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If \(y = f(x)\) is symmetrical graph about the line \(x = 2\), then
Since the graph is symmetrical about .
If \(f(x).f\left( {\frac{1}{x}} \right) = f(s) + f\left( {\frac{1}{x}} \right)\) and \(f(4) = 64,\) then \(f(6)\) is equal to
This condition satisfy for
If \(f(x) = {(a – {x^n})^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 n}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$n$}}}}\), where \(a > 0,n \in N\), then \(f0f(x)\) is equal to
If \(f:(2,3) \to (0,1)\) is defined by \(f(x) = x – [x],\) then \({f^{ – 1}}(x)\) is
Domain of \({\cos ^{ – 1}}\left( {\frac{{x – 3}}{2}} \right) – {\log _{10}}(4 – x)\) is
\(\)\therefore D_1 \cap D_2 \in [1,4) /latex]
If W denotes the words in the English dictionary define by \(R = \{ (x,y) \in W \times W:\) the words \(x\& y\)have atleast one letter in common\(\} \) then R is
It is reflexive.
it is symmetric.
It is not transitive.
Range of \(f\) defined by \(f(x) = \frac{{{x^2}}}{{{x^2} + 7}}\)
is defined as
Can’t be or negative
Range of \(\) f=[0,1) /latex]
If \(f\left( {{\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}}} \right) = \sqrt {{p^2} – {q^2}} ,\) fore any \({\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}} \in Q\), then \(I.f\left( {{\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}}} \right)\) is real for each \({\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}} \in Q\).\(II.f\left( {{\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}}} \right)\) is complete for each \({\raise0.7ex\hbox{$p$} \!\mathord{\left/
{\vphantom {p q}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$q$}} \in Q\)
If is not real.
If is not complex.
Hence both statements are incorrect.
If \(f:R \to R\) is given by \(f(x) = \{ – 1,\) when \(x\) is rational\(\} \)
\(f(x) = \{ 1,\)when \(x\) is irrational\(\} \) then \((f0f)(1 – \sqrt 3 )\) is equal to
If \(f:N \to N\) is defined by \(f(n) = \left\{ {\frac{{n + 1}}{2}} \right.,\) if \(n\) is odd.\(\} \)
\(f(n) = \left\{ {\frac{n}{2},} \right.\) if \(n\) is even\(\} \) then \(f0f( – 1)\) is
\(f:\left[ {0,3} \right] \to \left[ {1,29} \right]\) is defined by \(f(x) = 2{x^3} – 15{x^2} + 36x + 1,\) is
is increasing as well as decreasing. So, it is manyone function.
Now, and .
So, it is onto function.
If \(f:Z \to Z\) as \(f(n) = \left\{ {{\raise0.7ex\hbox{$n$} \!\mathord{\left/
{\vphantom {n 2}}\right.\kern\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}},n} \right.\) is even
\(f(n) = 0,n\) is odd. Then \(f\) is
It is a manyone function as for every odd value of , it will give zero. Also, it is onto as for every even value of , we will get a set of integer. So it is surjective but not injective.
If \(f:C \to R\) defined by \(f(z) = \left z \right\) is
such that
is many one as modulus of and have same values.
It is not onto function as is always nonnegative real numbers.
If \(f:N \to N\) defined by \(f(x) = {x^2} + x + 1,x \in N,\) then \(f\) is
such that
or
So, is oneone but not onto.
If \(f:R \to R\) is given by \(f(x) = {x^3} – 8\) is
Let
So, is oneone. It is also onto as .
The function is bijective.
If \(f:R \to R\) defined by \(f(x) = (x – 1)(x – 2)(x – 3)\) is
Range of \(f(x) = 3\sin \sqrt {\frac{{{\pi ^2}}}{{16}} – {x^2}} \) is
For
For
If \(A = \{ 1,2,3,4,5\} \& R\) be a relation on \(A\) defined by \(\{ (a,b):a,b \in A,a \times b\) is an odd number\(\} \). the rng \(R\) is
Range of \(f:\left[ {0,1} \right] \to R,f(x) = {x^3} – {x^2} + 4x + 2{\sin ^{ – 1}}x\)
The period of \(f(x) = \left {\sin 2x} \right + \left {\cos 8x} \right\) is
Period of of period of & period of
The even function is
The period of \(f(\theta ) = 4 + 4{\sin ^3}\theta – 3\sin \theta \) is
Period of
If \(f(x) = \sec \left[ {\log \left( {x + \sqrt {1 + {x^2}} } \right)} \right]\) then \(f(x)\) is
Domain of \(f(x) = \sqrt {1 + {{\log }_e}\left( {1 – x} \right)} \) is