Find the biggest negative value of a , for which the maximum of f ( x ) = s i n ( 24 x + π a 100 ) is at x 0 = π The answer is a = − 150 , but I don't understand the solving way. I would appreciate if you'd help me please

Question

Find the biggest negative value of a , for which the maximum of  f  (  x  )  =  s  i  n  (  24  x  +                    π            a        100    ) is at       x    0    =      π  The answer is   a  =  −  150 , but I don&#039;t understand the solving way. I would appreciate if you&#039;d help me please

Colin Moran · Accepted Answer

Step 1The problem involves determining the closest maximum of   sin  ⁡  (  24  x  ) to       x    0    =  π and then shift the curve to the right amount to put it in       x    0    =  πMaximums are at the points   x  =      1    48    (  4  π  ⋅  n  +  π  ) so you have that the closest maximum is at   n  =  11 so       x          c      l      o      s      e      s      t        =      45    48    πThis means that you have to shift the curve to the right by       3    48    π to respect the condition that the maximum is in       x    0    =  π Can you finish from here? (you can use   x  =  y  −      3    48    π to shift the curve )

Layla Velazquez · Accepted Answer

Step 1Recall that the general solution of   sin  ⁡  θ  =  1 is:  θ  =      π    2    +  2  π  n  ,  where   n  ∈      Z  So the general solution for finding all maximum values of f(x) is:                    24        x        +                              π            a                    100                                    =                  π          2                +        2        π        n        ,        where         n        ∈                  Z                                    24        x                            =                  π          2                −                              π            a                    100                +        2        π        n        ,        where         n        ∈                  Z                                    x                            =                  1          24                          (                      π            2                    −                                    π              a                        100                    +          2          π          n          )                ,        where         n        ∈                  Z                                        24        x        +                              π            a                    100                                    =                  π          2                +        2        π        n        ,        where         n        ∈                  Z                                    24        x                            =                  π          2                −                              π            a                    100                +        2        π        n        ,        where         n        ∈                  Z                                    x                            =                  1          24                          (                      π            2                    −                                    π              a                        100                    +          2          π          n          )                ,        where         n        ∈                  Z                                        24        x        +                              π            a                    100                                    =                  π          2                +        2        π        n        ,        where         n        ∈                  Z                                    24        x                            =                  π          2                −                              π            a                    100                +        2        π        n        ,        where         n        ∈                  Z                                    x                            =                  1          24                          (                      π            2                    −                                    π              a                        100                    +          2          π          n          )                ,        where         n        ∈                  Z                    In particular, we know that there is some       n    0    ∈      Z   such that       x    0    =  π so:                    π                            =                  1          24                          (                      π            2                    −                                    π              a                        100                    +          2          π                      n            0                    )                                    24        π                            =                  π          2                −                              π            a                    100                +        2        π                  n          0                                                          π            a                    100                                    =                  π          2                −        24        π        +        2        π                  n          0                                    a                            =        −        2350        +        200                  n          0                    Since   a  &amp;lt;  0 , we know that       n    0    &amp;lt;      2350    200    =  11.75 . Rounding down to the nearest integer, we find that       n    0    =  11 so that   a  =  −  2350  +  200  (  11  )  =  −  150 , as desired.

Find the biggest negative value of a , for which the maximum of f ( x ) = s i

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