If roots of the equation `f(x)=x^6-12 x^5+bx^4+cx^3+dx^2+ex+64=0`are positive, then
remainder when `f(x)` is divided by `x-1` is

Options

A. 2
B. 1
C. 3
D. 10

Answer

B

Explanation

Let roots of equation `x^(6) - 12x^(5) + bx^(4) + cx^(3) + dx^(2) + ex + 64 = 0`
be `x_(i), I = 1,2…..6` Now,
`x_(1) + x_(2) + x_(3) + x_(4) + x_(5) + x_(6) = 12`
and `x_(1) x_(2) x_(4) x_(5) x_(6) = 64`
Thus,
`(x_(1) + x_(2) …. + x_(6))/(6) = 2` and `(x_(1) x_(2) x_(3) x_(5) x_(6))^(1//6) = 2`
`implies a.M = G.M`
`implies x_(1) = x_(2) = x_(3) - x_(4) = x_(5) = x_(6) = 2`
Hence, the given equation is equivalent to
`(x - 2)^(6) = 0`
or `x^(6) - 12 x^(5) + 60x^(4) - 160 x^(3) + 240x^(2) - 192 x - 64 = 0`
`:. f(1) = 1 - 12 + 60 - 160 + 240 - 192 + 64 = 1`