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Question

Consider the cubic equation $x cubed minus left parenthesis 1 plus cos space theta plus sin space theta right parenthesis x squared plus left parenthesis cos space theta sin space theta plus cos space theta plus sin space theta right parenthesis x minus sin space theta cos space theta equals 0$ whose roots are x₁, x₂, and x₃
The value of $x subscript 1 superscript 2 plus x subscript 2 superscript 2 plus x subscript 3 superscript 2$ equals

1
2
2 cos $θ$
$\sin θ (\sin θ + \cos θ)$

hint Hint:

In this question, we have given a cubic equation. which is $x cubed minus left parenthesis 1 plus c o s theta plus s i n theta right parenthesis x squared plus left parenthesis c o s theta s i n theta plus c o s theta plus s i n theta right parenthesis x minus s i n theta c o s theta equals 0$ .
We have given roots is $x 1 comma space x 2 comma x 3 space$ . We have to find the $x 1 squared plus x 2 squared plus x 3 squared$ . For that use the do $left parenthesis x 1 plus x 2 plus x 3 right parenthesis squared$ and find the product of root and sum of root then substitute them into it.

The correct answer is: 2

Here we have to find the $x 1 squared plus x 2 squared plus x 3 squared$ .
We know that
$left parenthesis x 1 plus x 2 plus x 3 right parenthesis squared equals x 1 squared plus x 2 squared plus x 3 squared plus 2 left parenthesis x 1 x 2 plus x 2 x 3 plus x 3 x 1 right parenthesis$
We have cubic equation that is ,
$x cubed minus left parenthesis 1 plus c o s theta plus s i n theta right parenthesis x squared plus left parenthesis c o s theta s i n theta plus c o s theta plus s i n theta right parenthesis x – s i n theta. c o s theta equals 0$ In this equation,
The sum of root = $negative b over a$
$left parenthesis space x 1 space plus space x 2 space plus space x 3 space right parenthesis space equals space 1 plus cos theta plus sin theta space minus negative negative negative negative left parenthesis 1 right parenthesis$
And product of roots = $c over a$
$x 1 x 2 space plus space x 2 x 3 space plus space x 3 x 1 space equals space cos theta space sin theta space plus space cos theta space plus space sin theta space minus negative negative negative negative negative left parenthesis 2 right parenthesis$
we have,
$left parenthesis x 1 plus x 2 plus x 3 right parenthesis squared equals x 1 squared plus x 2 squared plus x 3 squared plus 2 left parenthesis x 1 x 2 plus x 2 x 3 plus x 3 x 1 right parenthesis$
$equals greater than left parenthesis 1 plus c o s theta plus s i n theta right parenthesis squared equals x 1 squared plus x 2 squared plus x 3 squared plus 2 left parenthesis c o s theta s i n theta plus c o s theta plus s i n theta right parenthesis$
$equals greater than 1 plus c o s squared theta plus s i n squared theta plus 2 c o s theta plus 2 c o s theta. s i n theta plus 2 s i n theta equals x 1 squared plus x 2 squared plus x 3 squared plus 2 c o s theta s i n theta plus 2 c o s theta plus 2 s i n theta$
$equals greater than x 1 squared plus x 2 squared plus x 3 squared equals 1 plus left parenthesis c o s squared theta plus s i n squared theta right parenthesis plus left parenthesis 2 c o s theta plus 2 c o s theta. s i n theta plus 2 s i n theta right parenthesis – left parenthesis 2 c o s theta s i n theta plus 2 c o s theta plus 2 s i n theta right parenthesis$
$equals greater than x 1 squared plus x 2 squared plus x 3 squared equals 1 plus 1 plus 0$
$equals greater than x 1 squared plus x 2 squared plus x 3 squared equals 2$
Therefore, the value of $x 1 squared plus x 2 squared plus x 3 squared i s 2.$
The correct answer is 2.

In this question, we have to given the cubic equation and its root x1,x2,x3 and we have to find the
$x 1 squared plus x 2 squared plus x 3 squared$ .Use, $left parenthesis x 1 plus x 2 plus x 3 right parenthesis squared equals x 1 squared plus x 2 squared plus x 3 squared plus 2 left parenthesis x 1 x 2 plus x 2 x 3 plus x 3 x 1 right parenthesis$ and sum of roots is $fraction numerator negative c over denominator a end fraction$ and product of roots is b/a.

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The ratio of the gases obtained on dehydration ofHCOOH and $H subscript 2 C subscript 2 O subscript 4 space b y space c o n c. H subscript 2 S O subscript 4$

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If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

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If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

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If A, B, C, are three matrices, then A^T + B^T + C^T is -

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If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

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If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

Maths-General

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Consider the cubic equation whose roots are x1, x2, and x3The value of equals

122 cos

In this question, we have given a cubic equation. which is.We have given roots is . We have to find the . For that use the do and find the product of root and sum of root then substitute them into it.

The correct answer is: 2

Here we have to find the .We know that We have cubic equation that is ,In this equation,The sum of root = And product of roots = we have,Therefore, the value of The correct answer is 2.

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1
2
2 cos $θ$
$\sin θ (\sin θ + \cos θ)$

If A is a square matrix, then A– $straight A to the power of straight prime$ is -

If A is a square matrix, then A– $straight A to the power of straight prime$ is -

If A is symmetric matrix and B is a skew- symmetric matrix, then for n $element of$ N, false statement is -

If A is symmetric matrix and B is a skew- symmetric matrix, then for n $element of$ N, false statement is -

If A $negative A to the power of straight prime$ = 0, then $straight A to the power of straight prime$ is -

If A $negative A to the power of straight prime$ = 0, then $straight A to the power of straight prime$ is -

If A is a matrix of order 3 × 4, then both AB^Tand B^TA are defined if order of B is -

If A is a matrix of order 3 × 4, then both AB^Tand B^TA are defined if order of B is -

The ratio of the gases obtained on dehydration ofHCOOH and $H subscript 2 C subscript 2 O subscript 4 space b y space c o n c. H subscript 2 S O subscript 4$

The ratio of the gases obtained on dehydration ofHCOOH and $H subscript 2 C subscript 2 O subscript 4 space b y space c o n c. H subscript 2 S O subscript 4$

If A = $open square brackets table row 3 x row y 0 end table close square brackets$ and A = A^T, then -

If A = $open square brackets table row 3 x row y 0 end table close square brackets$ and A = A^T, then -

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row a b row b a end table close square brackets$ , then |A +A^T| equals -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A = $open square brackets table row 3 2 row 1 4 end table close square brackets$ and B = $open square brackets table row cell negative 1 end cell 2 row cell negative 1 end cell 1 end table close square brackets$ , then correct statement is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A, B, C, are three matrices, then A^T + B^T + C^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A and B are matrices of order m × n and n × m respectively, then the order of matrix B^T (A^T)^T is -

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is-