If the equation $5 x^{5} - 25 x^{4} + a x^{3} + b x^{2} + c x - 5 = 0$ has five positive roots, then the value of $2 a + 3 b + 2 c$ is

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300

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cannot be determine

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Solution

The correct option is C 0
$\begin{matrix} If this has 5 roots then it is differentiable differentiable 5 times \end{matrix}$

$f = 5 x^{4} - 25 x^{4} + a x^{3} + b x^{2} + c x - 5$
$f^{'} = 25 x^{4} - 100 x^{3} + 3 a x^{2} + 2 b x + c = 0$
$f^{''} = 100 x^{3} - 300 x^{2} + 60 x + 2 b = 0$
$f^{'''} = 300 x^{2} - 600 x + 6 a = 0$
$f^{''''} = 600 x - 600 = 0 \Rightarrow x = 1$
$f^{'''} \Rightarrow 300 - 600 + 6 a = 0 \Rightarrow a = 50$
$f^{''} = 100 - 300 - 200 + 2 b = 0 \Rightarrow b = - 50$
$f^{'} (x) + 25 x^{4} - 100 x^{3} + 3 a x^{2} + 2 b x + c = 0$
$25 - 100 + 150 - 100 + c - 0 \Rightarrow c = 25$
$2 a + 3 b + 2 c \Rightarrow 2 (50) + 3 (- 50) + 2 (25) = 0$

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