Limit Continuity And Differentiability Ques 19
- 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)=\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| $
then $F^{\prime}(x)$ at $x=a$ is $…….$
(1985, 2M)
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Answer:
Correct Answer: 19.$(0)$
Solution: Given, $F(x)=\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right|$
$\therefore \quad F^{\prime}(x)=\left|\begin{array}{lll}f_1^{\prime}(x) & f_2^{\prime}(x) & f_3^{\prime}(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right|$ $+\left|\begin{array}{lll}f_1(x) & f_2(x) & f_3(x) \\ g_1^{\prime}(x) & g_2^{\prime}(x) & g_3^{\prime}(x) \\ h_1(x) & h_2(x) & h_3(x)\end{array}\right|+\left|\begin{array}{ccc}f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1{ }^{\prime}(x) & h_2^{\prime}(x) & h_3^{\prime}(x)\end{array}\right|$
$\Rightarrow \quad F^{\prime}(a)=0+0+0=0$
$\left[\because \quad f_r(a)=g_r(a)=h_r(a) ; r=1,2,3\right]$
BETA
