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Diketahui  dan  untuk . NIlai  yang memenuhi adalah...

Pertanyaan

Diketahui begin mathsize 14px style straight f open parentheses straight x close parentheses equals cos space 2 straight x end style dan begin mathsize 14px style f apostrophe open parentheses straight p minus straight pi over 3 close parentheses equals square root of 3 end style untuk begin mathsize 14px style 0 less or equal than straight p less or equal than straight pi end style. NIlai begin mathsize 14px style straight p end style yang memenuhi adalah...

Pembahasan Soal:

Substiktusikan begin mathsize 14px style x equals p minus straight pi over 3 end style ke fungsi begin mathsize 14px style f open parentheses x close parentheses end style sebagai berikut;

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 cos space 2 x end cell row blank rightwards double arrow cell f open parentheses p minus straight pi over 3 close parentheses equals cos space 2 open parentheses straight p minus straight pi over 3 close parentheses end cell row blank left right double arrow cell f open parentheses p minus straight pi over 3 close parentheses equals cos open parentheses 2 straight p minus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end cell end table end style 

Kemudian, turunkan fungsi tersebut dengan menggunakan aturan rantai begin mathsize 14px style straight g equals p minus straight pi over 3 end style, dimana sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight d over dp open parentheses f open parentheses straight g close parentheses close parentheses end cell equals cell straight d over dg open parentheses f italic left parenthesis g italic right parenthesis close parentheses cross times straight d over dp open parentheses straight g close parentheses end cell row cell straight d over dp open parentheses f open parentheses straight p minus straight pi over 3 close parentheses close parentheses end cell equals cell straight d over dg open parentheses f open parentheses straight p minus straight pi over 3 close parentheses close parentheses cross times straight d over dp open parentheses straight p minus straight pi over 3 close parentheses end cell row blank equals cell straight d over dg open parentheses cos space 2 straight g close parentheses cross times straight d over dp open parentheses straight p minus straight pi over 3 close parentheses end cell row blank equals cell open parentheses negative 2 space sin space 2 straight g close parentheses cross times 1 end cell row blank equals cell negative 2 space sin space 2 straight g end cell row blank equals cell negative 2 space sin space 2 open parentheses straight p minus straight pi over 3 close parentheses end cell end table end style 

Setelah itu substitusikan hasil turunan tersebut ke fungsi turunan yang telah diketahui, sebagai berikut:

begin mathsize 14px style f apostrophe open parentheses p minus straight pi over 3 close parentheses equals square root of 3 minus 2 space sin space 2 open parentheses p minus straight pi over 3 close parentheses equals square root of 3 space sin space 2 open parentheses p minus straight pi over 3 close parentheses equals negative fraction numerator square root of 3 over denominator 2 end fraction sin open parentheses 2 p minus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals negative fraction numerator square root of 3 over denominator 2 end fraction sin open parentheses straight pi minus open parentheses 2 straight p minus straight pi over 3 close parentheses close parentheses equals sin fraction numerator 4 straight pi over denominator 3 end fraction sin open parentheses 2 p minus straight pi over 3 close parentheses equals sin fraction numerator 4 straight pi over denominator 3 end fraction end style 

Persamaan di atas deselesaikan dengan 2 cara, sebagai berikut:

Penyelesaian pertama:

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight p minus straight pi over 3 end cell equals cell fraction numerator 4 straight pi over denominator 3 end fraction plus straight k times 2 straight pi end cell row cell 2 straight p end cell equals cell fraction numerator 5 straight pi over denominator 3 end fraction plus straight k times 2 straight pi end cell row straight p equals cell fraction numerator 5 straight pi over denominator 6 end fraction plus straight k times straight pi end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight k 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 0 rightwards double arrow cell straight p equals fraction numerator 5 straight pi over denominator 6 end fraction end cell end table end style

Untuk begin mathsize 14px style straight k greater than 0 end style hasilnya akan lebih dari begin mathsize 14px style straight pi end style,maka nilai yang memenuhi untuk penyelesaian pertama adalah begin mathsize 14px style fraction numerator 5 straight pi over denominator 6 end fraction end style.

Penyelesaian kedua:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight p minus fraction numerator 2 straight pi over denominator 3 end fraction end cell equals cell open parentheses straight pi minus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses plus straight k times 2 straight pi end cell row cell 2 straight p minus fraction numerator 2 straight pi over denominator 3 end fraction end cell equals cell negative straight pi over 3 plus straight k times 2 straight pi end cell row cell 2 straight p end cell equals cell straight pi over 3 plus straight k times 2 straight pi end cell row straight p equals cell straight pi over 6 plus straight k times straight pi end cell end table end style 

begin mathsize 14px style straight k equals 0 rightwards double arrow straight p equals straight pi over 6 end style 

Untuk undefined hasilnya akan lebih dari undefined,maka nilai yang memenuhi untuk penyelesaian kedua adalah begin mathsize 14px style straight pi over 6 end style.

Sehingga, nilai undefined yang memenuhi adalah begin mathsize 14px style open curly brackets straight pi over 6 comma fraction numerator 5 straight pi over denominator 6 end fraction close curly brackets end style .

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 30 Maret 2021

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