A) 2
B) 3
C) At most 2
D) At most 3
Correct Answer: C
Solution :
[c] Let \[f(x)={{a}_{0}}{{x}^{2}}+{{a}_{1}}x+{{a}_{2}}\] \[g(x)={{b}_{0}}{{x}^{2}}+{{b}_{1}}x+{{b}_{2}}\] \[h(x)={{c}_{0}}{{x}^{2}}+{{c}_{1}}x+{{c}_{2}}\] Then, \[\Delta (x)\left| \begin{matrix} f(x) & g(x) & h(x) \\ 2{{a}_{0}}x+{{a}_{1}} & 2{{b}_{0}}x+{{b}_{1}} & 2{{c}_{0}}x+{{c}_{1}} \\ 2{{a}_{0}} & 2{{b}_{0}} & 2{{c}_{0}} \\ \end{matrix} \right|\] \[=x\left| \begin{matrix} f(x) & g(x) & h(x) \\ 2{{a}_{0}} & 2{{b}_{0}} & 2{{c}_{0}} \\ 2{{a}_{0}} & 2{{b}_{0}} & 2{{c}_{0}} \\ \end{matrix} \right|+\left| \begin{matrix} f(x) & g(x) & h(x) \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ 2{{a}_{0}} & 2{{b}_{0}} & 2{{c}_{0}} \\ \end{matrix} \right|\] \[=0+2\left| \begin{matrix} f(x) & g(x) & h(x) \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{0}} & {{b}_{0}} & {{c}_{0}} \\ \end{matrix} \right|\] \[=2[({{b}_{1}}{{c}_{0}}-{{b}_{0}}{{c}_{1}}]f(x)-({{a}_{1}}{{c}_{0}}-{{a}_{0}}{{c}_{1}})g(x)\] \[+({{a}_{1}}{{b}_{0}}-{{a}_{0}}{{b}_{1}})h(x)]\] Hence degree of \[\Delta (x)\le 2\].
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