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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 3
Let $$S$$ be set of all rational numbers. The functions $$f:R\rightarrow R,\ g:R\rightarrow R$$ are
defined as
$$f(x)=\begin{cases}
0, & x \in S \\
1, & x \notin S
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\in S \\
0 & x\notin S
\end{cases}$$
then, $$(fog) (\pi)+(gof)(e)=$$
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$$-1$$
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$$0$$
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$$1$$
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$$2$$
Explanation
$$f(x)=\begin{cases}
0, & x \text{ is rational} \\
1, & x \text{ is irrational}
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\text { is rational} \\
0& x\text { is irrational }
\end{cases}$$
$$fog(\pi )=f(g(\pi ))=f(0)=0$$
so $$gof(e)=g(f(e))=g(1)=-1$$
$$\therefore fog(\pi )+gof(e)=-1$$
Set $$A$$ has $$n$$ elements. The number of functions that can be defined from $$A$$ into $$A$$ is:
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$$n^2$$
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$$n!$$
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$$n^n$$
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$$n$$
Explanation
The number of functions that can be defined from a set into another set can be found by
$${ \text{(Number of elements in co-domain)} }^{ \text{Number of elements in domain} }$$
Since, set $$A$$ has $$n$$ elements, we have
Number of elements in co-domain $$=$$ Number of elements in domain $$= n (A) = n$$.
Therefore, number of functions that can be defined from $$A$$ into $$A$$ $$=n(A)^{n(A)}= n^{n}$$.
If $$n\geq 1$$ is any integer, $$\mathrm{d}(n)$$ denotes the number of positive factors of $$n$$, then for any prime number $$\mathrm{p},\ \mathrm{d}(\mathrm{d}(\mathrm{d}(\mathrm{p}^{7})))=$$
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
For a number being $$p^{n}$$ the total number of positive factors excluding $$1$$ and the number itself will be $$n-1$$.
Therefore total number of positive factors will be $$n-1+2$$
$$=n+1$$
Hence for $$p^{7}$$ we will have $$8$$ factors.
Now for $$8=2^{3}$$.
Hence $$8$$ will have $$4$$ factors.
Now $$4=2^{2}$$, hence the number of factors will be $$3$$.
Let $$\displaystyle f\left( x \right)={ x }^{ 2 }-x+1,x\ge \left( \frac { 1 }{ 2 } \right) $$ then the solution of the equation $$f(x)=f^{-1}(x)$$ is
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$$x=1$$
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$$x=2$$
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$$\displaystyle x=\frac{1}{2}$$
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None of these
Explanation
$$\displaystyle y={ x }^{ 2 }-x+1\Rightarrow { x }^{ 2 }-x+\left( 1-y \right) =0$$
Here, $$a=1,b=-1$$ and $$c=1-y$$
$$\displaystyle x=\frac { 1\pm \sqrt { 1-4\left( 1-y \right) } }{ 2 } \quad \quad \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] $$
$$\displaystyle \because x>\frac { 1 }{ 2 } ,\quad \therefore x=\frac { 1 }{ 2 } +\sqrt { y-\frac { 3 }{ 4 } } $$
$$\displaystyle \Rightarrow f^{ -1 }\left( x \right) =\frac { 1 }{ 2 } +\sqrt { x-\frac { 3 }{ 4 } } $$
Now, $$\displaystyle { x }^{ 2 }-x+1=\frac { 1 }{ 2 } +\sqrt { x-\frac { 3 }{ 4 } } $$
Since the graphs of the original and inverse functions can intersect only on the straight line $$y=x,$$ therefore
$$x=f\left( x \right) \quad \quad \Rightarrow x={ x }^{ 2 }-x+1$$
$$\Rightarrow { x }^{ 2 }-2x+1=0$$
$$\Rightarrow { \left( x-1 \right) }^{ 2 }=0$$
$$\Rightarrow x=1$$
lf $$f:[-6,6]\rightarrow \mathbb{R}$$ is defined by $$f(x)=x^{2}-3$$ for $$x\in \mathbb{R}$$ then
$$(fofof)(-1)+(fofof)(0)+(fofof)(1)=$$
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$$f(4\sqrt{2})$$
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$$f(3\sqrt{2})$$
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$$f(2\sqrt{2})$$
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$$f(\sqrt{2})$$
Explanation
$$fofof(-1)=fof(-2)=f(1)=-2$$
$$fofof(1)=fof(-2)=f(1)=-2$$
$$fofof(0)=fof(-3)=f(0)=33$$
$$\therefore fofof(-1)+fofof(1)+fofof(0)=29$$
$$=32-3$$
$$=(4\sqrt{2})^{2}-3$$
$$\Rightarrow f(4\sqrt{2})=(4\sqrt{2})^{2}-3$$
lf $$f$$ : $$R\rightarrow R$$ is defined by
$$f(x)=\left\{\begin{array}{l}x+4 & x<-4\\3x+2 & -4\leq x<4\\x-4 & x\geq 4\end{array}\right.$$
then the correct matching of list I to List II is.
List - I
List - II
$$\mathrm{A}) f(-5)+f(-4)=$$
$$\mathrm{i}) 14$$
$$\mathrm{B}) f(|f(-8)|)=$$
ii $$) 4$$
$$\mathrm{C}) f(f(-7)+f(3))=$$
$$\mathrm{i}\mathrm{i}\mathrm{i})-11$$
$$\mathrm{D}) f(f(f(f(0)))+1=$$
$$\mathrm{i}\mathrm{v})-1$$
v) $$1$$
vi) $$0$$
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A-iii , B-vi , C-ii , D- v
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A-iii , B-iv , C-ii , D- vi
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A-iv , B-iii , C-ii , D- i
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A-ii , B-vi , C-v , D- ii
Explanation
$$f(-5)=-5+4=-1 ; f(-4)=3(-4)+2=-10$$
$$\therefore f(-5)+f(4)=-11$$
$$f(|f(-8)|)=f((-8+4))=f(4)=4-4=0$$
$$f(f(-7)) tf(3) = f(-7+4)+3(3)+2$$
$$=3(-3)+2+3(3)+2$$
$$=4$$
$$f(f(f(f(0))))+1=f(f(f(2)))+1=f(f(8))+1$$
$$=f(4)+1=0+1=1$$
lf $$g(f(x)) =|\sin \mathrm{x}|,f(g(x)) =(\sin\sqrt{\mathrm{x}})^{2}$$, 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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$${f}, {g}$$ cannot be determined
Explanation
$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$ .......(i)
$$f(g(x))=\sin^{2}\sqrt{x}$$ .......(ii)
Comparing $$i$$ and $$ii$$
$$\Rightarrow$$
$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$
$$\Rightarrow$$
$$g(\sin^{2}x)=\sqrt{\sin^{2}x}=g(f(x))$$
And
$$\Rightarrow$$
$$f(g(x))=\sin^{2}\sqrt{x}$$
$$\Rightarrow$$
$$f(\sqrt{x})=\sin^{2}\sqrt{x}=f(g(x))$$
Therefore
$$\Rightarrow$$
$$g(x)=\sqrt{x}$$
$$\Rightarrow$$
$$f(x)=\sin^{2}x$$
lf $$ f(x)=x-x^{2}+x^{3}-x^{4}+\ldots..\infty$$ when $$|x|<1$$, then the ascending order of the following is
a) $$f(1/2)$$
b) $$f^{-1}(1/2)$$
c) $$ f(-1/2)$$
d) $$f^{-1}(-1/2)$$
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a, b, c, d
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c, d, a, b
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b, a, d, c
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d, c, a, b
Explanation
$$f(x)=x-x^2+x^3-x^4$$
$$f(x)=\dfrac{x}{1+x}(x)< 1$$
$$f\left ( \dfrac{1}{2} \right )=\dfrac{1}{3}$$
$$f\left ( \dfrac{-1}{2} \right )=-1$$
$$f(x)=\dfrac{x}{1+x}$$
$$y=\dfrac{x}{1+x}$$
$$y+xy=x$$
$$\dfrac{y}{(u-1)}=x$$
$$f^{-1}x=\dfrac{-x}{x-1}$$
$$f^{-1}\left ( \dfrac{1}{2} \right )=1$$
$$f^{-1}\left ( \dfrac{-1}{2} \right )=\dfrac{-1}{3}$$
Let $$f$$ be an injective function with domain $$\{x, y, z\}$$and range $$\{1,2,3\}$$ such that exactly one of the follwowing statements is correct and the remaining are false :
$${f}({x})=1,{f}({y})\neq 1$$,
$${f}({z})\neq 2$$,
then the value of $${f}^{-1}(1)$$ is
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$$x$$
0%
$$y$$
0%
$$z$$
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none
Explanation
It gives three cases
Case (1) When $$f(x)=1$$ is true.
In this case remaining two are false
$$f(y)=1$$ and $$f(z)=2$$
This means $$x$$ and $$y$$ have the same image, so $$f(x)$$ is not an injective, which is a contradiction.
Case (2) when $$f\left( y \right)\neq 1$$ is true
If $$f\left( y \right)\neq 1$$is true then the remaining statements are false.
$$\therefore f\left( x\right)\neq 1$$ and $$f(z)=2$$
i.e., both $$x$$ and $$y$$ are not mapped to $$1.$$
So, either both associate to $$2$$ or $$3,$$
Thus, it is not injective.
Case (3) When $$f(z) \neq 2$$ is true
If $$f(z) \neq2$$ is true then remaining statements are fales
$$\therefore$$ If $$f(x) \neq1$$ and $$f(y)=1$$
But $$f$$ is injective
Thus, we have $$f(x)=2, f(y)=1$$ and $$f(z)=3$$
Hence, $$\displaystyle f^{ -1 }\left( 1 \right)=y$$
If $$ f : R \rightarrow R$$ is defined by $$f(x)=2x-2,$$ then $$(f\circ f) (x) + 2 =$$
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$$f(x)$$
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$$2f(x)$$
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$$3f(x)$$
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$$-f(x)$$
Explanation
Let $$f : R \rightarrow R$$ is defined by $$f(x)=2x-2$$.
Then
$$(f\circ f)(x)+2= f[f(x)]+2$$
$$=f(2x-2)+2$$
$$=2(2x-2)-2+2$$
$$=2f(x)$$
If $$f(x) =\displaystyle \frac{x}{\sqrt{1-x^2}}, g(x)=\frac{x}{\sqrt{1+x^2}}$$ then $$(f\circ g)(x) =$$
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$$\displaystyle \frac{x}{\sqrt{1-x^2}}$$
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$$\displaystyle \frac{x}{\sqrt{1+x^2}}$$
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$$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$$
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$$x$$
Explanation
Let $$x= \tan \theta$$
$$(f\circ g)(x)=f[g(x)] = f[g(\tan \theta)]$$.
Since $$\displaystyle g(x) = \frac{x}{\sqrt{1+x^2}}$$, we have
$$f[g(\tan \theta)]=\displaystyle f \left [ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} \right ] = f \left [ \frac{\tan \theta}{\sec \theta} \right ]=f (\sin \theta)$$
Also since $$f(x)=\cfrac{x}{\sqrt{1-x^2}}$$ , we have
$$\displaystyle f (\sin \theta) = \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$$
$$=\tan \theta =x$$
If $$f(x)=\log x, g(x) = x^3$$ then $$f[g(a)]+f[g(b)]= $$
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$$f[g(a)+g(b)]$$
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$$f[g(ab)]$$
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$$g[f(ab)]$$
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$$g[f(a)+f(b)]$$
Explanation
Given that $$g(x)=x^3$$. Therefore,
$$f[g(a)]+f[g(b)]=f(a^3)+f(b^3)$$.
Since $$f(x)=\log x$$, we have
$$f(a^3)+fb^3=\log a^3+\log b^3$$
$$=\log (a^3b^3)=\log ((ab)^{3})$$
$$=f[g(ab)]$$
If $$X=\left \{ 1, 2, 3, 4, 5 \right \}$$ and $$Y=\left \{ 1, 3, 5, 7, 9 \right \}$$, determine which of the following sets represent a relation and also a mapping.
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$$R_{1}=\left \{ (x, y):y=x+2, x\in X, y\in Y \right \}$$
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$$R_{2}=\left \{ (1, 1),(2, 1),(3, 3),(4, 3),(5, 5) \right \}$$
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$$R_{3}=\left \{ (1, 1),(1, 3),(3, 5),(3, 7),(5, 7) \right \}$$
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$$R_{4}=\left \{ (1, 3),(2, 5),(4, 7),(5, 9),(3, 1) \right \}$$
Explanation
$$R_1={(1,3),(2,4),(3,5),(4,6),(5,7)}$$
Since $$4$$ and $$6$$ do not belong to $$Y$$
$$(2,4),(4,6)∉R_1$$
$$R_1={(1,3),(3,5),(5,7)}⊂A×B$$
Hence $$R_1$$ is a relation but not a mapping as the elements $$2$$ and $$4$$ do not have any image.
$$R_2$$ : It is certainly a mapping and since every mapping is a relation, it is a relation as well.
$$R_3$$: It is a relation being a subset of $$A×B$$ but the elements $$1$$ and $$3$$ do not have a unique image and hence it is not mapping.
$$R_4$$ : It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection
If $$f:R \rightarrow R$$ and $$g : R \rightarrow R$$ are defined by $$f(x)=2x+3$$ and $$g(x)=x^2+7$$, then the values of $$x$$ such that $$g(f(x)) =8$$ are:
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$$1, 2$$
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$$-1, 2$$
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$$-1, -2$$
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$$1, -2$$
Explanation
Since $$f(x)=2x+3$$,
$$g[f(x)]=8 \Rightarrow g(2x+3)=8$$.
Also since $$g(x)=x^2+7$$,
$$g(2+3x)=8\Rightarrow (2x+3)^2+7=8$$
$$\Rightarrow 2x+3 = \pm 1$$
$$\Rightarrow x=-1$$ or $$-2$$.
If $$f(x) = \dfrac{2x+5}{x^{2} + x + 5}$$, then $$f\left [ f(- 1 ) \right ]$$ is equal to
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$$\dfrac{149}{155}$$
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$$\dfrac{155}{147}$$
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$$\dfrac{155}{149}$$
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$$\dfrac{147}{155}$$
Explanation
Given expression is $$ f(x)=\dfrac { 2x+5 }{ { x }^{ 2 }+x+5 } $$
$$ \therefore f(-1)=\dfrac { 2\times (-1)+5 }{ (-1)^{ 2 }+(-1)+5 } =\dfrac { 3 }{ 5 } $$
$$ \therefore f\left( f\left( -1 \right) \right) =\dfrac { 2\times \dfrac { 3 }{ 5 } +5 }{ \left( \dfrac { 3 }{ 5 } \right) ^{ 2 }+\dfrac { 3 }{ 5 } +5 } =\dfrac { 155 }{ 149 }$$
Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A into B is :
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144
0%
12
0%
24
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64
Explanation
If $$ABC\sim PQR$$, then AB:PQ =
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$$AC:PB$$
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$$AC:PR$$
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$$AB:PR$$
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$$AC:RQ$$
If $$f : R \rightarrow R$$ and $$g :R \rightarrow R$$ are defined by $$f(x) = x -[x]$$ and $$g(x) = [x]$$ for $$x \in R$$, where $$[x]$$ is the greatest integer not exceeding $$x$$, then for every $$x \in R, f(g(x)) =$$
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$$x$$
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$$0$$
0%
$$f(x)$$
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$$g(x)$$
Explanation
Since $$f(x)=x-[x]$$, we have
$$f[g(x)] =g(x)-[g(x)]$$
Also since $$g(x)=[x]$$,we have
$$g(x)-[g(x)]=[x] - [[x]]$$.
Here, $$[x]$$ is the greatest integer not exceeding $$x$$. Therefore,
$$[[x]]=[x]$$ and hence,
$$[x]-[[x]]=[x]-[x]=0$$
If $$y=f(x) = \dfrac{2x-1}{x-2}$$, then $$f(y)=$$
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$$x$$
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$$y$$
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$$2y-1$$
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$$y-2$$
Explanation
$$\displaystyle f(y) = \frac{2y-1}{y-2} = \frac{2 \left ( \dfrac{2x-1}{x-2} \right ) -1}{\left ( \dfrac{2x-1}{x-2} \right ) -2}$$
$$=\displaystyle \frac{4x-2-x+2}{2x-1-2x+4} = \frac{3x}{3} = x$$
If $$f(g(x))$$ is one-one function, then
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g(x) must be one-one
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f(x) must be one-one
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f(x) may not be one-one
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g(x) may not be one-one
Explanation
It is fundamental concept that, If $$f(g(x))$$ is one-one function then,
$$f(x)$$ must be one-one function and $$g(x)$$ may be one-one or many one
Which of the following functions are one-one?
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$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$
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$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$
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$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$
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$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$
Explanation
For all even powered function,
$$f(x_{1})=f(x_{2})$$ where $$x_{1}=\pm(x_{2})$$
Hence, the only possible one-one function is $$f(x)=x^3+4$$
$$f(x)$$ is a strictly increasing function. So, it is one-one.
A mapping function $$f:X\rightarrow Y$$ is one-one, if
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$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$
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$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$
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$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$
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none of these
Explanation
For a function to be one one
$$f(x_{1})=f(x_{2})$$
Implies that
$$x_{1}=x_{2}$$
Hence the answer is
option B
Let $$\displaystyle f:R\rightarrow A=\left \{ y: 0\leq y< \dfrac{\pi}{2} \right \}$$ be a function such that $$\displaystyle f(x)=\tan^{-1}(x^{2}+x+k),$$ where $$k$$ is a constant. The value of $$k$$ for which $$f$$ is an onto function is
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$$1$$
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$$0$$
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$$\displaystyle \frac{1}{4}$$
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none of these
Explanation
For $$f$$ to be an onto function, Range and co-domain of $$f$$ should be equal
$$\Rightarrow f(x)\ge 0 \forall x\in R$$
$$\Rightarrow \tan^{-1}(x^2+x+k)\ge 0$$
For the above equation to be valid for all $$x$$
We must have, the discriminant of $$x^2+x+k=0$$, is zero
$$ b^2-4ac=0$$
$$\Rightarrow 1-4k=0\Rightarrow k =\dfrac{1}{4}$$
If $$f:R\rightarrow R$$ given by $$f(x)={ x }^{ 3 }+({ a+2)x }^{ 2 }+3ax+5$$ is one-one, then $$a$$ belongs to the interval
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$$(-\infty ,1)$$
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$$(1 ,\infty)$$
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$$(1 ,4)$$
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$$(4 ,\infty)$$
Explanation
If $$ f(x) $$ is one one then $$ f'\left( x \right) $$ should be either positive or negative for all $$ x $$.
$$f'\left( x \right) = 3{ x }^{ 2 }+2(a+2)x+3a$$
$$\Rightarrow D = 4{ (a+2) }^{ 2 }-4\times3\times3a<0$$
$$\Rightarrow$$ $${ a }^{ 2 }-5a+4<0$$
$$\Rightarrow$$ $$(a-1)(a-4)<0$$
Hence, $$a\in \left( 1,4 \right) $$
Let R be the relation in the set N given by $$=\left \{ (a, b):a=b-2, b >6 \right \}$$. Choose the correct answer.
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$$(2, 4)\in R$$
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$$(3, 8)\in R$$
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$$(6, 8)\in R$$
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$$(8, 7)\in R$$
Explanation
$$(a,b)∈R$$ only if $$a=b−2$$ and $$b>6$$
$$(6,8)∈R$$ as $$6=8−2$$ and $$8 >6$$
Hence $$(c)$$ is the correct alternative
Let $$\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \}$$ be a one-one function and only one of the conditions $$(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a$$ is true then the function $$f$$ is given by the set
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$$\displaystyle \left \{ (x,a),(y,b),(z,c)\right \}$$
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$$\displaystyle \left \{ (x,a),(y,c),(z,b)\right \}$$
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$$\displaystyle \left \{ (x,b),(y,a),(z,c)\right \}$$
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$$\displaystyle \left \{ (x,c),(y,b),(z,a)\right \}$$
Explanation
$$f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\} $$ is a one-one function
$$ \Rightarrow$$ each element in $$ \left\{ x,y,z \right\} $$ will have exactly one image in $$\left\{ a,b,c \right\}$$
and no two elements of $$ \left\{ x,y,z \right\}$$ will have same image in $$\left\{ a,b,c \right\} $$
Coming to the given 3 conditions, only one is true.
1) if $$f\left( x \right) \neq b$$ is true then $$f\left( y \right) =b$$ is false which makes $$f\left( z \right) \neq a$$ true $$\Longrightarrow f\left( x \right) \neq b$$ is false.
2) if $$f\left( y \right) =b$$ is true then $$f\left( x \right) \neq b$$ will also be true $$\Longrightarrow f\left( y \right) =b$$ is false
$$ \therefore f\left( z \right) \neq a$$ is the true condition and remainig two are false conditions.
$$\therefore f\left( x \right) =b, f\left( y \right) =a, f\left( z \right) =c$$
hence $$f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right) \right\} $$
Which of the following function is one-one?
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$$f:R\rightarrow R$$ given by$$ f(x)=|x-1|$$ for all $$x\in R$$
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$$g:\left[ -\dfrac{\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \rightarrow R$$ given by $$g(x)=|sinx|$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$
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$$h:\left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] \in R$$ given by $$ h(x)=sinx$$ for all $$ x\in \left[ \dfrac{ -\pi }{ 2 },\dfrac{ \pi }{ 2 } \right] $$
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$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$ x\in R$$
Explanation
Option A - Consider $$x = 2 $$ and $$ x =0 $$ , the value of $$f(x)$$ is same. Hence it is not one-one
Option B - If we replace $$x$$ by $$-x$$, then the value of $$g(x)$$ remains the same. Hence it is not one-one
Option C - $$h(x)$$ is an increasing function for the given values of $$x$$. Hence it is one-one function
Option D -$$f(x)$$ is an even function. So it is not one-one for the given values of $$x$$
If $$f$$ and $$g$$ are one-one functions from $$R\to R$$, then
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$$f+g$$ is one-one
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$$fg$$ is one-one
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$$fog$$ is one-one
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none of these
Explanation
Let $${x_1},{x_2} \in R$$ be two distinct elements, then $$g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$$, as $$g$$ is one-one function. Similarly $$f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$$ as $$f$$ is also one-one function.
Hence, $$f\circ g$$ is one-one function.
Note that $$f+g$$ and $$f\cdot g$$ may not one-one functions even if $$f$$ and $$g$$ are one one. For example consider $$f\left( x \right) = x$$ and $$g\left( x \right) =- x$$, then $$f+g$$ and $$f\cdot g$$ are not one-one.
If $$f:R^+\rightarrow A$$, Where $$A=\{x:-5<x<\infty\}$$ be defined by $$f(x)=x^2-5$$. Then $$f^{-1}(7)=$$
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only $$-2{\sqrt3}$$
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only $$2{\sqrt3}$$
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both options A and B
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none
Explanation
The value $$f^{-1}(7)$$ will be the roots of the equation
$$x^2-5= 7\Rightarrow x^2 = 12\Rightarrow x = \pm 2\sqrt{3}$$
But only positive. $$\therefore x = 2\sqrt{3}$$
If the function $$f:[1,\infty)\rightarrow [1,\infty)$$ is defined by $$\displaystyle f(x)=3^{x(x-1)}$$ then $$f^{-1}(x) $$ is
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$$\displaystyle \left ( \frac{1}{2} \right )^{x(x-1)}$$
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$$\displaystyle \frac{1}{2}\left ( 1+\sqrt{1+4\log_{3}x}\right)$$
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$$\displaystyle \frac{1}{2}\left ( 1-\sqrt{1+4\log_{3}x}\right)$$
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not defined
Explanation
For the given domain, the function is one one.
Thus
,
$$f(x)=2^{x(x-1)}$$
$$y=2^{x^{2}-x}$$
$$lny=(x^{2}-x)ln2$$
$$log_{2}y=x^{2}-x$$
$$log_{2}y=(x-\dfrac{1}{2})^{2}-\dfrac{1}{4}$$
$$log_{2}y+\dfrac{1}{4}=(x-\dfrac{1}{2})^{2}$$
$$\dfrac{\pm 1}{2}(\sqrt{1+4log_{2}y})=x-\dfrac{1}{2}$$
Since $$f:[1,\infty)\rightarrow[1,\infty)$$
Thus domain is $$[1,\infty)$$
Hence domain is positive so negative sign will be ignored. $$So$$
$$x-\dfrac{1}{2}=\dfrac{1}{2}(\sqrt{1+4log_{2}y})$$
$$x=\dfrac{1}{2}[1+\sqrt{1+4log_{2}y}]$$
Replacing x and y, gives us
$$f^{-1}(x)=\dfrac{1}{2}[1+\sqrt{1+4log_{2}x}]$$
If the function $$f:R\rightarrow R$$ be such that $$\displaystyle f(x)=x-[x],$$ where $$[y]$$ denotes the greatest integer less than or equal to $$y$$, then $$f^{-1}(x)$$ is
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$$\displaystyle \frac{1}{x-[x]}$$
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$$\displaystyle [x]-x$$
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not defined
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none of these
Explanation
$$f\left( x \right) = x - \left[ x \right] = \left\{ x \right\}$$ is not one-one function from R to R. Hence $${f^{ - 1}}\left( x \right)$$ can not be defined.
If $$\displaystyle f:\left ( 3,4 \right )\rightarrow \left ( 0,1 \right )$$ is defined by $$\displaystyle f\left ( x \right )=x-\left [ x \right ]$$ where $$\displaystyle [x] $$denotes the greatest integer function then $$ f^{-1}(x)$$ is
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$$\displaystyle \frac{1}{x-\left [ x \right ]}$$
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$$\displaystyle [x]-x$$
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$$\displaystyle x-3$$
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$$\displaystyle x+3$$
Explanation
Note that $$\displaystyle f\left ( x \right )=x-\left [ x \right ] =\left\{ x \right\}$$. Hence every number $$x \in \left( {3,4} \right)$$ is mapped to its fractional part.
Hence, the inverse of the function can be written by adding the integer part of the number i.e., $$3$$ to each of the fraction part.
Hence the function $${f^{ - 1}}\left( x \right) = x + 3$$.
Let $$\displaystyle f:(-\infty,1]\rightarrow (-\infty,1]$$ such that $$\displaystyle f(x)=x(2-x).$$ Then $$f^{-1}(x)$$ is
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$$1+\sqrt{1-x}$$
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$$1-\sqrt{1-x}$$
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$$\sqrt{1-x}$$
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none of these
Explanation
We have, $$f(x)=x(2-x)$$
$$y=x(2-x)$$
$$-y=x(x-2)$$
$$-y=x^{2}-2x$$
$$1-y=x^{2}-2x+1$$
$$1-y=(x-1)^{2}$$
$$\pm\sqrt{1-y}=x-1$$
Now $$f:(-\infty,1] \rightarrow (-\infty,1]$$
$$x-1=-\sqrt{1-y}$$
$$x=1-\sqrt{1-y}$$
Hence $$f^{-1}(x)=1-\sqrt{1-x}$$
If $$f(x)=ax+b$$ and $$g(x)=cx+d$$, then $$f(g(x))=g(f(x))$$ implies
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$$f(a)=g(c)$$
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$$f(b)=g(b)$$
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$$f(d)=g(b)$$
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$$f(c)=g(a)$$
Explanation
$$f(g(x))=a(cx+d)+b$$
$$=acx+ad+b$$...(i)
$$g(f(x))=c(ax+b)+d$$
$$=ac(x)+bc+d$$ ...(ii)
$$f(g(x))=g(f(x))$$
From (i) and (ii)
$$ac(x)+bc+d=acx+ad+b$$
$$cb+d=ad+b$$
$$g(b)=f(d)$$
If $$\displaystyle f(x)=\frac{1}{1-x},x\neq 0,1$$ then the graph of the function $$\displaystyle y=f\left \{ f(f(x)) \right \},x> 1,$$ is
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a circle
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an ellipse
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a straight line
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a pair of straight lines
Explanation
$$f(f(x))=\dfrac{1}{1-\dfrac{1}{1-x}}$$
$$=-\dfrac{1-x}{x} =g(x)$$
$$f(g(x))=\dfrac{1}{1+\dfrac{1-x}{x}}$$
$$=\dfrac{x}{x+1-x}$$
$$=x$$
$$y=x$$ is an equation of straight line.
Hence, 'C' is correct.
If $$\displaystyle f(x)=3x-5$$ then $$f^{-1}(x)=$$
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$$\displaystyle \frac{1}{3x-5}$$
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$$\displaystyle \frac{x+5}{3}$$
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does not exist because $$f$$ is not one-one
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does not exist because $$f$$ is not onto
Explanation
$$f(x)=3x-5$$
Since $$f(x)$$ is a purely linear function (straight line), it is therefore a one-one function.
Hence $$f(x)$$ is revertible for all real $$x.$$
$$y=3x-5$$
$$y+5=3x$$
$$x=\dfrac{y+5}{3}$$
Replacing $$x$$ by $$y$$, gives us
$$f^{-1}(x)=\dfrac{x+5}{3}$$
If $$f$$ and $$g$$ are two functions such that $$\displaystyle \left ( fg \right )\left ( x \right )=\left ( gf \right )\left ( x \right )$$ for all $$x$$. Then $$f $$ and $$g$$ may be defined as
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$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$
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$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$
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$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$
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$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers
Explanation
A$$)f(x)=$$$$\sqrt{x},g(x)=cosx then (fg)(x) = \sqrt{\cos x} and (gf)(x) = \cos \sqrt{x}$$.They are not equal.
B)$$f(x)=x^{3},g(x)=x+1 then (fg)(x) = (x+1)^{3} and (gf)(x)= x^{3}+1$$.They are not equal.
C)$$f(x)=x1,g(x)=x^{2}+1 then (fg)(x)=x^{2} and (gf)(x)=x^{2}-2x$$. They are not equal.
D)$$f(x)=x^{m}, g(x)=x^{n} then (fg)(x)=x^{n+m} and (gf)(x)=x^{n+m}$$.Hence $$(fg)(x)= (gf)(x)$$
If $$\displaystyle f(x)=x^{n},n\in N$$ and $$(gof)(x)=ng(x)$$ then $$g(x)$$ can be
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$$n\:|x|$$
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$$3.\sqrt[3]{x}$$
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$$e^{x}$$
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$$\log\:|x|$$
Explanation
$$f(x)=x^{n}$$
$$g(f(x))=ng(x)$$ ...$$(i)$$
$$\log(f(x))=n\log(x)$$ ...$$(ii)$$
Taking $$\log(|x|)$$ as $$g(x)$$ the above expression is reduced to eq $$i$$.
Hence
$$g(x)=\log(|x|)$$.
The composite mapping $$fog$$ of the map $$f: R\rightarrow R,f(x)=\sin x$$ and $$g: R\rightarrow R, g(x)=x^2$$ is
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$$x^2 \sin x$$
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$$(\sin x)^2$$
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$$\sin x^2$$
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$$\dfrac{ \sin x}{x^2}$$
Explanation
Composite mapping will be $$f(g(x))$$
$$=f(x^2)$$
$$=\sin(x^2)$$
Hence option $$'C'$$ is the answer.
Let $$S$$ be a set containing $$n$$ elements. Then the total number of binary operations on $$S$$ is
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$$n^n$$
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$${2^n}^{2}$$
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$$n^{n^2}$$
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$$n^2$$
Explanation
No of binary operations imply
$$S\times S\rightarrow S$$
$$=n(S)^{[n(S\times S)]}$$
$$=n^{n\times n}$$
$$=n^{n^2}$$
Let $$f: R\rightarrow R$$ be defined by $$f(x)=3x-4$$ then $$f^{-1}(x)$$ is
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$$\dfrac{1}{3}(x+4)$$
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$$\dfrac{1}{3}(x-4)$$
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$$3x+4$$
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not defined
Explanation
$$f(x)$$ is invertible for all real values of $$x$$
Hence
$$y=3x-4$$
$$y+4=3x$$
$$\dfrac{y+4}{3}=x$$
$$f^{-1}(x)=\dfrac{x+4}{3}$$
Let $$\displaystyle f(x)=\frac{ax}{x+1}$$, where $$\displaystyle x\neq -1$$. Then for what value of $$\displaystyle a$$ is $$\displaystyle f( f(x))=x$$ always true
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$$\displaystyle \sqrt{2}$$
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$$\displaystyle -\sqrt{2}$$
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$$1$$
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$$-1$$
Explanation
$$\displaystyle f\left( {f\left( x \right)} \right) = \dfrac{{a\dfrac{{ax}}{{x + 1}}}}{{\dfrac{{ax}}{{x + 1}} + 1}} = \dfrac{{\dfrac{{{a^2}x}}{{x + 1}}}}{{\dfrac{{ax + x + 1}}{{x + 1}}}} = \dfrac{{{a^2}x}}{{ax + x + 1}}$$
Since, $$\displaystyle f\left( {f\left( x \right)} \right) =x$$, we have,
$$\displaystyle\dfrac{{{a^2}x}}{{ax + x + 1}}=x$$.
Simplifying the equation we get,
$${a^2}x = \left( {a + 1} \right){x^2} + x$$
$$\therefore \left( {a + 1} \right){x^2} + \left( {1 - {a^2}} \right)x = 0$$
or $$\left( {a + 1} \right)x\left( {x + 1 - a} \right) = 0$$
Hence the only possible value is $$a=-1$$
A function $$y=f\left ( x \right )$$ is invertible only when
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$$y=f\left ( x \right )$$ is monotonic increasing
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$$y=f\left ( x \right )$$ is bijective
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$$y=f\left ( x \right )$$ is monotonic decreasing
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invertible
Explanation
Invertible function is defined as, the function $$f$$ applied to an input $$x$$ gives a result of $$y$$, then applying its inverse function $$g$$ to $$y$$ gives the result $$x$$, and vice versa. i.e., $$f\left ( x \right )=y$$ $$\Rightarrow g\left ( y \right )$$ = $$x$$.
And bijective function has the same definition as that of an Invertible function
Let $$ f:R \rightarrow R$$ and $$g:R \rightarrow R$$ be defined by $$f(x)=x^2+2x-3,g(x)=3x-4$$ then $$(gof) (x)=$$
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$$3x^2+6x-13$$
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$$3x^2-6x-13$$
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$$3x^2+6x+13$$
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$$-3x^2+6x-13$$
Explanation
Given that, $$f(x)=x^2+2x-3$$ and $$g(x)=3x-4$$
$$(gof) (x)=g\left(f(x)\right)$$
$$=3(x^2+2x-3)-4$$
$$(gof) (x)=3x^2+6x-13$$
Hence, option A.
If $$\displaystyle f:[1,+\infty ]\rightarrow [2,+\infty )$$ is given by $$f(x)=x+\dfrac{1}{x}$$ then $$f^{-1}(x)$$ equals
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$$\displaystyle \frac{x+\sqrt{x^{2}+4}}{2}$$
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$$\displaystyle \frac{x}{1+x^{2}}$$
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$$\displaystyle \frac{x+\sqrt{x^{2}-4}}{2}$$
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$$\displaystyle 1+\sqrt{x^{2}-4}$$
Explanation
Let $$y = x+\cfrac{1}{x}$$
$$\Rightarrow x^2-yx+1=0\Rightarrow x = \cfrac{y+\sqrt{y^2-4}}{2}=f^{-1}(y)$$
Hence $$f^{-1}(x) = \cfrac{x+\sqrt{x^2-4}}{2}$$
Note: -ve term in step two is discarded using given domain and co-domain.
$$f:R\rightarrow R$$ is a function defined by $$f(x)=10x-7$$. If $$g=f^{-1}$$, then $$g(x)$$ is equals
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$$\dfrac{1}{10x-7}$$
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$$\dfrac{1}{10x+7}$$
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$$\dfrac{x+7}{10}$$
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$$\dfrac{x-7}{10}$$
Explanation
$$y=10x-7$$
It is a one one function over the given domain.
$$y+7=10(x)$$
$$x=\dfrac{y+7}{10}$$
Hence
$$f^{-1}$$ will be $$g(x)$$
$$g(x)=\dfrac{x+7}{10}$$
Hence, option 'C' is correct.
Set $$A$$ has $$3$$ elements and set $$B$$ has $$4$$ elements. The number of injections that can be defined from $$A$$ to $$B$$ is
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$$144$$
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$$12$$
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$$24$$
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$$64$$
Explanation
If $$f:R\rightarrow R$$ such that $$f(x)=log_3 x$$ then $$f^{-1} x$$ is equal to
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$$log x^3$$
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$$3^x$$
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$$3^{-x}$$
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$$3^{1/x}$$
Explanation
$$f(x)$$ is invertible for all positive $$x$$.
$$y=log_{3}(x)$$
Taking antilogarithm on both sides, we get
$$3^{y}=x$$.
Hence,
$$f^{-1}(x)=3^{x}$$
Hence, option 'B' is correct.
The inverse of $$\displaystyle f\left ( x \right )=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$ is
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$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{1+x}{1-x} \right )$$
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$$\displaystyle \frac{1}{6}\log_{10}\left ( \frac{x}{1-x} \right )$$
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$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1+x}{1-x} \right )$$
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$$\displaystyle \frac{1}{6}\log_{e}\left ( \frac{1-x}{1+x} \right )$$
Explanation
$$\displaystyle y=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$
$$\Rightarrow \displaystyle y=\frac{e^{6x}-1}{e^{6x}+1}$$
$$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{e^{6x}-1+e^{6x}+1}{e^{6x}-1-e^{6x}-1}$$
$$\Rightarrow \displaystyle \frac{y+1}{y-1}=\frac{2e^{6x}}{-2}$$
$$\Rightarrow \displaystyle \frac{1+y}{1-y}=e^{6x}$$
Taking $$\log_e$$ we get $$\displaystyle \log_{e}\left(\frac{1+y}{1-y}\right)=6x$$
$$\displaystyle x=\frac{1}{6}\left(\log_{e}\left(\frac{1+y}{1-y}\right)\right)$$
Hence $$f^{-1}$$$$\displaystyle =\frac{1}{6}\left(\log_{e}\left(\frac{1+x}{1-x}\right)\right)$$
If $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}$$ and $$\displaystyle Y= \left \{ 1,3,5,7,9 \right \}$$ then which of the following sets are relation from $$X$$ to $$Y$$
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$$\displaystyle R_{1}= \left \{ (x,a):a= x+2,x\in X,a\in Y \right \} $$
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$$\displaystyle R_{2}= \left \{ (1,1),(2,1),(3,3),(4,3),(5,5) \right \} $$
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$$\displaystyle R_{3}= \left \{ (1,1),(1,3),(3,5),(3,7),(5,7) \right \} $$
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$$\displaystyle R_{4}= \left \{ (1,3),(2,5),(4,7),(5,9),(3,1) \right \} $$
Explanation
$${ R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 2,4 \right) ,\left( 3,5 \right) ,\left( 4,6 \right) ,\left( 5,7 \right) \right\} $$
since $$4$$ and $$6$$ do not belong to $$y$$.
$$\therefore \left( 2,4 \right) ,\left( 4,6 \right) \notin { R }_{ 1 }$$
$$\therefore { R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 3,5 \right) ,\left( 5,7 \right) \right\} \subset X\times Y$$
Hence, $${ R }_{ 1 }$$ is a relation but not a mapping as the elements $$2$$ and $$4$$ do not have any image.
$${ R }_{ 2 }:$$ It is certainly a mapping and since every mapping is a relation, it is a relation as well.
$${ R }_{ 3 }:$$ It is a relation being a subset of $$X\times Y$$.
$${ R }_{ 4 }:$$ It is both mapping and a relation. Each element in $$X$$ has a unique image. It is also one-one and on-to mapping and hence a bijection.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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