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Inverse Trigonometric Functions

Question
CBSEENMA12036012
Wired Faculty App

Consider f(x) = tan-1 open parentheses square root of fraction numerator 1 space plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses comma space straight x space element of space open parentheses 0 comma space straight pi over 2 close parentheses. A normal to y = f (x) at x = π/6 also passes through the point

  • (0,0)

  • (0, 2π/3)

  • (π/6 ,0)

  • (π/4, 0)

Solution

B.

(0, 2π/3)


We space have comma space straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent square root of fraction numerator 1 plus space sin space straight x over denominator 1 minus space sin space straight x end fraction end root comma space straight x element of space open parentheses 0 comma space straight pi over 2 close parentheses rightwards double arrow space straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent space square root of open parentheses cos begin display style straight x over 2 end style plus space sin begin display style straight x over 2 end style close parentheses squared over open parentheses cos straight x over 2 minus space sin straight x over 2 close parentheses squared end root space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator cos space begin display style straight x over 2 end style space plus space sin space begin display style straight x over 2 end style over denominator cos space begin display style straight x over 2 end style space minus space sin space begin display style straight x over 2 end style end fraction close parentheses open square brackets therefore space cos space straight x over 2 greater than space sin space straight x over 2 space for space 0 space less than thin space straight x over 2 less than straight x over 4 close square brackets equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 1 plus tan space begin display style straight x over 2 end style over denominator 1 minus space tan space begin display style straight x over 2 end style end fraction close parentheses equals space tan to the power of negative 1 end exponent space open square brackets tan space open parentheses straight pi over 4 plus straight pi over 2 close parentheses close square brackets space equals space straight pi over 4 plus straight pi over 2 rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 half rightwards double arrow space straight f apostrophe open parentheses straight pi over 6 close parentheses equals 1 half Now comma space equation space of space normal space at space straight x space equals space straight pi over 6 space is space given space by open square brackets straight y minus straight f open parentheses straight pi over 6 close parentheses close square brackets space equals space minus 2 open parentheses straight x minus straight pi over 6 close parentheses rightwards double arrow open parentheses straight y minus straight pi over 3 close parentheses space equals space minus space 2 open parentheses straight x minus straight pi over 6 close parentheses open square brackets therefore comma space straight f open parentheses straight pi over 6 close parentheses space equals space straight pi over 4 plus straight pi over 12 equals fraction numerator 4 straight pi over denominator 12 end fraction space equals straight pi over 3 close square brackets Whcih space passes space through space the space point space open parentheses 0 comma space fraction numerator 2 straight pi over denominator 3 end fraction close parentheses

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