Let S be the set of all real numbers. A relation R has been defin

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 Multiple Choice QuestionsMultiple Choice Questions

291.

In the group G = {1, 5, 7, 11} under multiplication modulo 12, the solution of 7- 1 12 x 12 11 = 5 is equals :

  • 5

  • 1

  • 7

  • 11


292.

If f : R  R is defined by f(x) = lxl, then

  • f-1(x) = - x

  • f-11x

  • the function f-1(x) does not exist

  • f-1x = 1x


293.

On the set of all natural numbers N, which one of the following * is a binary operation?

  • a * b = ab

  • a * b = a - ba + b

  • a * b = a + 3b

  • a * b = 3a - 4b


294.

On the set of integers Z, define f : Z  Z as f(n) = n2, n is even0,  n is odd, then 'f' is

  • injective but not surjective

  • neither injective nor surjective

  • surjective but not injective

  • bijective


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295.

The inverse of 2010 in the group Q* of all positive rational under the binary operation * defined by a * b = ab2010, a, b  Q+ is

  • 2009

  • 2011

  • 1

  • 2010


296.

Define a relation R on A = {1, 2, 3, 4} as xRy if x divides y. R is

  • reflexive and transitive

  • reflexive and symmetric

  • symmetric and transitive

  • equivalence


297.

On the set of all non-zero reals, an operation * is defined as a * b = 3ab2. In this group, a solution of (2 * x) * 3-1 = 4-1 is

  • 6

  • 1

  • 1/6

  • 3/2


298.

If A and B have n elements in common, then the numberofelements common to A x B and B x A is

  • n

  • 2n

  • n2

  • 0


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299.

Which of the following is false ?

  • (N, *) is a group

  • (N, +) is a semi-group

  • (Z, +) is a group

  • Set of even integers is a group under usual addition


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300.

Let S be the set of all real numbers. A relation R has been defined on S by aRb  a - b  1, then R is

  • symmetric and transitive but not reflexive

  • reflexive and transitive but not symmetrIc

  • reflexive and symmetric but not transitive

  • an equivalence relation


C.

reflexive and symmetric but not transitive

Given, aRb  a - b  1For ReflexiveaRa = a - a =  1So, it is reflexive.For SymmetricaRb  a - b  1         b - a  1i.e.,             aRb  bRaSo, it is symmetric.ForTransitiveTake a = 1, b = 2 and c = 3Now, a - b = 1 - 2 = 1and    b - c = 2 - 3 = 1But     a - c = 1 - 3 = 2 > 1, which is not true.So, it is not transitive.


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