Maths-
General
Easy

Question

The solution of the equation 2 c o s to the power of negative 1 end exponent invisible function application x equals s i n to the power of negative 1 end exponent invisible function application open parentheses 2 x square root of 1 minus x to the power of 2 end exponent end root close parentheses

  1. [–1, 0]    
  2. [0, 1]    
  3. [–1, 1]    
  4. open square brackets fraction numerator 1 over denominator square root of 2 end fraction comma 1 close square brackets    

hintHint:

space R a n g e space o f space cos to the power of negative 1 end exponent space x space i s space space space 0 space less or equal than space cos to the power of negative 1 end exponent x space less or equal than space straight pi

and space space space Range space of space sin to the power of negative 1 end exponent space straight x space is space space minus space straight pi over 2 space less or equal than space sin to the power of negative 1 end exponent straight x space less or equal than space straight pi over 2

The correct answer is: open square brackets fraction numerator 1 over denominator square root of 2 end fraction comma 1 close square brackets


     Given : 2 c o s to the power of negative 1 end exponent invisible function application x equals s i n to the power of negative 1 end exponent invisible function application open parentheses 2 x square root of 1 minus x to the power of 2 end exponent end root close parentheses
    To find : Values of x satisfying the equation
    Detailed Solution
    LHS
    2 c o s to the power of negative 1 end exponent invisible function application x

L e t space cos to the power of negative 1 end exponent space x space equals space theta space o r space space x space equals space cos theta space

W e space k n o w space t h a t space space r a n g e space o f space cos to the power of negative 1 end exponent space x space i s space space space 0 space less or equal than space cos to the power of negative 1 end exponent x space less or equal than space straight pi

Range space of space 2 space cos to the power of negative 1 end exponent space straight x space is space space space 0 space less or equal than space 2 cos to the power of negative 1 end exponent straight x space less or equal than space 2 straight pi

rightwards double arrow 0 space less or equal than space 2 straight theta space less or equal than space 2 straight pi space space space space left parenthesis space cos to the power of negative 1 end exponent space straight x space equals space straight theta space right parenthesis space space..................... space left parenthesis 1 right parenthesis

Now space in space RHS

sin to the power of negative 1 end exponent left parenthesis 2 straight x square root of 1 minus straight x squared end root right parenthesis

Substitute space straight x space equals space cosθ space in space above space equation

rightwards double arrow sin to the power of negative 1 end exponent left parenthesis 2 cosθ square root of 1 minus cos squared straight theta end root right parenthesis

We space know space that space space range space of space sin to the power of negative 1 end exponent space straight x space is space space space minus straight pi over 2 space less or equal than space sin to the power of negative 1 end exponent straight x space less or equal than space straight pi over 2

rightwards double arrow space minus straight pi over 2 space less or equal than space sin to the power of negative 1 end exponent left parenthesis 2 cosθ square root of 1 minus cos squared straight theta end root right parenthesis space less or equal than space straight pi over 2

rightwards double arrow space minus straight pi over 2 space less or equal than space sin to the power of negative 1 end exponent left parenthesis 2 cosθsinθ right parenthesis space less or equal than space straight pi over 2 space space space space space space left parenthesis space sin squared straight theta space plus space cos squared straight theta space equals space space 1 right parenthesis

rightwards double arrow space minus straight pi over 2 space less or equal than space sin to the power of negative 1 end exponent left parenthesis sin 2 straight theta space right parenthesis space less or equal than space straight pi over 2 space space space space space space space space space space space space left parenthesis space 2 sinθcosθ space equals space space sin 2 straight theta right parenthesis

rightwards double arrow space minus straight pi over 2 space less or equal than space 2 straight theta space less or equal than space straight pi over 2 space space space space space space space space space space space space space space space space space space space space space space left parenthesis space space sin to the power of negative 1 end exponent left parenthesis sinθ space right parenthesis space equals space space straight theta right parenthesis space........ space left parenthesis 2 right parenthesis space space space space space space

From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis

rightwards double arrow 0 space less or equal than space 2 straight theta space less or equal than space straight pi over 2 space space

From space above space range space of space space space straight theta space space will space be space colon

rightwards double arrow 0 space less or equal than space straight theta space less or equal than space straight pi over 4 space space space
From space above space range space of space space cos space straight theta space space will space be space colon

rightwards double arrow cos space 0 space less or equal than cos space straight theta space less or equal than space cos straight pi over 4 space space space

rightwards double arrow 1 space less or equal than cos space straight theta space less or equal than space fraction numerator 1 over denominator square root of 2 end fraction space

rightwards double arrow space fraction numerator 1 over denominator square root of 2 end fraction space space less or equal than space straight x space space less or equal than space 1 space space S i n c e space cos to the power of negative 1 end exponent space x space i s space a space d e c r e a sin g space f u n c t i o n space s o comma space i n e q u a l i t y space w i l l space g e t space r e v e r s e d.
Thus comma space the space value space of space straight x space satisfying space the space equation space will space be space space fraction numerator 1 over denominator square root of 2 end fraction space space less or equal than space straight x space space less or equal than space 1 space

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