Maths-
General
Easy

Question

The area (in square units)bounded by the curves y=square root of x comma blank 2 y minus x plus 3 equals 0 comma blank x minusa x is, and lying in the first quadrant is

  1. 36    
  2. 18    
  3. 27/4    
  4. 9    

The correct answer is: 9


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    Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius  2 (inradius)

       

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    The complete solution set of the equation open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar is


    table row cell open parentheses x to the power of 2 end exponent minus x close parentheses left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x element of left parenthesis negative infinity comma negative 3 right square bracket union left square bracket 0 , 1 right square bracket end cell end table

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    table row cell open parentheses x to the power of 2 end exponent minus x close parentheses left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x element of left parenthesis negative infinity comma negative 3 right square bracket union left square bracket 0 , 1 right square bracket end cell end table
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    The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0,straight pi ] is/are –

    space space space space sin space 5 x. space cos space 3 x space equals sin space 6 x. space cos space 2 x
rightwards double arrow 1 half open parentheses sin space 8 x plus sin space 2 x close parentheses equals 1 half open parentheses sin space 8 x plus sin space 4 x close parentheses
rightwards double arrow sin space 2 x space minus sin space 4 x space equals 0
rightwards double arrow sin left parenthesis 3 x minus x right parenthesis minus sin left parenthesis 3 x plus x right parenthesis equals 0
rightwards double arrow negative 2 space sin space x. space cos space 3 x space equals 0
rightwards double arrow sin space x equals 0 space o r space cos space 3 x space equals 0
i. e. space x space equals n straight pi left parenthesis straight n element of straight I right parenthesis comma space or space 3 straight x equals 2 kπ plus-or-minus straight pi over 2 left parenthesis straight k element of straight k right parenthesis
So comma space straight x element of left square bracket 0 comma space straight pi right parenthesis
space then space given space equation space is space satisfied space of space straight x equals 0 comma space straight pi comma space straight pi over 6 comma space straight pi over 2 comma space fraction numerator 5 straight pi over denominator 6 end fraction.

    The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0,straight pi ] is/are –

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    space space space space sin space 5 x. space cos space 3 x space equals sin space 6 x. space cos space 2 x
rightwards double arrow 1 half open parentheses sin space 8 x plus sin space 2 x close parentheses equals 1 half open parentheses sin space 8 x plus sin space 4 x close parentheses
rightwards double arrow sin space 2 x space minus sin space 4 x space equals 0
rightwards double arrow sin left parenthesis 3 x minus x right parenthesis minus sin left parenthesis 3 x plus x right parenthesis equals 0
rightwards double arrow negative 2 space sin space x. space cos space 3 x space equals 0
rightwards double arrow sin space x equals 0 space o r space cos space 3 x space equals 0
i. e. space x space equals n straight pi left parenthesis straight n element of straight I right parenthesis comma space or space 3 straight x equals 2 kπ plus-or-minus straight pi over 2 left parenthesis straight k element of straight k right parenthesis
So comma space straight x element of left square bracket 0 comma space straight pi right parenthesis
space then space given space equation space is space satisfied space of space straight x equals 0 comma space straight pi comma space straight pi over 6 comma space straight pi over 2 comma space fraction numerator 5 straight pi over denominator 6 end fraction.
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    The equation 4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0 has

    sin to the power of 4 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application xsin invisible function application x plus 2 sin to the power of 2 end exponent invisible function application x plus sin invisible function application x equals 0
    table row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus 1 plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x close square brackets equals 0 end cell row cell s i n to the power of 2 end exponent invisible function application x equals 0 text end text text o end text text r end text text end text s i n to the power of 2 end exponent invisible function application x plus s i n invisible function application x plus 2 equals 0 end cell end table
    (not possible for real x)
    text or  end text sin invisible function application x equals 0
    Hence, the solutions are x = 0, p, 2p, 3p.

    The equation 4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0 has

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    sin to the power of 4 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application xsin invisible function application x plus 2 sin to the power of 2 end exponent invisible function application x plus sin invisible function application x equals 0
    table row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus 1 plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x close square brackets equals 0 end cell row cell s i n to the power of 2 end exponent invisible function application x equals 0 text end text text o end text text r end text text end text s i n to the power of 2 end exponent invisible function application x plus s i n invisible function application x plus 2 equals 0 end cell end table
    (not possible for real x)
    text or  end text sin invisible function application x equals 0
    Hence, the solutions are x = 0, p, 2p, 3p.
    General
    maths-

    s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x if

    table row cell s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x end cell row cell rightwards double arrow s i n to the power of 2 end exponent invisible function application x minus open parentheses 1 minus 2 s i n to the power of 2 end exponent invisible function application x close parentheses equals 2 minus 2 s i n invisible function application x c o s invisible function application x end cell row cell rightwards double arrow 3 s i n to the power of 2 end exponent invisible function application x plus 2 s i n invisible function application x c o s invisible function application x equals 3 end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text 1 end text text end text colon c o s invisible function application x not equal to 0 end cell row cell therefore 3 t a n to the power of 2 end exponent invisible function application x plus 2 t a n invisible function application x equals 3 open parentheses 1 plus t a n to the power of 2 end exponent invisible function application x close parentheses rightwards double arrow t a n invisible function application x equals fraction numerator 3 over denominator 2 end fraction end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text I end text text I end text text end text colon c o s invisible function application x equals 0 end cell row cell therefore 3 left parenthesis 1 right parenthesis plus 2 left parenthesis plus-or-minus 1 right parenthesis left parenthesis 0 right parenthesis equals 3 text end text text w end text text h end text text i end text text c end text text h end text text end text text i end text text s end text text end text text t end text text r end text text u end text text e end text text end text therefore x equals left parenthesis 2 n plus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell end table

    s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x if

    maths-General
    table row cell s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x end cell row cell rightwards double arrow s i n to the power of 2 end exponent invisible function application x minus open parentheses 1 minus 2 s i n to the power of 2 end exponent invisible function application x close parentheses equals 2 minus 2 s i n invisible function application x c o s invisible function application x end cell row cell rightwards double arrow 3 s i n to the power of 2 end exponent invisible function application x plus 2 s i n invisible function application x c o s invisible function application x equals 3 end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text 1 end text text end text colon c o s invisible function application x not equal to 0 end cell row cell therefore 3 t a n to the power of 2 end exponent invisible function application x plus 2 t a n invisible function application x equals 3 open parentheses 1 plus t a n to the power of 2 end exponent invisible function application x close parentheses rightwards double arrow t a n invisible function application x equals fraction numerator 3 over denominator 2 end fraction end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text I end text text I end text text end text colon c o s invisible function application x equals 0 end cell row cell therefore 3 left parenthesis 1 right parenthesis plus 2 left parenthesis plus-or-minus 1 right parenthesis left parenthesis 0 right parenthesis equals 3 text end text text w end text text h end text text i end text text c end text text h end text text end text text i end text text s end text text end text text t end text text r end text text u end text text e end text text end text therefore x equals left parenthesis 2 n plus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell end table
    parallel

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