A) proportional to \[{{x}^{2}}\]
B) proportional to \[x\]
C) proportional to \[{{x}^{3}}\]
D) a constant
Correct Answer: B
Solution :
We have \[f(x)=\left| \begin{matrix} {{x}^{3}} & {{x}^{4}} & 3{{x}^{2}} \\ 1 & -6 & 4 \\ p & {{p}^{2}} & {{p}^{3}} \\ \end{matrix} \right|\] \[\Rightarrow \]\[f(x)={{x}^{3}}(-6{{p}^{3}}-4{{p}^{2}})-{{x}^{4}}({{p}^{3}}-4p)\] \[+\,3{{x}^{2}}({{p}^{2}}+6p)\] \[\Rightarrow \]\[\frac{d}{dx}f(x)=3{{x}^{2}}(-6{{p}^{3}}-4{{p}^{2}})-4{{x}^{3}}\] \[\times \,({{p}^{3}}-4p)+6x({{p}^{2}}+6p)\] \[\Rightarrow \] \[\frac{{{d}^{2}}}{d{{x}^{2}}}f(x)=6x(-6{{p}^{3}}-4{{p}^{2}})\] \[-12{{x}^{2}}({{p}^{3}}-4p)+6({{p}^{2}}+6p)\] \[\Rightarrow \]\[\frac{{{d}^{3}}}{d{{x}^{3}}}f(x)=6(-6{{p}^{3}}-4{{p}^{2}})-24x({{p}^{3}}-4p)\] \[\Rightarrow \] \[\frac{{{d}^{3}}}{d{{x}^{3}}}f(x)\]is proportional to \[x.\]
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