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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 $F(x)=\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| . \quad$ Then $F^{\prime}(a)$ is equal to
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We have $F(x)=\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right|$
$\begin{aligned} \therefore F^{\prime}(x)=\left|\begin{array}{lll}f'_1(x) & f'_2(x) & f'_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| & +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1^{\prime}(x) & g'_2(x) & g_3^{\prime}(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1^{\prime}(x) & h_2^{\prime}(x) & h_3^{\prime}(x)\end{array}\right|\end{aligned}$
$\Rightarrow F(a)=0\left(\right.$ since $\left.f_n(a)=g_n(a)=h_n(a), \quad n=1,2,3\right)$
Therefore two rows in each determinant become identical on putting $x=a$.
$\begin{aligned} \therefore F^{\prime}(x)=\left|\begin{array}{lll}f'_1(x) & f'_2(x) & f'_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| & +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1^{\prime}(x) & g'_2(x) & g_3^{\prime}(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right| +\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1^{\prime}(x) & h_2^{\prime}(x) & h_3^{\prime}(x)\end{array}\right|\end{aligned}$
$\Rightarrow F(a)=0\left(\right.$ since $\left.f_n(a)=g_n(a)=h_n(a), \quad n=1,2,3\right)$
Therefore two rows in each determinant become identical on putting $x=a$.
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