A) Proportional to \[{{x}^{2}}\]
B) Proportional to x
C) Proportional to \[{{x}^{3}}\]
D) A constant
Correct Answer: D
Solution :
\[f(x)=\left| \,\begin{matrix} {{x}^{3}} & {{x}^{2}} & 3{{x}^{2}} \\ 1 & -6 & 4 \\ p & {{p}^{2}} & {{p}^{3}} \\ \end{matrix}\, \right|\] Þ\[f(x)={{x}^{3}}(-6{{p}^{3}}-4{{p}^{2}})-{{x}^{2}}({{p}^{3}}-4p)+3{{x}^{2}}({{p}^{2}}+6p)\] Þ \[f(x)=-6{{p}^{3}}{{x}^{3}}-4{{p}^{2}}{{x}^{3}}-{{x}^{2}}{{p}^{3}}+4p{{x}^{2}}+3{{p}^{2}}{{x}^{2}}+18p{{x}^{2}}\] \\[\frac{d}{dx}f(x)=-18{{p}^{3}}{{x}^{2}}-12{{p}^{2}}{{x}^{2}}-2x{{p}^{3}}+8px+6{{p}^{2}}x+36px\] and\[\frac{{{d}^{2}}}{d{{x}^{2}}}\,f(x)=-36{{p}^{3}}x-24{{p}^{2}}x-2{{p}^{3}}+8p+6{{p}^{2}}+36p\] and \[\frac{{{d}^{3}}f(x)}{d{{x}^{3}}}=-36{{p}^{3}}-24{{p}^{2}}\] = a constant.
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