Question

Let |(x, 2, x)(x², x, 6)(x, x, 6)| = ax⁴ + bx³ + cx² + dx + e then, 5a + 4b + 3c + 2d + e is equal to

(a)  0

(b)  –16

(c)  16

(d)  -11

Answer 1

Correct option  (d)  -11

Explanation:

R₃ → R₃ – R₂ gives

∴ a = 1, b = –1, c = –12, d = 12, e = 0

Put the values to get 5a + 4b + 3c + 2d + e = –11