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НоВыЙ13
@НоВыЙ13
July 2022
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Найдите количество целых чисел, принадлежащих множеству значений функции:
[tex]f(x)=16log_{ \frac{1}{6} } \frac{sinx+cosx+3 \sqrt{2} }{ \sqrt{2} } [/tex]
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oganesbagoyan
Verified answer
Найдите количество целых чисел, принадлежащих множеству значений функции: f(x) =16Log(1/6) (sinx +cosx +3
√2) /√2 .
----------------------------------
f(x
) =16Log(1/6) (sinx +cosx +3√2) /√2
=16Log(1/6) ( (sinx +cosx)/√2 +3)
.
(sinx +cosx) / √2 =(1/√2) *sinx + (1/√2) *cosx) =
cos(π/4) *sinx + sin(π/4) *cosx = sin(π/4+x )
следовательно -1 ≤ (sinx +cosx) /√2 ≤
1 ;
2 ≤ (sinx +cosx) /√2 +3 ≤ 4
т.к. 0 < 1/6 < 1
, то
Log(1/6) 2 ≥ Log(1/6) ( ( sinx +cosx)√2 +3 ) ≥
Log(1/6)
4 ;
16*Log(1/6) 2 ≥
16* Log(1/6) ( ( sinx +cosx)√2 +3 )
≥ 16* Log(1/6) 2² ;
32*Log(1/6) 2 ≤
f(x
) ≤ 16* Log(1/6) 2 ;
-32*Log(6) 2 ≤
f(x)
≤ -16*
Log(6) 2 ;
-32/(1+Log(2) 3) ≤ f(x) ≤ - 16 /
(1+Log(2) 3 ) ;
{ -12 ; -11; -10 ; -9 ; -8 ; -7 }
ответ : 6 .
------------
2 votes
Thanks 1
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Verified answer
Найдите количество целых чисел, принадлежащих множеству значений функции: f(x) =16Log(1/6) (sinx +cosx +3√2) /√2 .----------------------------------
f(x) =16Log(1/6) (sinx +cosx +3√2) /√2 =16Log(1/6) ( (sinx +cosx)/√2 +3) .
(sinx +cosx) / √2 =(1/√2) *sinx + (1/√2) *cosx) =
cos(π/4) *sinx + sin(π/4) *cosx = sin(π/4+x )
следовательно -1 ≤ (sinx +cosx) /√2 ≤ 1 ;
2 ≤ (sinx +cosx) /√2 +3 ≤ 4
т.к. 0 < 1/6 < 1 , то
Log(1/6) 2 ≥ Log(1/6) ( ( sinx +cosx)√2 +3 ) ≥ Log(1/6) 4 ;
16*Log(1/6) 2 ≥16* Log(1/6) ( ( sinx +cosx)√2 +3 ) ≥ 16* Log(1/6) 2² ;
32*Log(1/6) 2 ≤ f(x) ≤ 16* Log(1/6) 2 ;
-32*Log(6) 2 ≤ f(x) ≤ -16*Log(6) 2 ;
-32/(1+Log(2) 3) ≤ f(x) ≤ - 16 / (1+Log(2) 3 ) ;
{ -12 ; -11; -10 ; -9 ; -8 ; -7 }
ответ : 6 .
------------