Question
Question: 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...
If fr(x),gr(x),hr(x),r=1,2,3 are polynomials in x such that fr(a)=gr(a)=hr(a),r=1,2,3 and
F(x)=f1(x)g1(x)h1(x)f2(x)g2(x)h2(x)f3(x)g3(x)h3(x), then find F′(x) at x=a
A
0
B
f1(a)g2(a)h3(a)
C
1
D
None of these
Answer
0
Explanation
Solution
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|$$
\+$\left| \begin{matrix}
f_{1}(x) & f_{2}(x) & f_{3}(x) \\
g_{1}(x) & g_{2}(x) & g_{3}(x) \\
h_{1}^{'}(x) & h_{2}^{'}(x) & h_{3}^{'}(x)
\end{matrix} \right|$
**∴** $\mathbf{F}^{\mathbf{'}}\mathbf{(a) =}\left| \begin{matrix}
\mathbf{f}_{\mathbf{1}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{f}_{\mathbf{2}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{f}_{\mathbf{3}}^{\mathbf{'}}\mathbf{(a)} \\
\mathbf{g}_{\mathbf{1}}\mathbf{(a)} & \mathbf{g}_{\mathbf{2}}\mathbf{(a)} & \mathbf{g}_{\mathbf{3}}\mathbf{(a)} \\
\mathbf{h}_{\mathbf{1}}\mathbf{(a)} & \mathbf{h}_{\mathbf{2}}\mathbf{(a)} & \mathbf{h}_{\mathbf{3}}\mathbf{(a)}
\end{matrix} \right|$ + $\left| \begin{matrix}
\mathbf{f}_{\mathbf{1}}\mathbf{(a)} & \mathbf{f}_{\mathbf{2}}\mathbf{(a)} & \mathbf{f}_{\mathbf{3}}\mathbf{(a)} \\
\mathbf{g}_{\mathbf{1}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{g}_{\mathbf{2}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{g}_{\mathbf{3}}^{\mathbf{'}}\mathbf{(a)} \\
\mathbf{h}_{\mathbf{1}}\mathbf{(a)} & \mathbf{h}_{\mathbf{2}}\mathbf{(a)} & \mathbf{h}_{\mathbf{3}}\mathbf{(a)}
\end{matrix} \right|$
**+**$\left| \begin{matrix}
\mathbf{f}_{\mathbf{1}}\mathbf{(a)} & \mathbf{f}_{\mathbf{2}}\mathbf{(a)} & \mathbf{f}_{\mathbf{3}}\mathbf{(a)} \\
\mathbf{g}_{\mathbf{1}}\mathbf{(a)} & \mathbf{g}_{\mathbf{2}}\mathbf{(a)} & \mathbf{g}_{\mathbf{3}}\mathbf{(a)} \\
\mathbf{h}_{\mathbf{1}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{h}_{\mathbf{2}}^{\mathbf{'}}\mathbf{(a)} & \mathbf{h}_{\mathbf{3}}^{\mathbf{'}}\mathbf{(a)}
\end{matrix} \right|$
**=.**$\mathbf{0 + 0 + 0 = 0}\mathbf{\lbrack}\mathbf{\because}\mathbf{f}_{\mathbf{r}}\mathbf{(a) =}\mathbf{g}_{\mathbf{r}}\mathbf{(a) =}\mathbf{h}_{\mathbf{r}}\mathbf{(a),r = 1,2,3\rbrack}$