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Diketahui fungsi h ( y ) = cos 2 y + 4 sin 2 y ​ .Jika h ( y ) = ( cos 2 y + 4 ) 2 p + q cos 2 y ​ , nilai dari p + q adalah ....

Diketahui fungsi . Jika , nilai dari   adalah ....

  1. begin mathsize 14px style 14 end style

  2. begin mathsize 14px style 12 end style

  3. begin mathsize 14px style 10 end style

  4. begin mathsize 14px style 6 end style

  5. begin mathsize 14px style negative 6 end style

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H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

Pembahasan

Ingat kembalijika ,maka Sehingga, jika dan , maka dan . Didapat hasil sebagai berikut. Dari soal diketahui . Maka, didapat dan , sehingga Jadi, jawaban yang tepat adalah C.

Ingat kembali jika begin mathsize 14px style h left parenthesis y right parenthesis equals fraction numerator u left parenthesis y right parenthesis over denominator v left parenthesis y right parenthesis space end fraction end style,  maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell h to the power of apostrophe left parenthesis y right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator u to the power of apostrophe left parenthesis y right parenthesis times v left parenthesis y right parenthesis minus u left parenthesis y right parenthesis times v to the power of apostrophe left parenthesis y right parenthesis over denominator left parenthesis v left parenthesis y right parenthesis right parenthesis squared end fraction end cell end table end style

Sehingga, jika begin mathsize 14px style u left parenthesis y right parenthesis equals sin space 2 y end style  dan begin mathsize 14px style v left parenthesis y right parenthesis equals cos space 2 y plus 4 end style, maka  begin mathsize 14px style u apostrophe left parenthesis y right parenthesis equals 2 space cos space 2 y end style dan  begin mathsize 14px style v apostrophe left parenthesis y right parenthesis equals negative 2 space sin space 2 y end style

Didapat hasil sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell h to the power of apostrophe left parenthesis y right parenthesis end cell equals cell fraction numerator u to the power of apostrophe left parenthesis y right parenthesis times v left parenthesis y right parenthesis minus u left parenthesis y right parenthesis times v to the power of apostrophe left parenthesis y right parenthesis over denominator left parenthesis v left parenthesis y right parenthesis right parenthesis squared end fraction end cell row blank equals cell fraction numerator left parenthesis 2 cos invisible function application space 2 y right parenthesis times left parenthesis cos invisible function application space 2 y plus 4 right parenthesis minus left parenthesis sin invisible function application space 2 y right parenthesis times left parenthesis negative 2 sin invisible function application space 2 y right parenthesis over denominator left parenthesis cos invisible function application space 2 y plus 4 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 2 cos squared invisible function application space 2 y plus 8 cos invisible function application space 2 y plus 2 sin squared invisible function application space 2 y over denominator left parenthesis c o s invisible function application space 2 y plus 4 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 8 cos invisible function application space 2 y plus 2 left parenthesis cos squared invisible function application space 2 y plus sin squared invisible function application space 2 y right parenthesis over denominator left parenthesis c o s invisible function application space 2 y plus 4 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 8 cos invisible function application space 2 y plus 2 over denominator left parenthesis c o s invisible function application space 2 y plus 4 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 2 plus 8 cos invisible function application space 2 y over denominator left parenthesis c o s invisible function application space 2 y plus 4 right parenthesis squared end fraction end cell end table end style 

Dari soal diketahui begin mathsize 14px style h to the power of apostrophe left parenthesis y right parenthesis equals fraction numerator p plus q cos invisible function application space 2 y over denominator left parenthesis c o s invisible function application space 2 y plus 4 right parenthesis squared end fraction end style. Maka, didapat begin mathsize 14px style p equals 2 end style dan begin mathsize 14px style q equals 8 end style, sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p plus q end cell equals cell 2 plus 8 end cell row blank equals 10 end table end style

Jadi, jawaban yang tepat adalah C.

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Salah satu titik stasioner dari fungsi f ( x ) = 2 ​ − cos x sin x ​ dengan 0 < x ≤ π adalah ....

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