Consider the polynomial fucntion `f(x) = |{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}|` a,b being positive integers. The constant term in f(x) is

Options

A. 2
B. 1
C. -1
D. 0

Answer

D

Explanation

Let
`|{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}|=A +Bx+Cx^(2)+…….`
Putting x=0 we get
`A= |{:(1,,1,,1),(1,,1,,1),(1,,1,,1):}|=0`
now differenting both sides with respect to x and putting x=0 we get
`B= |{:(a,,2b,,0),(1,,1,,1),(1,,1,,1):}|+|{:(1,,1,,1),(0,,a,,2b),(1,,1,,1):}|+|{:(1,,1,,1),(1,,1,,1),(2b,,0,,a):}|=0`
Hence ,coefficient of x is 0 .Since f(x)=0 and f(0)=0 ,x=0 is a repreating root of the equation f(x)=0