Differentiation Question 398
Question: If $ f _{n}(x) $ , $ g _{n}(x) $ , $ h _{n}(x),n=1,2,3 $ are polynomials in x such that $ f _{n}(a)=g _{n}(a)=h _{n}(a),n=1,2,3 $ and $ F(x)= \begin{vmatrix} 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{vmatrix} $ . Then $ {F}’(a) $ is equal to
Options:
A) 0
B) $ f_1(a)g_2(a)h_3(a) $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ F(x)= \begin{vmatrix} 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{vmatrix} $
$ \therefore F’(x)= \begin{vmatrix} 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{vmatrix} + \begin{vmatrix} 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{vmatrix} $
$ + \begin{vmatrix} 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{vmatrix} $
Therefore $ 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.
BETA
