The first part of the problem was to solve for a = 3 implies x &#x2208;<!-- ∈ -->

gnatopoditw 2022-06-25 Answered
The first part of the problem was to solve for a = 3 implies x [ 0 , 2 ]. This makes me think that I will somehow use this. I try to find the solution for the second part of the problem by squaring the inequality. I get 2 ( x 2 + 2 x ) a 2.
This is equal to two systems:
{ a < 2 x [ 0 , 2 ] and { a 2 4 x 2 8 x + a 2 4 a + 2 0
What do I do know and is this actually correct?
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Answers (2)

Christina Ward
Answered 2022-06-26 Author has 19 answers
Note that the function f ( x ) := x + 2 x is defined in [ 0 , 2 ], it is symmetric with respect to x = 1, i. e. f ( 1 + x ) = f ( 1 x ), and it is strictly increasing in [ 0 , 1 ] (because f > 0 in ( 0 , 1 )). Hence f attains its maximum value 2 at 1 and its minimum value 2 at the endpoints 0 and 2.
It follows that the inequality
x + 2 x a
has at least a solution if and only if a 2 and the set of solutions is the interval
[ 1 a 2 4 a 2 , 1 + a 2 4 a 2 ]
when a 2 , and it is the whole domain [ 0 , 2 ] if a 2 .
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Yahir Crane
Answered 2022-06-27 Author has 9 answers
Hint:
As a is real, 0 x 2
WLOG x : 2 sin 2 t , 0 t π 2
Now sin t + cos t = 2 cos ( 45 t )
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Also, by the points given:
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Using the fact that a is positive, I got the expressions:
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a m 2 + b m + c = 2 b m + c < 2
m = 4 a + 2 b + c 2 b + c < m
But I don't know how to solve it.