\(x^2 + 2cx + c^2 = 1………..(x+c)^2=1=1^2=(-1)^2\)
It is possible only when x+c=1 or x+c=-1
So two roots of the equation are
(i) x=1-c and
(ii) x=-1-c

One of the two is a and other b.

Thus,
a-b can be (1-c)-(-1-c)=2
Or
a-b can be -1-c-(1-c)=-2

Only II possible



B

Rewrite the equation in standard form, \(x^2 + 2cx + c^2 - 1 = 0\).

Note from the constant part, we have \(c^2 - 1 = (c - 1)(c + 1)\). Then also note if we add up \(c - 1\) and \(c + 1\) we get exactly \(2c\), the coefficeint of the x term. So we can actually factor this equation into \((x - c + 1)(x - c - 1) = 0\). Thus the solutions in terms of c would be \(x = c - 1\) and \(x = c + 1\).

Finally, if we subtract the two solutions from each other we can only get 2 or -2.

Ans: B
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