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If fr(x),g(r)(x),hr(x),r=1,2,3 are poly...

If `f_r(x),g_(r)(x),h_r(x),r=1,2,3` are polynomials in x such that `f_r(a) = g_r(a) = h_r (a) , r = 1,2,3` and `F(x) = |(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))|` then F' at x = a is ………

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To solve the problem, we need to find the derivative of the determinant \( F(x) \) at the point \( x = a \). The determinant is defined as: \[ F(x) = \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \] Given that \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), we can denote this common value as \( K_r \). Therefore, we have: - \( f_1(a) = g_1(a) = h_1(a) = K_1 \) - \( f_2(a) = g_2(a) = h_2(a) = K_2 \) - \( f_3(a) = g_3(a) = h_3(a) = K_3 \) ### Step 1: Differentiate the Determinant Using the property of determinants, the derivative of \( F(x) \) can be computed using the formula for the derivative of a determinant: \[ F'(x) = \sum_{i=1}^{n} \det(A_i) \] where \( A_i \) is the matrix formed by differentiating the \( i^{th} \) row of the determinant and keeping the others unchanged. ### Step 2: Construct the Derivatives We will differentiate each row of the determinant: 1. **Differentiate the first row:** \[ F'(x) = \begin{vmatrix} f_1'(x) & f_2'(x) & f_3'(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \] 2. **Differentiate the second row:** \[ + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1'(x) & g_2'(x) & g_3'(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \] 3. **Differentiate the third row:** \[ + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1'(x) & h_2'(x) & h_3'(x) \end{vmatrix} \] ### Step 3: Evaluate at \( x = a \) Now we substitute \( x = a \): \[ F'(a) = \begin{vmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1'(a) & g_2'(a) & g_3'(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1'(a) & h_2'(a) & h_3'(a) \end{vmatrix} \] Since \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), we have: \[ F'(a) = \begin{vmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ K_1 & K_2 & K_3 \\ K_1 & K_2 & K_3 \end{vmatrix} + \begin{vmatrix} K_1 & K_2 & K_3 \\ g_1'(a) & g_2'(a) & g_3'(a) \\ K_1 & K_2 & K_3 \end{vmatrix} + \begin{vmatrix} K_1 & K_2 & K_3 \\ K_1 & K_2 & K_3 \\ h_1'(a) & h_2'(a) & h_3'(a) \end{vmatrix} \] ### Step 4: Simplify the Determinants In each of these determinants, two rows are identical (the second and third rows in each case), which means that each determinant evaluates to zero. Therefore: \[ F'(a) = 0 + 0 + 0 = 0 \] ### Final Answer Thus, the derivative \( F'(a) \) is: \[ \boxed{0} \]
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Knowledge Check

  • 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)=|{:(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)):}| . Then, F' (a) is equal to

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