A) 3
B) 9
C) 6
D) 18
Correct Answer: D
Solution :
\[P(x)=0\] \[\Rightarrow \]\[f(x)=g(x)\] \[\Rightarrow \]\[a{{x}^{2}}+bx+c={{a}_{1}}{{x}^{2}}+{{b}_{1}}x+C,\] \[\Rightarrow \]\[(a-{{a}_{1}}){{x}^{2}}+(b-{{b}_{1}})x+(c-{{c}_{1}})=0.\] 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\[P(2)=f(2)-g(2)\] \[=4(a-{{a}_{1}})+2(b-{{b}_{1}})+(c-{{c}_{1}})\] \[=8+8+2=18\]
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