The number of values of the pair $$(a, b)$$  for which $$a{\left( {x + 1} \right)^2} + b\left( {{x^2} - 3x - 2} \right) + x + 1 = 0$$         is an identity in $$x$$ is

Solution :
$$\eqalign{ & {\text{We have,}} \cr & f\left( x \right) = a{\left( {x + 1} \right)^2} + b\left( {{x^2} - 3x - 2} \right) + x + 1 \cr & {\text{On expanding }}{\left( {x + 1} \right)^2}{\text{,}}\,{\text{we get}} \cr & f\left( x \right) = a\left( {{x^2} + 2x + 1} \right) + b\left( {{x^2} - 3x - 2} \right) + x + 1 = 0 \cr & f\left( x \right) = {x^2}\left( {a + b} \right) + x\left( {2a - 3b + 1} \right) + \left( {a - 2b + 1} \right) \cr & {\text{If the above quadratic equation }}f\left( x \right) = 0\forall x = 0,{\text{ then}} \cr & a + b = 0......\left( 1 \right) \cr & 2a - 3b + 1 = 0......\left( 2 \right) \cr & a - 2b + 1 = 0......\left( 3 \right) \cr & {\text{From equation }}\left( 1 \right),\,a = - b \cr & {\text{Put }}a = - b{\text{ in equation }}\left( 2 \right){\text{ and }}\left( 3 \right),{\text{ we get}} \cr & b = \frac{1}{5}{\text{ from equation }}\left( 2 \right){\text{ and}} \cr & b = \frac{1}{3}{\text{ from equation }}\left( 3 \right),{\text{ which is not possible}}{\text{.}} \cr & \therefore \,\left( {a,\,b} \right) \in \phi {\text{ for which }}f\left( x \right) = 0\forall x \in R \cr} $$