A) 0
B) \[{{f}_{1}}(a){{g}_{2}}(a){{h}_{3}}(a)\]
C) 1
D) None of these
Correct Answer: A
Solution :
We have \[F(x)=\left| \begin{matrix} {{f}_{1}}(x) & {{f}_{2}}(x) & {{f}_{3}}(x) \\ {{g}_{1}}(x) & {{g}_{2}}(x) & {{g}_{3}}(x) \\ {{h}_{1}}(x) & {{h}_{2}}(x) & {{h}_{3}}(x) \\ \end{matrix} \right|\] \[\therefore F'(x)=\left| \begin{matrix} f_{1}^{'}(x) & f_{2}^{'}(x) & f_{3}^{'}(x) \\ {{g}_{1}}(x) & {{g}_{2}}(x) & {{g}_{3}}(x) \\ {{h}_{1}}(x) & {{h}_{2}}(x) & {{h}_{3}}(x) \\ \end{matrix} \right|+\left| \begin{matrix} {{f}_{1}}(x) & {{f}_{2}}(x) & {{f}_{3}}(x) \\ g_{1}^{'}(x) & g_{2}^{'}(x) & g_{3}^{'}(x) \\ {{h}_{1}}(x) & {{h}_{2}}(x) & {{h}_{3}}(x) \\ \end{matrix} \right|\] \[+\left| \begin{matrix} {{f}_{1}}(x) & {{f}_{2}}(x) & {{f}_{3}}(x) \\ {{g}_{1}}(x) & {{g}_{2}}(x) & {{g}_{3}}(x) \\ h_{1}^{'}(x) & h_{2}^{'}(x) & h_{3}^{'}(x) \\ \end{matrix} \right|\] Þ \[F'(a)=0\](since \[{{f}_{n}}(a)={{g}_{n}}(a)={{h}_{n}}(a),\ \ \ n=1,\ 2,\ 3)\] Therefore two rows in each determinant become identical on putting x = a.
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