then f (x) is divisible by
n2 + n
(n + 1)!
(n + 2)!
n! (n2 + n + 1)
The value of is
20
-2
0
5
, then
Δ = 0
Δ = 2 abc
Δ = -abc
Δ = a2 + b2 + c2
2 Δ
3 Δ
6 Δ
α = 1
α = 0
α = -1
None of these
a + b + c - 3 abc
3 (a + b) (b + c) (c + a)
(a - b) (b - c) (c - a)
(a - b) (b - c) (c - a) (a + b + c)
The solution set of the equation is
{1, 2}
{-1, 2}
{1, -2}
{-1, -2}
The value of the determinant is
1
∞
ω
x2 + 2
2