The complete solution set of the equation |x^(2)-x|+|x+3|=|x^(2-Turito

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Maths-

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

Maths-General

$[1, \infty)$
$(- \infty, - 3] \cup [0, 1]$
$[- 3, 0] \cup [1, \infty)$
$(- \infty, - 3]$

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

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The 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

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.

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

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

The area of the region between the curve $blank x to the power of 2 end exponent plus y to the power of 2 end exponent equals$ 4 and x = 0; x =1 is….

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$fraction numerator 3 x squared plus x plus 1 over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction equals fraction numerator a over denominator left parenthesis x minus 1 right parenthesis end fraction plus fraction numerator b over denominator left parenthesis x minus 1 right parenthesis squared end fraction plus fraction numerator c over denominator left parenthesis x minus 1 right parenthesis cubed end fraction plus fraction numerator d over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction text then end text open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a b end cell row cell c d end cell end table close square brackets equals$

$fraction numerator 3 x squared plus x plus 1 over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction equals fraction numerator a over denominator left parenthesis x minus 1 right parenthesis end fraction plus fraction numerator b over denominator left parenthesis x minus 1 right parenthesis squared end fraction plus fraction numerator c over denominator left parenthesis x minus 1 right parenthesis cubed end fraction plus fraction numerator d over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction text then end text open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a b end cell row cell c d end cell end table close square brackets equals$