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Class 12 Maths
Relations And Functions
Quiz 3
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The relation R is defined on the set of natural numbers as {(a, b): a = 2b}. Then, R is given by
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{(2, 1), (4, 2), (6, 3),….}
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{(1, 2), (2, 4), (3, 6),….}
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R is not defined
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None of these
Explanation
{(1, 2), (2, 4), (3, 6),….}
The relation R = {(1,1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is
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Reflexive but not symmetric
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Reflexive but not transitive
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Symmetric and transitive
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Neither symmetric nor transitive
Explanation
Reflexive but not symmetric
Let P = {(x, y) | x² + y² = 1, x, y ∈ R]. Then, P is
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Reflexive
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Symmetric
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Transitive
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Anti-symmetric
Explanation
Symmetric
Let R be an equivalence relation on a finite set A having n elements. Then, the number of ordered pairs in R is
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Less than n
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Greater than or equal to n
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Less than or equal to n
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None of these
Explanation
Greater than or equal to n
For real numbers x and y, we write xRy ⇔ x – y + √2 is an irrational number. Then, the relational R is
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Reflexive
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Symmetric
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Transitive
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None of these
Explanation
Reflexive
Let R be a relation on the set N be defined by {(x, y) | x, y ∈ N, 2x + y = 41}. Then R is
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Reflexive
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Symmetric
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Transitive
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None of these
Explanation
None of these
Which one of the following relations on R is an equivalence relation?
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aR b ⇔ |a| = |b|
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aR b ⇔ a ≥ b
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aR b ⇔ a divides b
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aR b ⇔ a < b
Explanation
aR b ⇔ |a| = |b|
Let R be a relation on the set N of natural numbers denoted by nRm ⇔ n is a factor of m (i.e. n | m). Then, R is
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Reflexive and symmetric
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Transitive and symmetric
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Equivalence
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Reflexive, transitive but not symmetric
Explanation
Reflexive, transitive but not symmetric
A relation R in S= {1,2,3} is defined as R= {(1, 1), (2, 3), (2, 2), (3, 3)}. Which element(s) of relation R be removed to make R an equivalence relation?
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(1,1)
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(2,2)
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(3,3)
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(2,3)
If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is
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720
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120
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0
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100
R is a relation from {11,12,13} to {8,10,12} defined by y=x-3, then $R^{-1} is
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{(8,11), (10,13)}
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{(11,8), (13,10)}
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{(12,12),(13,10},(8,11)}
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None of these
Which of the following functions from Z into Z are bijections?
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$f(x) = x^3$
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f (x) = x + 2
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f (x) = 2x + 1
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$f(x)=x^2 + 1$
The relation is defined as R={(x,y) : x-y is an integer }, where $x,y \in Z$ The relation is
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reflexive
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symmetric
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transitive
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All the above
Let f : R → R be defined by $f(x) = 3x^2 -5$ and g : R → R by $g (x) =\frac {x}{x^2 + 1}$ Then g o f is
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$\frac {3x^2 -5}{9x^4 -30x^2 +26}$
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$\frac {3x^2 }{9x^4 -30x^2 +26}$
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$\frac {3x^2 +5}{9x^4 -30x^2 +26}$
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$\frac {3x^2 -5}{9x^4 -6x^2 +26}$
Let f : R → R be the function defined by f (x) = 2x – 3 for $x \in R$. then $f^{-1}$
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$\frac {x-3}{2}$
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$\frac {x+3}{2}$
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$\frac {x+2}{3}$
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None of the above
Let f: R -{n} -> R be a function defined by $f(x)= \frac {x-m}{x-n}$ where $m \ne n$. Then
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f is one-one onto
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f is one-one into
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f is many-one onto
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f is many-one into
if g(f(x)) = |sin x| and $f(g(x))= (sin \sqrt x)^2$,then
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$f(x) =sin^2 x$, $g(x) = \sqrt x$
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$f(x) =sin x$, $g(x) = | x|$
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$f(x) =x^2$, $g(x) = sin \sqrt x$
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f and g cannot be determined
Let f : R → R be defined by $f(x) = \frac {x}{\sqrt {1+x^2}}$ then (f o f o f) (x) is
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\frac {x}{\sqrt {1+3x^2}}$
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\frac {x}{\sqrt {1+2x^2}}$
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\frac {x}{\sqrt {1- x^2}}$
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\frac {x}{\sqrt {x^2 -1}}$
if $f(x) = \frac {x-1}{x+1} , x \ne -1$ The domain of f(f(x)) is
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R - {1,0}
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R - {1,0,-1}
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R - {-1,0}
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R - {1,0,2}
The value of f(f(x) is
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$\frac {-1}{x}$
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$\frac {1}{x}$
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$\frac {-1}{x-1}$
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$\frac {-1}{1+x}$
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