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 &lt; 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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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          &amp;lt;          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?

Christina Ward · Accepted Answer

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    ′    &amp;gt;  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    ≥  ahas 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  .

Yahir Crane · Accepted Answer

Hint:As   a is real,   0  ≤  x  ≤  2WLOG   x  :  2      sin    2    ⁡  t  ,  0  ≤  t  ≤            π      2      Now   sin  ⁡  t  +  cos  ⁡  t  =      2    cos  ⁡  (      45    ∘    −  t  )

The first part of the problem was to solve for a = 3 implies x ∈

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