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Maths-

General

Easy

Question

If [x] stands for the greatest integer less or equal to x, then in order than the set of equations x – 3y = 4 ; 5x + y = 2 ; [2 $pi$ ]x – [e]y = [2a] may be consistent, then ‘a’ should lie in -

$[3, \frac{7}{2})$
$(3, \frac{7}{3})$
$(3, \frac{7}{3}]$
None of these

The correct answer is: $open square brackets 3 comma fraction numerator 7 over denominator 2 end fraction close parentheses$

On solving x – 3y = 4 and 5x + y = 2, we get x = $fraction numerator 5 over denominator 8 end fraction$ , y = – $fraction numerator 9 over denominator 8 end fraction$ . As [2 $pi$ ] = 6 and [e] = 2.
So that the three equations are consistence if
This gives 6 $less or equal than$ 2a < 7 or 3 $less or equal than$ a < $fraction numerator 7 over denominator 2 end fraction$ .
Hence (A) is correct answer.

Related Questions to study

If the matrix $open square brackets table row 0 cell 2 beta end cell gamma row alpha beta cell negative gamma end cell row alpha cell negative beta end cell gamma end table close square brackets$ is orthogonal, then -

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If the system of equations x + 2y – 3z = 2, (k + 3)z = 3, (2k + 1)y + z = 2 is inconsistent, then value of k can be -

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If A is a 3 × 3 skew-symmetric matrix, then |A| is given by -

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The solubility of iodine in water is greatly increased by:

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chemistry-General

Let A = $open square brackets table row 2 3 row cell negative 1 end cell 5 end table close square brackets$ if A^–1 = xA + yI₂ then x and y are respectively -

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The value of a for which the system of equationsx + y + z = 0ax + (a + 1)y + (a + 2)z = 0 a³x + (a + 1)³y + (a + 2)³z = 0has a non-zero solution is -

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If A is non-singular matrix of order 3 × 3, then adj (adj A) is equal to -

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If A = $open square brackets table row a b c row x y z row p q r end table close square brackets$ , B = $open square brackets table row q cell negative b end cell y row cell negative p end cell a cell negative x end cell row r cell negative c end cell z end table close square brackets$ then -

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Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is -

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parallel

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

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$C l subscript 2 space i s space u s e d space i n space t h e space e x t r a c t i o n space o f colon$

chemistry-General

parallel

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