin{array}{ccc}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{array}\right|=\mathrm{K}(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$, then K is equal to

JEE Main · 2014 · 06 Apr 2014 · Offline · Determinants

α, β \neq = 0

, and

f (n) = α^{n} + β^{n}

and

3 1 + f (1) 1 + f (2) 1 + f (1) 1 + f (2) 1 + f (3) 1 + f (2) 1 + f (3) 1 + f (4) = K (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, then K is equal to

Answer: (C)

Replacing

f (n) = α^{n} + β^{n}

and factoring, the determinant becomes the square of the

3 \times 3

Vandermonde determinant built from

1, α, β

. That Vandermonde value is

(α - 1) (β - 1) (β - α)

, so the determinant equals

(1 - α)^{2} (1 - β)^{2} (α - β)^{2}

. Comparing with

K (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, we read off

K = 1

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option