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 Multiple Choice QuestionsMultiple Choice Questions

1.

If the function g(x) =open curly brackets table attributes columnalign left end attributes row cell straight k square root of straight x plus 1 end root comma space space 0 space less or equal than straight x space less or equal than 3 end cell row cell mx plus 2 space comma space 3 space less than straight x less or equal than 5 end cell end table close is differntiable, then the value of k+m is

  • 2

  • 16/5

  • 10/3

  • 10/3

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2.

If a ε R and the equation - 3(x-[x]2 + 2(x-[x] +a2 = 0(where,[x] denotes the greatest integer ≤ x) has no  integral solution, then all possible value of  lie in the interval

  • (-1,0) ∪ (0,1)

  • (1,2)

  • (-2,-1)

  • (-∞,-2) ∪ (2, ∞)

182 Views

3.

Consider the following relations:
R = {(x, y)| x, y are real numbers and x = wy for some rational number w}; S = {(m/p, p/q)| m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then

  • R is an equivalence relation but S is not an equivalence relation

  • neither R nor S is an equivalence relation

  • S is an equivalence relation but R is not an equivalence relation

  • S is an equivalence relation but R is not an equivalence relation

308 Views

4.

Let p(x) be a function defined on R such that limit as straight x space rightwards arrow infinity of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction = 1, p'(x)  p'(1-x),for all x∈[0,1] p(0) = 1 and p(1) = 41. Then integral subscript 0 superscript straight x space straight p space left parenthesis straight x right parenthesis space dx equals

  • √41

  • 21

  • 41

  • 41

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5.

The function straight f colon space straight R space rightwards arrow with space on top space open square brackets negative 1 half comma 1 half close square brackets space defined space as space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator 1 plus space straight x squared end fraction comma space is

  • neither injective nor surjective.

  • invertible

  • injective but not surjective.

  • injective but not surjective.


D.

injective but not surjective.



straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator 1 space plus space straight x squared end fraction semicolon space straight f colon space straight R space rightwards arrow open square brackets negative 1 half comma 1 half close square brackets
From above diagram of f(x), f(x) is surjective but not injective.
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6.

Let W denote the words in the English dictionary. Define the relation R by :
R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is

  • not reflexive, symmetric and transitive

  • reflexive, symmetric and not transitive

  • reflexive, symmetric and transitive

  • reflexive, symmetric and transitive

141 Views

7.

The graph of the function y = f(x) is symmetrical about the line x = 2, then

  • f(x + 2)= f(x – 2)

  • f(2 + x) = f(2 – x)

  • f(x) = f(-x)

  • f(x) = f(-x)

219 Views

8.

The domain of the function straight f left parenthesis straight x right parenthesis space equals space fraction numerator sin to the power of negative 1 end exponent space left parenthesis straight x minus 3 right parenthesis over denominator square root of 9 minus straight x squared end root end fraction

  • [2, 3]

  • [2, 3)

  • [1, 2]

  • [1, 2]

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9.

x  R: cosx  sinx  0, 3π2 is equal to

  • 0, π4  3π4, 3π2

  • 0, π4  π2, 3π2

  • 0, π4  5π4, 3π2

  • 0, 3π2


10.

If log0.2x - 1 > log0.04x + 5, then

  • - 1 < x < 4

  • 2 < x < 3

  • 1 < x < 4

  • 1 < x < 3


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