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If D=|1 1 1 1 1+x1 1 1 1+y|"f o r"x!=0, ...

If `D=|1 1 1 1 1+x1 1 1 1+y|"f o r"x!=0, y!=0` then D is (1) divisible by neither x nor y (2) divisible by both x and y (3) divisible by x but not y (4) divisible by y but not x

Text Solution

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`D=[[1,1,1],[1,1+x,1],[1,1,1+y]]`
`=(1+x)(1+y)-1-(1+y)+1+1-(1+x)`
`=1+x+y+xy-1-1-y+1+1-1-x`
`=3-3+xy`
`=xy`
option`2`
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Knowledge Check

  • If D = |(1,1,1),(1,1 +x,1),(1,1,1 +y)| " for " x!= 0, y != 0 , then D is divisible by

    A
    x but not y
    B
    y but not x
    C
    neither x nor y
    D
    both x and y

  • If the six digit number 15 x 1y 2 is divisible by 44 then (x + y) is equal to :

    A
    A) 7
    B
    B) 8
    C
    C) 6
    D
    D) 9

  • If f(y)=|(a,-1,0),(ay,a,-1),(ay^(2),ay,a)| , then f(2y)-f(y) is divisible by

    A
    `y^(3)`
    B
    0
    C
    `2a+3y`
    D
    `y^(2)`

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    Prove the following by the principle of mathematical induction: x^(2n-1)+y^(2n-1) is divisible by x+y for all n in N.

    Using mathematical induction, prove that for x^(2n-1)+y^(2n-1) is divisible by x+y for all n in N

    Let X={1, 2, 3, 4, 5, 6, 7, 8, 9} , Let R_1 be a relation on X given by R_1={(x , y): x-y is divisible by 3} and R_2 be another relation on X given by R_2={(x , y):{x , y}sub{1, 4, 7} or {x , y}sub{2, 5, 8} or {x , y}sub{3, 6, 9}}dot Show that R_1=R_2 .

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