If the equations $x^{2} + b^{2} = 1 - 2 b x$ and $x^{2} + a^{2} = 1 - 2 a x$ have one and only one common root, then the value of $| a - b |$ is

None of these

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Solution

The correct option is D $2$
Here $a \neq b$ ,
$∵$ For $a = b$ two equations become identical, so they have two roots in common.
Let $α$ be the common root of the given equations. Then,
$α^{2} + b^{2} = 1 - 2 b α \dots (1) α^{2} + a^{2} = 1 - 2 a α \dots (2)$
Using equation $(1)$ and $(2)$ ,
$b^{2} - a^{2} = 2 (a - b) α \Rightarrow α = - \frac{(a + b)}{2}$
Putting the value in equation $(1)$ ,
$\frac{(a + b)^{2}}{4} + b^{2} = 1 + 2 b \times \frac{a + b}{2} \Rightarrow (a + b)^{2} + 4 b^{2} = 4 + 4 b^{2} + 4 a b \Rightarrow (a + b)^{2} - 4 a b = 4 \Rightarrow (a - b)^{2} = 4 ∴ | a - b | = 2$

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