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If \(f\) and \(g\) be two function given by \(f = \{ (1,2),(2,3),(3,4),(4,5)\} \)
\(g = \{ (1,0),(2,1),( – 3, – 4),(4, – 5)\} \) then domain of \(f.g\) is given by
Domain for
and Domain of
Range of the function \(f(x) = 9 – \left {x – 7} \right\) is
Domain of
Range of
If \(A = \{ 1,2,3\} ,\) then \(A \times A \times A\) has ______number of elements.
The domain of the function \(f(x) = \sqrt {9 – x} + \frac{1}{{\sqrt {{x^2} – 9} }}\) is equal to
is defined, if or .
or
or and
Domain of
If \(f:R \to R\) and defined by \(f(x) = \frac{1}{{2 – \cos 3x}},\gamma x \in R,\) then range of \(f\) is
is defined
Range of
If \(s\) be the set of all real numbers. A relation \(R\) has been defined on \(s\) by \(aRb \Leftrightarrow \left {a – b} \right\left. \le \right\) the \(R\) is
it is reflexive
i.e.,
So it is symmetric.
Taking &
but
The relation is reflexive and symmetric not transitive.
If \(R\) is a relation on \(N\), defined by
\(\{ (x,y):2x – y = 10\} \), the \(R\) is
but .
So, it is not reflexive.
As and
So, it is not symmetric.
As, but
So it is not transitive.
If \(R\) is a relation defined \(aRb,iff\left {a – b} \right > 0,\) then the relation is
For which is not true.
So it is not reflexive.
This is symmetric.
Now
So this is transitive.
If \(R\) be a relation defined on \(N\), iff \(GCD\) of \(a\) and \(b\) is 2, then \(R\) is
Now, for , then of and is thus is not reflexive.
Again
So it is Symmetric.
Again for
If of is and is , then it is not necessary that is 2.
It is no transitive.
If \(R = \{ (3,3),(3,6),(3,9),(6,6),(9,9),(6,9),(9,6),(9,3),(6,3)\} \) is a relation on set \(A = \{ 3,6,9\} \), the relation is
is reflexive.
is symmetric.
is transitive.
Hnece is an equivalence relation.
The relation R defined on N as \(\{ (a,b):a\) differs from \(b\) by \(3\} \) is given by
The total number of injections (oneone & into) mappings from \(\{ {a_1},{a_2},{a_3}\} \) to \(\{ {b_1},{b_2},{b_3}\} \) is
Total number of oneone & into mappings
If \(A\& B\) be two equivalence relations defined on set C, then
If & be equivalence relation on , then is also an equivalence relation on .
If \(R\& S\) be two nonvoid relation on A. then which of the following statement is false?
Let
and are two transitive relation on .
Hence
So, is not transitive.
If R is a relation in Z given by \(aRb\) for \(a = {2^k}b,\) fore some \(k \in Z\), then R is
For
It is reflexive.
It is Symmetric.
It is transitive.
is an equivalence relation.
\(f:[0,\infty ]\) is defined by \(f(x) = \frac{{2x}}{{1 + 2x}}\) is
As is oneone.
As does not have preimage.
So is not onto.
\(A = \{ 1,2,3,4\} \& B = \{ 1,2,3,4,5,6\} \) are two sets \(f:A \to B\) is defined by \(f(x) = x + 2,\cancel{\gamma }x \in A\) then \(f\) is
do not have any preimage in .
is oneone and into.
If \(f:R \to C\) is defined by \(f(x) = {e^{2ix}}\), fore \(x \in R\) then \(f\) is
Ater some value of ,function will give same values, so function is not oneone. Also, function has minimum & maximum value i.e., & respectively.
So, is not onto.
If \(f:[2,3] \to R\) define by then \(rng\) \(f\) is contained in the interval.
is increasing function.
The period of function \(f(x) = \cos e{c^2}3x + \cot 4x\) is
Period of
Period of
Period of of
The function \(f(x) = {x^3} – 9x\) is
is an odd fuction.
Domain of the function \(f(x) = \frac{{\sqrt {4 – {x^2}} }}{{{{{\mathop{\rm Sin}\nolimits} }^{ – 1}}(2 – x)}}\) is
is defined as
Domain \(\)= [1,2) /latex]
If \(f:R \to R\) defined by \(f(x) = x – [x] – {\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}},x \in R\) where \(\left[ x \right]\) is the greatest integers not exceeding \(x,\) then \(\{ x \in R:f(x) = \frac{1}{2}\} \) is equal to
which is not possible.
If \(f,g:R \to R,f(x) = x – 3,g(x) = {x^2} + 1\), then the value of \(x\) fore which
\(g\{ f(x)\} = 10\) are
If \(f:[6,6] \to R\) is defined by \(f(x) = {x^2} – 3,x \in R,\) then \((f0f0f)( – 1) + (f0f0f)(0) + (f0f0f)(1)\) is
Now
If \(f(x) = \frac{{ax + b}}{{cx + d}}\), then \(f0f(x) = x\) provided that
Solving this we get