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Jika maka nilai dari adalah ....

Jika size 14px f begin mathsize 14px style left parenthesis x right parenthesis end style size 14px equals size 14px 1 over size 14px 3 blank size 14px cot size 14px invisible function application size 14px space size 14px 2 size 14px x maka nilai dari size 14px f to the power of size 14px apostrophe open parentheses size 14px pi over size 14px 3 close parentheses adalah .... 

  1. size 14px 8 over size 14px 9

  2. begin mathsize 14px style negative 4 over 3 end style

  3. size 14px 4 over size 14px 3

  4. begin mathsize 14px style negative 8 over 9 end style

  5. begin mathsize 14px style 9 over 8 end style

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H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

jawaban yang tepat adalah D.

Pembahasan

Ingat kembali bahwa jika ,maka . Perhatikan perhitungan berikut ini! Didapat . Kemudian, substitusikan . Dengan demikian, . Jadi, jawaban yang tepat adalah D.

Ingat kembali bahwa jika size 14px f begin mathsize 14px style left parenthesis x right parenthesis end style size 14px equals size 14px cot size 14px invisible function application size 14px space size 14px a size 14px x, maka size 14px f to the power of size 14px apostrophe open parentheses size 14px x close parentheses size 14px equals size 14px minus size 14px a size 14px space size 14px cosec to the power of size 14px 2 size 14px invisible function application size 14px space size 14px a size 14px x

Perhatikan perhitungan berikut ini!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 1 third blank cot invisible function application space 2 x end cell row cell f blank to the power of apostrophe open parentheses x close parentheses end cell equals cell 1 third times open parentheses negative 2 space cosec squared invisible function application space 2 x close parentheses end cell row cell f blank to the power of apostrophe open parentheses x close parentheses end cell equals cell negative 2 over 3 cosec blank squared invisible function application space 2 x end cell end table end style

Didapat begin mathsize 14px style f apostrophe left parenthesis x right parenthesis equals negative 2 over 3 cos e c squared space 2 x end style.

Kemudian, substitusikan begin mathsize 14px style x equals straight pi over 3 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f blank to the power of apostrophe open parentheses pi over 3 close parentheses end cell equals cell negative 2 over 3 cosec blank squared invisible function application space 2 open parentheses pi over 3 close parentheses end cell row blank equals cell negative 2 over 3 cosec blank squared invisible function application space fraction numerator 2 pi over denominator 3 end fraction end cell row blank equals cell negative 2 over 3 open parentheses cosec invisible function application space fraction numerator 2 pi over denominator 3 end fraction close parentheses blank squared end cell row blank equals cell negative 2 over 3 open parentheses 2 over 3 square root of 3 close parentheses blank squared end cell row blank equals cell negative 2 over 3 open parentheses 4 over 3 close parentheses end cell row blank equals cell negative 8 over 9 end cell end table end style    

Dengan demikian, begin mathsize 14px style f apostrophe open parentheses straight pi over 3 close parentheses equals negative 8 over 9 end style.

Jadi, jawaban yang tepat adalah D.

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