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If one root of the equation t^2-(12x)t-(f(x) 64x)=0 is twice of other, then find the maximum value of the function f(x),where x belong to R?
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If one root of the equation t^2-(12x)t-(f(x) 64x)=0 is twice of other,...
Problem:
Find the maximum value of the function f(x), where x belongs to R, if one root of the equation t^2 - (12x)t - f(x) 64x=0 is twice of other.

Solution:
Let the roots of the equation be a and 2a, then we have:
t^2 - (12x)t - f(x) 64x = 0
(t - a)(t - 2a) = 0
t^2 - (3a)t + 2a^2 = 0

Comparing the coefficients of t in both equations, we get:
3a = 12x
a = 4x

Substituting the value of a in the second equation, we get:
16x^2 - 12x(4x) + 2a^2 = 0
16x^2 - 48x^2 + 2a^2 = 0
2a^2 = 32x^2

Therefore, the two roots of the equation are:
a = 4x
2a = 8x

Sum of the roots = a + 2a = 12x

Using sum and product of roots formula, we get:
a + 2a = 12x = -b/a
a(2a) = 8ax^2 = c/a

Substituting the value of a in c/a, we get:
c/a = f(x)64x
8x^2 = f(x)64x
f(x) = x/8

Therefore, the maximum value of f(x) is 0.125, which occurs when x = 1.

Answer:
The maximum value of the function f(x) is 0.125, which occurs when x = 1.
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If one root of the equation t^2-(12x)t-(f(x) 64x)=0 is twice of other, then find the maximum value of the function f(x),where x belong to R?
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