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Question

he equation $open parentheses cos invisible function application p minus 1 close parentheses x to the power of 2 end exponent plus open parentheses cos invisible function application p close parentheses x plus sin invisible function application p equals 0$ in the variable $x$ has real roots. Then $p$ can take any value in the interval

$3$
$\frac{8 \sqrt{3}}{3}$
$5$
$\frac{10 \sqrt{3}}{3}$

The correct answer is: $fraction numerator 10 square root of 3 over denominator 3 end fraction$

$375 blank$ In triangle $A B C comma blank angle A B C equals 120 degree comma blank A B equals 3 blank$ and $blank B C equals 4.$ If perpendicular constructed to the side $A B blank$ at $blank A$ and to the side $B C blank$ at $blank C$ meets at $D$ , then $blank C D$ is equal to
Note that $increment P C D$ is right angle as shown in the figure
Now $blank tan invisible function application 30 degree equals fraction numerator x over denominator 10 end fraction$

$increment P A B blank$ is $blank 30 degree minus 60 degree minus 90 degree blank$ triangle, hence in $increment P A B comma blank P B equals fraction numerator 3 over denominator sin invisible function application 30 degree end fraction equals 6$
Therefore, in $blank increment P C D comma blank x equals 10 tan invisible function application 30 degree equals 10 divided by square root of 3 equals open parentheses 10 square root of 3 close parentheses divided by 3$
$open parentheses cos invisible function application p minus 1 close parentheses x to the power of 2 end exponent plus open parentheses cos invisible function application p close parentheses x plus sin invisible function application p equals 0$
For this equation to have real roots $D greater or equal than 0$
$rightwards double arrow cos to the power of 2 end exponent invisible function application p minus 4 sin invisible function application p blank open parentheses cos invisible function application p minus 1 close parentheses greater or equal than 0$
$rightwards double arrow cos to the power of 2 end exponent invisible function application p minus 4 sin invisible function application p cos invisible function application p plus 4 sin to the power of 2 end exponent invisible function application p plus 4 sin invisible function application p minus 4 sin to the power of 2 end exponent invisible function application p greater or equal than 0$
$rightwards double arrow open parentheses cos invisible function application p minus 2 sin invisible function application p close parentheses to the power of 2 end exponent plus 4 sin invisible function application p blank open parentheses 1 minus sin invisible function application p close parentheses greater or equal than 0$
For every real value of $p$ , we have
$open parentheses cos invisible function application p minus 2 sin invisible function application p close parentheses to the power of 2 end exponent greater or equal than 0$ and $sin invisible function application p left parenthesis 1 minus sin invisible function application p right parenthesis greater or equal than 0$
$therefore D greater or equal than 0 comma blank for all sin invisible function application p element of open parentheses 0 comma blank pi close parentheses$

Related Questions to study

he value of expression $square root of 3 c o s e c blank 20 degree minus sec invisible function application 20 degree$ is equal to

he value of expression $square root of 3 c o s e c blank 20 degree minus sec invisible function application 20 degree$ is equal to

$f cos invisible function application x equals fraction numerator 2 cos invisible function application y minus 1 over denominator 2 minus cos invisible function application y end fraction comma blank w h e r e blank x comma blank y element of open parentheses 0 comma blank pi close parentheses comma blank t h e n tan invisible function application fraction numerator x over denominator 2 end fraction cot invisible function application fraction numerator y over denominator 2 end fraction blank i s blank e q u a l blank t o$

$f cos invisible function application x equals fraction numerator 2 cos invisible function application y minus 1 over denominator 2 minus cos invisible function application y end fraction comma blank w h e r e blank x comma blank y element of open parentheses 0 comma blank pi close parentheses comma blank t h e n tan invisible function application fraction numerator x over denominator 2 end fraction cot invisible function application fraction numerator y over denominator 2 end fraction blank i s blank e q u a l blank t o$

he value of $cos invisible function application fraction numerator pi over denominator 7 end fraction plus cos invisible function application fraction numerator 2 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 3 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 4 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 5 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 6 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 7 pi over denominator 7 end fraction$ is

he value of $cos invisible function application fraction numerator pi over denominator 7 end fraction plus cos invisible function application fraction numerator 2 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 3 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 4 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 5 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 6 pi over denominator 7 end fraction plus cos invisible function application fraction numerator 7 pi over denominator 7 end fraction$ is

parallel

f $alpha plus beta equals pi divided by 2$ and $beta plus gamma equals alpha comma$ then $tan invisible function application alpha blank e q u a l s$

f $alpha plus beta equals pi divided by 2$ and $beta plus gamma equals alpha comma$ then $tan invisible function application alpha blank e q u a l s$

he smallest $plus v e blank x$ satisfying the equation $log subscript cos invisible function application x end subscript invisible function application sin invisible function application x plus log subscript sin invisible function application x end subscript invisible function application cos invisible function application x equals 2$ is

he smallest $plus v e blank x$ satisfying the equation $log subscript cos invisible function application x end subscript invisible function application sin invisible function application x plus log subscript sin invisible function application x end subscript invisible function application cos invisible function application x equals 2$ is

f $log subscript 4 end subscript invisible function application 5 equals a blank a n d log subscript 5 end subscript invisible function application 6 equals b comma$ then $log subscript 3 end subscript invisible function application 2$ is equal to

f $log subscript 4 end subscript invisible function application 5 equals a blank a n d log subscript 5 end subscript invisible function application 6 equals b comma$ then $log subscript 3 end subscript invisible function application 2$ is equal to

parallel

f in $increment A B C comma blank 8 R to the power of 2 end exponent equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent comma$ then the triangle $A B C$ is

f in $increment A B C comma blank 8 R to the power of 2 end exponent equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent comma$ then the triangle $A B C$ is

The total number of solutions of $open vertical bar cot invisible function application x close vertical bar equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction comma blank x element of open square brackets 0 comma blank 3 pi close square brackets$ is equal to

The total number of solutions of $open vertical bar cot invisible function application x close vertical bar equals cot invisible function application x plus fraction numerator 1 over denominator sin invisible function application x end fraction comma blank x element of open square brackets 0 comma blank 3 pi close square brackets$ is equal to

If the inequality $sin to the power of 2 end exponent invisible function application x plus a cos invisible function application x plus a to the power of 2 end exponent greater than 1 plus cos invisible function application x$ holds for any $x element of R comma$ then the largest negative integral value of $a$ is

If the inequality $sin to the power of 2 end exponent invisible function application x plus a cos invisible function application x plus a to the power of 2 end exponent greater than 1 plus cos invisible function application x$ holds for any $x element of R comma$ then the largest negative integral value of $a$ is

parallel

$fraction numerator sin invisible function application 3 theta plus sin invisible function application 5 theta plus sin invisible function application 7 theta plus sin invisible function application 9 theta over denominator cos invisible function application 3 theta plus cos invisible function application 5 theta plus cos invisible function application 7 theta plus cos invisible function application 9 theta end fraction blank i s blank e q u a l blank t o$

$fraction numerator sin invisible function application 3 theta plus sin invisible function application 5 theta plus sin invisible function application 7 theta plus sin invisible function application 9 theta over denominator cos invisible function application 3 theta plus cos invisible function application 5 theta plus cos invisible function application 7 theta plus cos invisible function application 9 theta end fraction blank i s blank e q u a l blank t o$

If $a to the power of 4 end exponent blank. b to the power of 5 end exponent equals 1 comma$ then the value of $log subscript a end subscript invisible function application open parentheses a to the power of 5 end exponent b to the power of 4 end exponent close parentheses$ equals

If $a to the power of 4 end exponent blank. b to the power of 5 end exponent equals 1 comma$ then the value of $log subscript a end subscript invisible function application open parentheses a to the power of 5 end exponent b to the power of 4 end exponent close parentheses$ equals

If $open parentheses x plus 1 close parentheses to the power of log subscript 10 end subscript invisible function application open parentheses x plus 1 close parentheses end exponent equals 100 open parentheses x plus 1 close parentheses comma$ then

If $open parentheses x plus 1 close parentheses to the power of log subscript 10 end subscript invisible function application open parentheses x plus 1 close parentheses end exponent equals 100 open parentheses x plus 1 close parentheses comma$ then

parallel

The sum of all the solution of $cot invisible function application theta equals sin invisible function application 2 theta comma blank open parentheses theta not equal to n pi comma n blank i n t e g e r close parentheses comma blank 0 less or equal than theta less or equal than pi blank$ is

The sum of all the solution of $cot invisible function application theta equals sin invisible function application 2 theta comma blank open parentheses theta not equal to n pi comma n blank i n t e g e r close parentheses comma blank 0 less or equal than theta less or equal than pi blank$ is

The number of solutions of the equation $sin to the power of 3 end exponent invisible function application x cos invisible function application x plus sin to the power of 2 end exponent invisible function application x cos to the power of 2 end exponent invisible function application x plus sin invisible function application x cos to the power of 3 end exponent invisible function application x equals 1 comma blank$ in the interval $open square brackets 0 comma blank 2 pi close square brackets comma blank$ is

The number of solutions of the equation $sin to the power of 3 end exponent invisible function application x cos invisible function application x plus sin to the power of 2 end exponent invisible function application x cos to the power of 2 end exponent invisible function application x plus sin invisible function application x cos to the power of 3 end exponent invisible function application x equals 1 comma blank$ in the interval $open square brackets 0 comma blank 2 pi close square brackets comma blank$ is

If $a comma blank b blank$ and $blank c$ are the sides of a triangle, then the minimum value of $fraction numerator 2 a over denominator b plus c minus a end fraction plus fraction numerator 2 b over denominator c plus a minus b end fraction plus fraction numerator 2 c over denominator a plus b minus c end fraction$ is

If $a comma blank b blank$ and $blank c$ are the sides of a triangle, then the minimum value of $fraction numerator 2 a over denominator b plus c minus a end fraction plus fraction numerator 2 b over denominator c plus a minus b end fraction plus fraction numerator 2 c over denominator a plus b minus c end fraction$ is

parallel

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