• # question_answer Let for $a\ne {{a}_{1}}\ne 0,$ $f(x)=a{{x}^{2}}+bx+c,g9x)={{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}$and$p(x)=f(x)-g(x).$ If p(x) = 0 only for x = -1 and p(-2) = 2, then the value of p is :     AIEEE  Solved  Paper (Held On 11 May  2011) A)  3 B)  9 C)  6 D)  18

$P(x)=0$ $\Rightarrow$$f(x)=g(x)$ $\Rightarrow$$a{{x}^{2}}+bx+c={{a}_{1}}{{x}^{2}}+{{b}_{1}}x+C,$ $\Rightarrow$$(a-{{a}_{1}}){{x}^{2}}+(b-{{b}_{1}})x+(c-{{c}_{1}})=0.$ It has only one solution x = - 1 $\Rightarrow$$b-{{b}_{1}}=a-{{a}_{1}}+c-{{c}_{1}}$                                                                                                   ?.(1) Vertex (-1,0)$\Rightarrow$$\frac{b-{{b}_{1}}}{2(a-{{a}_{1}})}=-1$$\Rightarrow$$b-{{b}_{1}}=2(a-{{a}_{1}})$                                                ?.(2) $\Rightarrow$$f(-2)-g(-2)=2$ $\Rightarrow$$4a-2b+c-4{{a}_{1}}+2{{b}_{1}}-{{c}_{1}}=2$ $\Rightarrow$$4(a-{{a}_{1}})-2(b-{{b}_{1}})+(c-{{c}_{1}})=2$                                                                                ?.(3) By (1), (2) and (3) $(a-{{a}_{1}})=(c-{{c}_{1}})=\frac{1}{2}(b-{{b}_{1}})=2$ Now$P(2)=f(2)-g(2)$                 $=4(a-{{a}_{1}})+2(b-{{b}_{1}})+(c-{{c}_{1}})$                 $=8+8+2=18$