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Let A = ( [ [ x + 1 ] [ x + 2 ] [ x + 2 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] ] )  where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval.

Option 1 -

[60, 61)

Option 2 -

[68, 69)

Option 3 -

[62, 63)

Option 4 -

[65, 66)

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    | A | = | [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] | = | [ x ] + 1 [ x ] + 2 [ x ] + 3 [ x ] [ x ] + 3 [ x ] + 3 [ x ] [ x ] + 2 [ x ] + 4 |

    R 1 → R 1 − R 3 & R 2 → R 2 − R 3

    | 1 0 − 1 0 1 − 1 [ x ] [ x ] + 2 [ x ] + 4 | = 1 9 2

    ⇒ [ x ] = 6 2

    ⇒ x ∈ [ 6 2 , 6 3 )              

     

Similar Questions for you

If the matrix A = [ 1 0 0 0 2 0 3 0 − 1 ]  satisfies the equation A20 = aA19 + βA = [ 1 0 0 0 4 0 0 0 1 ] for some real numbers a and b, then b - a is equal to…………
A
alok kumar singh

  A 2 = [ 1 0 0 0 2 0 3 0 − 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 2 0 0 0 1 ]

A 3 = [ 1 0 0 0 2 2 0 0 0 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 3 0 3 0 − 1 ]        

A 4 [ 1 0 0 0 2 3 0 3 0 − 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 4 0 0 0 1 ]          

Similarly we get A19 =   = [ 1 0 0 0 2 1 9 0 3 0 − 1 ] & A 2 0 = [ 1 0 0 0 2 2 0 0 0 0 1 ]

=   [ 1 0 0 0 4 0 0 0 1 ]

⇒ 1 + α + β = 1 g i v e s     α + β = 0 . . . . . . . . ( i )        

  2 2 0 + ( 2 1 9 − 2 ) α = 4 f r o m ( i )

⇒ α = 4 − 2 2 0 2 1 9 − 2 = 4 ( 1 − 2 1 8 ) − 2 ( 1 − 2 1 8 ) = − 2          

So, b = 2

Hence b - a = 4

 

Consider the following system of equations:

x + 2y – 3z = a                     

2x + 6y – 11z = b

x – 2y + 7z = c,

where a, b and c are real constants. Then the system of equations:
A
alok kumar singh

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here    Δ = | 1 2 − 3 2 6 − 1 1 1 − 2 7 | = ( 4 2 − 2 2 ) − 2 ( 1 4 + 1 1 ) − 3 ( − 4 − 6 ) = 2 0 − 5 0 + 3 0 = 0                                                                                

For infinite solution 

20a – 8b – 4c = 0 Þ 5a = 2b + c

Let A = { n ∈ N : H . C . F . ( n , 4 5 ) = 1 } and Let B = { 2 k : k ∈ { 1 ,   2 , . . . . . . ,   1 0 0 } } . Then the sum of all the elements of   A ∪ B     i s . . . . . . . . . .  
R
Raj Pandey

Sum of all elements of A ∩ B = 2   [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

= 2 [ 1 0 0 × 1 0 1 2 − 3 ( 3 3 × 3 4 2 ) − 5 ( 2 0 × 2 1 2 ) + 1 5 ( 6 × 7 2 ) ]

= 10100 – 3366 – 2100 + 630

              = 5264

Let  A = ( 2 − 1 0 2 ) . If B = I − 5 ​ C 1 ( a d j A ) + 5 ​ C 2 ( a d j A ) 2 − . . . 5 ​ C 5 ( a d j A ) 5 ,  then the sum of all elements of the matrix B is
A
alok kumar singh

Kindly go through the solution

B = (I – adjA)5

Let M = [ 0 − α α 0 ] , where a is a non-zero real number an N = ∑ k = 1 4 9 M 2 k . If (I – M2)N = -2I, then the positive integral value of a is…………….
A
alok kumar singh

N = M 2 + M 4 + . . . . . + M 9 8

= ( − α 2     I ) + ( − α 2     I ) 2 + . . . . + ( − α 2     I ) 4 9

= I ( − α 2 + α 4 − α 6 + . . . . − α 9 8 )

N =  − I ( α 2 − α 4 + α 6 . . . . . . . + α 9 8 )

= − I α 2 ( 1 − ( − α 2 ) 4 9 ) 1 − ( − α 2 )

N =  − I   α 2 ( 1 + α 9 8 ) 1 + α 2  

Now  ( I − m 2 ) N = − 2 I

( I + α 2 I ) ( − I α 2 ⋅ ( 1 + α 9 8 ) 1 + α 2 = − 2 I  

-> a100 + a2 = 2

->a = ± 1

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