[ 28.If f(x)=|[0,x-a,x-bx+a,0,x-cx+b,x+c,0]| then,[ 1) f(a)=0, 2) f(b)=0]]

Options

A. `f(a)=0`
B. `f(b)=0`
C. `f(0)=0`
D. `f(1)=0`

Answer

C

Explanation

We have
`f(x)=|(0,x-a,x-b),(x+a,0,x-c),(x+b,x+c,0)|`
`implies f(a)=|(0,0,a-b),(2a,0,a-c),(a+b,a+c,0)|`
`=[(a-b)}2a.(a+c)}]!=0`
`:. f(b)=|(0,b-a,0),(b+a,0,b-c),(2b,b+c,0)|`
`=-(b-a)[2b(b-c)]`
`=-2b(b-a)(b-c)!=0`
`:.f(0)=|(0,-a,-b),(a,0,-c),(b,c,0)|`
`=a(bc)-b(ac)`
`=abc-abc=0`