`f(x) = |{:(x+c_(1),,x+a,,x+a),(x+b,,x+c_(2),,x+a),(x+b,,x+b,,x+c_(3)):}| " and " g(x)= (C_(1) -x)(c_(3)-x)`
Coefficient of x in f(x) is

Options

A. `(g(a)-f(b))/(b-a)`
B. `(g(-a)-g(-b))/(b-a)`
C. `(g(a)-g(b))/(b-a)`
D. none of these

Answer

C

Explanation

In given determinant ,applying `C_(2) to C_(2) -C_(1) " and " C_(3) to C_(3) -C_(2)` we get
`f(x) = |{:(x+c_(1),,a-c_(1),,0),(x+b,,c_(2)-b,,a-c_(2)),(x+b,,0,,c_(3)-b):}|`
`=x |{:(1,,a-c_(1),,0),(1,,c_(2)-b,,a-c_(2)),(1,,0,,c_(3)-b):}|+|{:(c_(1),,a-c_(1),,0),(b,,c_(2)-b,,a-c_(2)),(b,,0,,c_(3)-b):}|`
So f(x) is linear Let f(x) = Px +Q. Then
f(-a) =-aP+Q , f(-b) =-bP+Q`
Then . f(0)=0 xxP +Q`
`rArr Q= (bf (-a) -af(-b))/((b-a)) =f(0)`
Also f(-a) `=|{:(c_(1)-a,,0,,0),(b-a,,c_(2)-a,,0),(b-a,,b-a,,c_(3)-a):}|`
`=(c_(1) -a)(c_(1) -a) (c_(3) -a)`
similarly
`f(-b) =(c_(1)-b) (c_(2) -b) (c_(3) -b)`
now `g(x) (c_(1) -x) (c_(2) -x)(c_(3) -x)`
`rArr g(a) =f(-a) "and "g(b) =f(-b)`
Now from (1) we get
`f(0) =(bg (a)-ag(b))/((b-a))`