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प्रश्न
Find the values of x, when the marginal function of y = x3 + 10x2 – 48x + 8 is twice the x.
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उत्तर
y = x3 + 10x2 – 48x + 8
Marginal function, `"dy"/"dx"` = 3x2 + 10(2x) – 48
= 3x2 + 20x – 48
Given that, the marginal function is twice the x.
Therefore, 3x2 + 20x – 48 = 2x
3x2 + 18x – 48 = 0
Divide throughout by 3, x2 + 6x – 16 = 0
(x + 8) (x – 2) = 0
x = -8 (or) x = 2
The values of x are -8, 2.
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