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Maths-

General

Easy

Question

The area of circle circumscribing ΔABC is

$\frac{π}{8}$
$\frac{π}{4}$
$\frac{π}{2}$
$π$

The correct answer is: $pi over 4$

Related Questions to study

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)

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)

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

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

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 –

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.

parallel

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

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

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

parallel

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

Because of the greater size of Br, Br exert greater repulsive force. Hence Br in the equitorial position add maximum stability.

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

Because of the greater size of Br, Br exert greater repulsive force. Hence Br in the equitorial position add maximum stability.

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

parallel

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

6 A g N O subscript 3 end subscript plus A s H subscript 3 end subscript plus 3 H subscript 2 end subscript O ⟶

6 A g stack stack downwards arrow with _ below plus H subscript 3 end subscript A s O subscript 3 end subscript with _ below plus 6 H N O subscript 3 end subscript

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

6 A g N O subscript 3 end subscript plus A s H subscript 3 end subscript plus 3 H subscript 2 end subscript O ⟶

6 A g stack stack downwards arrow with _ below plus H subscript 3 end subscript A s O subscript 3 end subscript with _ below plus 6 H N O subscript 3 end subscript

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

Because of the partial positive charge on adjacent N-atoms there is a repulsion along N  N bond, hence N  N bond distance become higher.

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

Because of the partial positive charge on adjacent N-atoms there is a repulsion along N  N bond, hence N  N bond distance become higher.

parallel

If a,b,c,d are consective binomial coefficients of $left parenthesis 1 plus x right parenthesis to the power of n$ then $fraction numerator a plus b over denominator a end fraction comma fraction numerator b plus c over denominator b end fraction comma fraction numerator c plus d over denominator c end fraction$ are is

If a,b,c,d are consective binomial coefficients of $left parenthesis 1 plus x right parenthesis to the power of n$ then $fraction numerator a plus b over denominator a end fraction comma fraction numerator b plus c over denominator b end fraction comma fraction numerator c plus d over denominator c end fraction$ are is

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

B a open parentheses N subscript 3 end subscript close parentheses subscript 2 end subscript stack ⟶ with capital delta on top B a plus 3 N subscript 2 end subscript

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

B a open parentheses N subscript 3 end subscript close parentheses subscript 2 end subscript stack ⟶ with capital delta on top B a plus 3 N subscript 2 end subscript

If $T subscript 0 comma T subscript 1 comma T subscript 2 comma horizontal ellipsis. T subscript n$ represent the terms is $left parenthesis x plus a right parenthesis to the power of n$ then $left parenthesis T subscript 0 minus T subscript 2 plus T subscript 4 minus T subscript 6 plus midline horizontal ellipsis right parenthesis squared plus$ $left parenthesis T subscript 1 minus T subscript 3 plus T subscript 5 minus horizontal ellipsis right parenthesis squared$ is

If $T subscript 0 comma T subscript 1 comma T subscript 2 comma horizontal ellipsis. T subscript n$ represent the terms is $left parenthesis x plus a right parenthesis to the power of n$ then $left parenthesis T subscript 0 minus T subscript 2 plus T subscript 4 minus T subscript 6 plus midline horizontal ellipsis right parenthesis squared plus$ $left parenthesis T subscript 1 minus T subscript 3 plus T subscript 5 minus horizontal ellipsis right parenthesis squared$ is

parallel

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