+91-8076753736[email protected]
logo
logo
-->

Inverse Trigonometric Functions

Question
CBSEGJMA12036305
Wired Faculty App

sin to the power of negative 1 end exponent left parenthesis 1 minus x right parenthesis minus 2 space sin to the power of negative 1 end exponent space x comma space equals straight pi over 2 comma space then x is equal to
  • 0 comma 1 half
  • 1 comma 1 half

  • 0

  • 1 half

Solution

C.

0

Here sin-1(1 - x)-2 sin-1 x=straight x over 2
therefore space space space minus 2 space sin to the power of negative 1 end exponent space straight x equals straight pi over 2 minus sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space space space space rightwards double arrow space minus 2 space sin to the power of negative 1 end exponent space straight x equals cos to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis rightwards double arrow space space cos left parenthesis negative 2 space sin to the power of negative 1 end exponent space straight x right parenthesis equals 1 minus straight x space space space space space space space space space space space space space space rightwards double arrow space cos left parenthesis 2 space sin to the power of negative 1 end exponent space straight x right parenthesis equals 1 minus straight x rightwards double arrow space space left square bracket 1 minus 2 sin squared left parenthesis sin to the power of negative 1 end exponent space straight x right parenthesis equals 1 minus straight x space space space space space space space space space space space space space space space left square bracket because space cos space 2 space straight theta equals 1 minus 2 space sin squared space straight theta space right square bracket rightwards double arrow space space 1 minus 2 straight x squared equals 1 minus straight x space space space rightwards double arrow space 2 space straight x squared minus straight x equals 0 rightwards double arrow space space space straight x left parenthesis 2 space straight x minus 1 right parenthesis equals 0 space space rightwards double arrow space straight x equals 0 comma space 1 half
But 1 half does not satisfy the given equation.
                                 open square brackets because space space sin to the power of negative 1 end exponent open parentheses 1 minus 1 half close parentheses minus 2 space sin to the power of negative 1 end exponent open parentheses 1 half close parentheses equals straight pi over 6 minus 2 cross times straight pi over 6 not equal to straight pi over 2 close square brackets therefore space space space space space space space space space space space space space space space x equals 0
                         

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation