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CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Functions
Quiz 9
Let f : $$R\rightarrow R$$ be a function defined by f(x) = $${ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .$$ Then f is.
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one-one & onto
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one-one & into
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many one & onto
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many one & into
Explanation
$$f(x)=x^{3}+x^{2}+3 x+\sin x$$
$$Suppose$$
$$f(x_{1})=f(x_{2})$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$$
$$Equating\;above\;equations\;to \;0$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$$
$$Solving\;them$$
$$It\; will\; not\; prove\; that$$
$$x_{1}=x_{2}$$
$$So,\;f(x)\;is\;one-one$$
$$Range\;f(x)=Real$$
$$Co-domain\;f(x)=Real$$
$$Range=Co-domain$$
$$Hence,\; f(x) \;is \;onto$$
$$So,\;option\; C \;is\;correct.$$
The domain of the definition of the function
$$f(x)=\cfrac { 1 }{ 4-{ x }^{ 2 } } +\log _{ }{ \left( { x }^{ 3 }-x \right) } $$ is
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$$\left( 1,2 \right) \cup \left( 2,\infty \right) $$
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$$\left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 3,\infty \right) $$
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$$\left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 2,\infty \right) $$
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$$\left( -2,-1 \right) \cup \left( -1,0 \right) \cup \left( 2,\infty \right) $$
Explanation
$$4-{ x }^{ 2 }\neq 0;{ x }^{ 3 }-x>0\quad $$
$$x=\pm 2;x(x-1)(x+1)>0$$
$$\therefore { D }_{ f }\in \left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 2,\infty \right) $$
If working set window is too large _____________________.
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it will not encompass entire locality
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it may overlap several localities
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it will cause memory problems
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none of the mentioned
Time complexity to check if an edge exists between two vertices would be __________.
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O(V*V)
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O(V+E)
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O(1)
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O(E)
What is the definition for Ackermann's function?
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A(1,i) = i+1 for i>=1
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A(i,j) = i+j for i>=j
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A(i,j) = i+j for i = j
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A(1,i) = i+1 for i<1
Let $$f:R\rightarrow R$$ be defined by $$f(x)=\left|\begin{matrix} 2x & x > 3\\ x^2 & 1 < x \leq 3\\ 3x & x\leq 1\end{matrix}\right.$$. Then $$f(-1)+f(2)+f(4)$$ is?
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$$9$$
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$$14$$
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$$5$$
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$$10$$
Explanation
$$f(-1)+f(2)+f(4)$$
$$=-3+4+8=9$$.
Following code snippet is the function to insert a string in a trie. Find the missing line.
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node = node.children[index];
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node = node.children[str.charAt(i + 1)];
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node = node.children[index++];
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node = node.children[index+++];
Function which is used to read strings is __________.
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getch()
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getc()
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getstr()
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gets()
A function inside another function is called a _____ function.
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Nested
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Sum
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Text
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None of these
LetN be the set of natural numbers and two functions f and g be defined as $$ f, g: N \rightarrow N $$ such that$$f ( n ) = \left\{ \begin{array} { l l } { \frac { n + 1 } { 2 } } & { \text { if } n \text { is odd } } \\ { \frac { n } { 2 } } & { \text { if } n \text { is even } } \end{array} \right.$$
and $$ g ( n ) = n - ( - 1 ) ^ { n } . $$ Then fog is :
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onto but not one-one.
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one-one but not onto.
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both one-one and onto.
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neither one-one nor onto.
A function $$f$$ from the set of natural numbers to integers defined by $$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$$ is
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neither one-one nor onto
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one-one but not onto
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onto but not one-one
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one-one and onto both
Explanation
$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be any two elements in the domain $$(N).$$
$$Case\,I:$$ When both $$x_1$$ and $$x_2$$ are even.
Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$-x_1=-x_2$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,II:$$ When both $$x_1$$ and $$x_2$$ are odd.
Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$$
$$\Rightarrow$$ $$x_1-1=x_2-1$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,III:$$ When $$x_1$$ be even and $$x_2$$ be odd.
Then, $$f(x_1)=\dfrac{-x_1}{2}$$ and $$f(x_2)=\dfrac{x_2-1}{2}$$
Then clearly,
$$\Rightarrow$$ $$x_1\ne x_2$$
$$\Rightarrow$$ $$f(x_1)\ne f(x_2)$$
From, all the cases, we can say that, $$f$$ is one-one.
$$onto$$ test of $$f:$$
Co-domain of $$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$$
Range of $$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$$
Range of $$f=\{...,-2,1,-1,0,0,-1,1,..\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
The function $$f:A\rightarrow B$$ defined by $$f(x)=-{ x }^{ 2 }+6x-8$$ is a bijection, if
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$$A=(-\infty ,3];B=(-\infty ,1]$$
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$$A=[-3,\infty );B=(-\infty ,1]$$
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$$A=(-\infty ,3];B=[1,\infty )$$
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$$A=[3,\infty );B=[1,\infty )$$
Explanation
$$f(x)=-x^2+6x-8$$
We can see function is a polynomial and the domain of polynomial function is real number.
$$\therefore$$ $$x\in R$$
$$f(x)=-x^2+6x-8$$
$$=-(x^2-6x+8)$$
$$=-(x^2-6x+9-1)$$
$$=-(x-3)^2+1$$
Maximum value of $$-(x-3)^2$$ would be $$0$$
$$\therefore$$ Maximum value of $$-(x-3)^2+1$$ would be $$1.$$
$$\therefore$$ $$f(x)\in(-\infty,1]$$
We can see from the given graph that function is symmetrical about $$x=3$$ and the given function is bijective.
So, $$x$$ would be either $$(-\infty,3]$$ or $$[3,\infty)$$
The correct option which satisfy $$A$$ and $$B$$ both is:
$$A=(-\infty,3]$$ and $$B=(-\infty,1]$$
Which of the following functions from $$Z$$ to itself 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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$$f(x)={X}^{2}+x$$
Explanation
$$(A)$$ $$f(x)=x^3$$
$$one-one$$ test:
Let $$x_1$$ and $$x_2$$ be the elements in the domain such that,
$$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$x_1^3=x_2^3$$
$$\Rightarrow$$ $$x_1=x_2$$
$$\therefore$$ $$f$$ is one-one.
$$onto$$ test :
Let $$y$$ be an element in the co-domain $$(Z),$$ such that,
$$f(x)=y$$
$$\Rightarrow$$ $$x^3=y$$
Consider $$y=3$$
$$\Rightarrow$$ $$x^3=3$$
$$\Rightarrow$$ $$x=\sqrt[3]{3}\notin Z$$
$$\therefore$$ $$f$$ is not onto.
$$\therefore$$ $$f$$ is not bijection.
$$(B)$$ $$f(x)=x+2$$
$$one-one$$ test:
Let $$x_1$$ and $$x_2$$ be the elements in the domain such that,
$$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$x_1+2=x_2+2$$
$$\Rightarrow$$ $$X_1=x_2$$
$$\therefore$$ $$f$$ is one-one.
$$onto$$ test :
Let $$y$$ be an element in the co-domain $$(Z),$$ such that,
$$f(x)=y$$
$$\Rightarrow$$ $$x+2=y$$
$$\Rightarrow$$ $$x=y-2\in Z(Domain)$$
$$\therefore$$ $$f$$ is onto
$$\therefore$$ $$f$$ is bijection.
$$(C)$$ $$f(x)=2x+1$$
$$one-one$$ test:
Let $$x_1$$ and $$x_2$$ be the elements in the domain such that,
$$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$2x_1+1=2x_2+1$$
$$\Rightarrow$$ $$2x_1=2x_2$$
$$\Rightarrow$$ $$x_1=x_2$$
$$\therefore$$ $$f$$ is one-one.
$$onto$$ test :
Let $$y$$ be an element in the co-domain $$(Z),$$ such that,
$$f(x)=y$$
$$\Rightarrow$$ $$2x+1=y$$
Consider $$y=4$$
$$\Rightarrow$$ $$2x+1=4$$
$$\Rightarrow$$ $$2x=3$$
$$\Rightarrow$$ $$x=\dfrac{3}{2}\notin Z$$
$$\therefore$$ $$f$$ is not onto.
$$\therefore$$ $$f$$ is not bijection.
$$(D)$$ $$f(x)=x^2+x$$
$$\Rightarrow$$
$$f(0)=(0)^2+0=0$$
$$\Rightarrow$$
$$f(-1)=(-1)^2+(-1)=1-1=0$$
We can see that $$0$$ and $$-1$$ have the same image.
$$\therefore$$ $$f$$ is not on-one.
$$\therefore$$ $$f$$ is not bijection.
Let $$M$$ be the set of all $$2\times 2$$ matrices with entries from the set $$R$$ of real numbers. Then the function $$f:M\rightarrow R$$ defined by $$f(A)=\left| A \right| $$ for every $$A\in M$$, is
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one-one and onto
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neither one-one nor onto
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one-one but not onto
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onto but not one-one
Explanation
$$M=\left\{A=\begin{bmatrix} a & b\\ c& d\end{bmatrix}: a,b,c,d\in R\right\}$$
$$f:M\rightarrow R$$ is given by $$f(A)=|A|$$
$$Injectivity:$$
$$f\left(\begin{bmatrix} 0 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 0&0\\0&0\end{vmatrix}=0$$
and
$$f\left(\begin{bmatrix} 1 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 1&0\\0&0\end{vmatrix}=0$$
$$\Rightarrow$$ $$f\left(\begin{bmatrix} 0&0\\0&0\end{bmatrix}\right)=f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)=0$$
So, $$f$$ is not one-one.
$$Surjectivity:$$
Let
$$f(A)=y,\,A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
$$\Rightarrow$$ $$\begin{vmatrix} a& b\\c&d\end{vmatrix}=y$$
$$\Rightarrow$$ $$ad-bc=y$$
$$\Rightarrow$$ $$a,b,c,d\in R$$ implies $$y\in R$$
$$\Rightarrow$$ CoDomain=Range.
$$\therefore$$ $$f$$ is onto.
If a function $$f:(2,\infty )\rightarrow B$$ defined by $$f(x)={ x }^{ 2 }-4x+5$$ is a bijection, then $$B=$$
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$$R$$
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$$[1,\infty)$$
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$$(0,1]$$
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$$[0,1)$$
Explanation
Since, $$f$$ is a bijection,
Co-domain of $$f=$$ Range of $$f$$
$$\Rightarrow$$ $$B=$$ range of $$f$$
$$\Rightarrow$$ $$f(x)=x^2-4x+5$$
Let $$f(x)=y$$
$$\Rightarrow$$ $$y=x^2-4x+5$$
$$\Rightarrow$$ $$x^2-4x+(5-y)=0$$
$$\because,\,D=b^2-4ac\ge 0,$$
$$\Rightarrow$$ $$(-4)^2-4\times 1\times (5-y)\ge 0$$
$$\Rightarrow$$ $$16-20+4y\ge 0$$
$$\Rightarrow$$ $$4y\ge 4$$
$$\Rightarrow$$ $$y\ge 1$$
$$\Rightarrow$$ $$y\in [1,\infty)$$
$$\Rightarrow$$ Range of $$f=[1,\infty)$$
$$\Rightarrow$$ $$B=[1,\infty)$$
Let $$f:Z\rightarrow Z$$ be given by $$f(x)=\begin{cases} \cfrac { x }{ 2 } ,\quad \text{if}\ x \ \text{is even} \\ 0,\quad \text{if }\ x \ \text{is odd} \end{cases}$$. Then, $$f$$ is
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onto but not one-one
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one-one but not onto
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one-one and onto
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neither one-one nor onto
Explanation
$$Injectivity:$$
Let $$x_1$$ and $$x_2$$ be two elements in the domain $$(Z)$$, such that,
$$f(x_1)=f(x_2)$$
$$Case\,I:$$
Let both $$x_1$$ and $$x_2$$ be even.
Then, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,II:$$
Let both $$x_1$$ and $$x_2$$ be odd.
Then, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$0=0$$
Here, we cannot determine whether $$x_1=x_2.$$
So, $$f$$ is not one-one.
$$Surjectivity:$$
Let $$y$$ be an element in the co-domain $$(Z)$$, such that
Co-domain of $$f=Z=\{0,\pm 1,\pm 2,\pm 3,\pm 4,...\}$$
Range of $$f=\left\{0,0,\dfrac{\pm 2}{2},0,\dfrac{\pm 4}{2},...\right\}=\{0,\pm 1,\pm 2,....\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
.
Which of the following functions from $$A=\left\{ x:-1\le x\le 1 \right\} $$ to itself are bijections?
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$$f(x)=\cfrac { x }{ 2 } $$
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$$g(x)=\sin { \left( \cfrac { \pi x }{ 2 } \right) } $$
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$$h(x)=\left| x \right| $$
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$$k(x)={ x }^{ 2 }$$
Explanation
$$(A)$$ Range of $$f=\left[\dfrac{-1}{2},\dfrac{1}{2}\right]\ne A$$
So, $$f$$ is not a bijection.
$$(B)$$ Range $$=\left[\sin\left(\dfrac{-\pi}{2}\right),\sin\left(\dfrac{\pi}{2}\right)\right]=[-1,1]=A$$
So, $$g$$ is a bijection.
$$(C)$$ $$h(-1)=|-1|=1$$
And $$h(1)=|1|=1$$
$$\Rightarrow$$ $$-1$$ and $$1$$ have the same images
So, $$h$$ is not a bijection.
$$(D)$$ $$k(-1)=(-1)^2=1$$
And $$k(1)=(1)^2=1$$
$$\Rightarrow$$ $$-1$$ and $$1$$ have the same images.
So, $$k$$ is not a bijection.
The function $$f:\left[ -\dfrac {1}{2},\dfrac {1}{2} \right] \rightarrow \left[ -\dfrac {\pi }{2},\dfrac {\pi }{2} \right] $$ defined by $$f(x)=\sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } $$ is
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bijection
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injection but not a surjection
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surjection but not an injection
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neither injection nor a surjection
Explanation
$$f(x)=\sin^{-1}(3x-4x^3)$$
$$f(x)=3\sin^{-1}x$$
$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be two elements in the domain $$\left[\dfrac{-1}{2},\dfrac{1}{2}\right],$$ such that,
$$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$3\sin^{-1}x=3\sin^{-1}x_2$$
$$\Rightarrow$$ $$\sin^{-1}x=\sin^{-1}x_2$$
$$\Rightarrow$$ $$x_1=x_2$$
$$\therefore$$ $$f$$ is one-one.
$$onto$$ test :
Let $$y$$ be element in the co-domain, such that
$$f(x)=y$$
$$\Rightarrow$$ $$3sin^{-1}(x)=y$$
$$\Rightarrow$$ $$\sin^{-1}(x)=\dfrac{y}{3}$$
$$\Rightarrow$$ $$x=\sin\dfrac{y}{3}\in\left[\dfrac{-1}{2},\dfrac{1}{2}\right]$$
$$\therefore$$ $$f$$ is onto.
Since it is one-one and onto both then, $$f$$ is a bijection.
If $$g(x)={ x }^{ 2 }+x-2$$ and $$\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2$$, then $$f(x)$$ is equal to
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$$2x-3$$
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$$2x+3$$
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$$2{x}^{2}+3x+1$$
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$$2{x}^{2}-3x-1$$
Explanation
We will solve this problem by the trial and error method.
Let us check option $$A$$ first.
If $$f(x)=2x-3$$
$$g(x)=x^2+x-2$$ [ Given ]
$$\Rightarrow$$ $$\dfrac{1}{2}(g\circ f)(x)=g[f(x)]$$
$$=\dfrac{1}{2}g(2x-3)$$
$$=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]$$
$$=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]$$
$$=\dfrac{1}{2}[4x^2-10x+4]$$
$$=2x^2-5x+2$$
The given condition is satisfied by $$A.$$
A function $$f$$ from the set of natural numbers to the set of integers defined by
$$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$$
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neither one-one nor onto
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one-one but not onto
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onto but not one-one
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one-one and onto both
Explanation
$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be any two elements in the domain $$(N).$$
$$Case\,I:$$ When both $$x_1$$ and $$x_2$$ are even.
Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$-x_1=-x_2$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,II:$$ When both $$x_1$$ and $$x_2$$ are odd.
Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$$
$$\Rightarrow$$ $$x_1-1=x_2-1$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,III:$$ When $$x_1$$ be even and $$x_2$$ be odd.
Then, $$f(x_1)=\dfrac{-x_1}{2}$$ and $$f(x_2)=\dfrac{x_2-1}{2}$$
Then clearly,
$$\Rightarrow$$ $$x_1\ne x_2$$
$$\Rightarrow$$ $$f(x_1)\ne f(x_2)$$
From, all the cases, we can say that, $$f$$ is one-one.
$$onto$$ test of $$f:$$
Co-domain of $$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$$
Range of $$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$$
Range of $$f=\{...,-2,1,-1,0,0,-1,1,..\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
If $$n \geq 2$$ then the number of surjections that can be defined from $$\{1, 2, 3, ....... n\}$$ onto $$\{1, 2\}$$ is
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$$2n$$
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$$^nP_2$$
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$$2^n$$
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$$2^{n}-2$$
Explanation
Let $$A=\{1,2,3,...,n\} $$ and $$B=\{1,2\} $$ .
Therefore, total number of mappings from $$A$$ to $$B$$ is $$2^n$$ of which two functions $$f(x)=1$$ for all $$x\in A$$ and $$g(x)=2$$ for all $$x\in A$$ are not surjective.
Thus, the total number of surjections from $$A$$ to $$B$$ is $$2^n-2$$.
The function $$f:R\rightarrow R$$ defined by $$f(x)=(x-1)(x-2)(x-3)$$ is
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one-one but not onto
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onto but not one-one
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both one and onto
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neither one-one nor onto
Explanation
$$f(x)=(x-1)(x-2)(x-3)$$
$$one-one$$ test:
$$\Rightarrow$$ $$f(1)=(1-1)(1-2)(1-3)=0$$
$$\Rightarrow$$ $$f(2)=(2-1)(2-2)(2-3)=0$$
$$\Rightarrow$$ $$f(3)=(3-1)(3-1)(3-3)=0$$
$$\Rightarrow$$ $$f(1)=f(2)=f(3)=0$$
We can see, $$1,2,3$$ has same image $$0.$$
$$\therefore$$ $$f$$ is not one-one.
$$onto$$ test:
Let $$y$$ be an element in the co-domain $$R,$$ such that
$$y=f(x)$$
$$\Rightarrow$$ $$y=(x-1)(x-2)(x-3)$$
Since, $$y\in R$$ and $$x\in R$$.
$$\therefore$$ $$f$$ is onto.
If $$g(x)=x^2+x-1$$ and
$$(gof)(x)=4x^2-10x+5$$, then
$$f\left(\dfrac{5}{4}\right)$$ is equal to:
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$$\dfrac{3}{2}$$
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$$\dfrac{1}{2}$$
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$$-\dfrac{3}{2}$$
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$$-\dfrac{1}{2}$$
Explanation
$$g(x)=x^2+x-1$$
$$g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}$$
$$g\left(f\left(\dfrac{5}{4}\right)\right)=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$
$$-\dfrac{5}{4}=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$
$$f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0$$
$$\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0$$
$$\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}$$
Let $$S$$ be the set of all real roots of the equation, $${ 3 }^{ x }\left( { 3 }^{ x }-1 \right) +2=\left| { 3 }^{ x }-1 \right| +\left| { 3 }^{ x }-2 \right| $$. Then $$S$$:
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contains at least four elements
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is a singleton
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is an empty set
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contains exactly two elements
Explanation
$$3^x (3^x - 1) + 2 = |3^x - 1| + |3^x - 2|$$
Let $$3^x = t$$
So, $$t (t - 1) + 2 = | t - 1| + |t - 2|$$
$$\Rightarrow t^2 - t + 2 = |t - 1|+ |t - 2|$$
$$f(t)=t^2-t+2 \quad\quad\text{[green curve]}$$
$$g(t)=|t-1|+|t+2|\quad\quad\text{[blue curve]}$$
we plot $$f(t)=t^2-t+2$$ and $$g(t)=|t-1|+|t+2|$$
$$\text{these two curves intersect at two points but only one root is positive.}$$
$$\text{As} \,3^x \,\text{is always positive, therefore only positive value of} \,t\, \text{will be the solution.}$$
$$\text{Therefore, we have only one solution.}$$
If $$f(x) = \dfrac{x+1}{x-1}$$, then the valueof $$f(f(x))$$ is equal to
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$$x$$
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$$0$$
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$$-x$$
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$$1$$
Explanation
$$f(x)=\dfrac{x+1}{x-1}$$
$$\therefore f(f(x))=f\left(\dfrac{x+1}{x-1}\right)$$
$$\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}$$
$$=\dfrac{x+1+x-1}{x+1-x+1}$$
$$=\dfrac{2x}{2}$$
$$=x$$
Let $$f : x \rightarrow y $$ be such that $$f(1) = 2$$ and $$f(x + y) = f(x) f(y)$$ for all natural numbers x and y. If $$\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)$$ , then a is equal to
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$$3$$
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$$4$$
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$$5$$
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$$6$$
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$$7$$
Explanation
We have,
$$f(1)=2$$ and $$f(x+y)=f(x).f(y)$$
Now, $$f(2)=f(1+1)=f(1).f(1)=2.2=2^2$$
$$f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3$$
and so on
$$\therefore f(x)=2^n$$......(i)
Now, we have
$$\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)$$
$$\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)$$
$$\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)$$
$$\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)$$
$$\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)$$
$$\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)$$
$$\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8$$
$$\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]$$
$$\Rightarrow 2^a=2^3=a=3$$
If $$f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}$$ then $$(f o f)(x)=?$$
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0%
$$x$$
0%
$$(2x-3)$$
0%
$$\dfrac{4x-6}{3x+4}$$
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None of these
If $$f(x)=\sqrt[3]{3-x^3}$$ then $$(f o f)(x)=?$$
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0%
$$x^{1/3}$$
0%
$$x$$
0%
$$(1-x^{1/3})$$
0%
None of these
Let $$ f : R \rightarrow R : f(x) =x +1 $$ and $$ g : R \rightarrow R : g(x) = 2x -3 $$.
Find $$(f +g) (x)$$.
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0%
$$3x -2$$
0%
$$4x -5$$
0%
$$3x -4$$
0%
$$2x -3$$
Explanation
Given ,
$$f(x)=x+1,g(x)=2x-3$$
$$\implies (f+g)x=f(x)+g(x)=x+1+2x-3=3x-2$$
If $$\displaystyle f(x) = | x - 2 |$$ and $$ g(x) = fof\,(x) $$ , then for $$ x > 20 , {g}\,'(x) = $$
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0%
$$ 2 $$
0%
$$ 1 $$
0%
$$ 3 $$
0%
None of these
If $$\displaystyle {f}'(x) = g\,(x) $$ and $$\displaystyle {g}'(x) = - f\,(x) $$ for all $$ x $$ and $$ f\,(2) = 4 = {f}'(2) $$ then $$\displaystyle f^{2}\,(19) + g^{2} \,(19) $$ is
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0%
$$ 16 $$
0%
$$ 32 $$
0%
$$ 64 $$
0%
None of these
The value of f(0), so that the function
f(x) = $$ \dfrac{2x-sin^{-1}x}{2x+tan^{-1}x} $$ is continuous at each point in its domain, is equal to
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0%
2
0%
1/3
0%
2/3
0%
-1/3
Explanation
The function f is clearly continuous at each point in its domain except possibly at x=0 Given that f(x) is continuous at x=0
Therfore,f (0) = $$ \underset{x\rightarrow 0}{lim}f(x) $$
$$ =\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x} $$
$$ = \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}$$
$$\dfrac{2-1}{2+1}=\dfrac13$$
let $$f(x) = sin^2 x/2 + cos ^2 x/2 $$ and $$g(x) = sec^2 x - tan ^2 x.$$ The two functions are equal over the set
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0%
$$\phi$$
0%
$$R$$
0%
$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$
0%
None of these
Let $$f(n)$$ denote the number of different ways in which the positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$. For example, $$f(4) = 5$$, since $$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$. Note that order of $$1s$$ and $$2s$$ is important.
$$f : N\rightarrow N$$ is
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0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
Many-one and into
Explanation
$$6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)$$
Number of $$2s$$
Number of $$1s$$
Number of permutations
$$0$$
$$6$$
$$1$$
$$1$$
$$4$$
$$\dfrac {5!}{4!} = 5$$
$$2$$
$$2$$
$$\dfrac {4!}{2!2!} = 6$$
$$3$$
$$0$$
$$\dfrac {3!}{3!} = 1$$
$$Total = 13$$
$$\therefore f(6) = 13$$
Now, $$f(f(6)) = f(13)$$
Number of $$1s$$
Number of $$2s$$
Number of permutations
$$13$$
$$0$$
$$1$$
$$11$$
$$1$$
$$\dfrac {12!}{11!} = 12$$
$$9$$
$$2$$
$$\dfrac {11!}{9!2!} = 55$$
$$7$$
$$3$$
$$\dfrac {10!}{7!3!} = 120$$
$$5$$
$$4$$
$$\dfrac {9!}{5!4!} = 126$$
$$3$$
$$5$$
$$\dfrac {8!}{3!5!} = 56$$
$$1$$
$$6$$
$$\dfrac {7!}{6!} = 7$$
$$Total = 377$$
$$\therefore f(f(6)) = f(13) = 377$$
$$f(1) = 1(1)$$
$$f(2) = 2 (1, 1$$ or $$2)$$
$$f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)$$
$$f(4) = 5$$ (explained in the paragraph)
By taking higher value of $$n$$ in $$f(n)$$, we always get more value of $$f(n)$$. Hence, $$f(x)$$ is one-one. Clearly, $$f(x)$$ is into.
The domain of the function $$\sqrt { (\log\ 5x) } $$ is
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0%
$$(1, \infty )$$
0%
$$(0, \infty )$$
0%
$$(0, 1 )$$
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$$(0.5 , 1 )$$
The function $$f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0 $$ , is continuous at $$x=0$$. Then the value of $$f(0)$$ is
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0%
2log 3
0%
$$ (\log_{e}3)^{2} $$
0%
$$ \log_{e} 6 $$
0%
None of these
Explanation
Given f(x) is continuous at $$x=0$$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0) $$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$$
$$ \Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$$ $$ (\log_e3)^{2} $$
The domain of the function $$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$$ is
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0%
$$- \infty < x < \infty$$
0%
$$1 \leqslant x \leqslant 4$$
0%
$$ 4 \leqslant x \leqslant 16$$
0%
$$ -1 \leqslant x \leqslant 1$$
Explanation
We have $$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$$
From (1), clearly f(x) is defined for those values of x for
which $$log_{10} \left [\frac{5x - x^{2}} {4} \right] \geq 0 $$
$$ \Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 10^{0}$$
$$ \Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 1$$
$$\Rightarrow x^{2} - 5x + 4 \leq 0$$
$$\Rightarrow (x - 1) (x - 4) \leq 0$$
Hence the domain of the function is [1, 4]
The domain of $$f(x) = \log| \log {x}|$$ is
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0%
$$\left(0, \infty\right )$$
0%
$$\left(1, \infty\right )$$
0%
$$\left(0, 1\right )\cup \left(1, \infty \right )$$
0%
$$\left(-\infty, 1\right )$$
Explanation
$$f(x) = \log| \log {x}|$$, f(x) is defined if $$|log \space x| > 0$$ and x > 0, i.e.,
if x > 0 and $$ x \neq 1$$ $$(\therefore |log \space x| > 0 \space if \space x \neq 1)$$
$$ \Rightarrow x \epsilon (0, 1) \cup (1, \infty)$$
Which of the following is not true about $$h_1(x)$$?
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It is periodic function with period $$\pi$$
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Range is [0,1]
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Domain if R
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None of these
If f: $$R\rightarrow R$$ be given by $$f(x) = 3 + 4x$$ and $$a_n = A + Bx$$, then which of the following is not true?
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A + B + 1 = $$2^{2n + 1}$$
0%
| A - B| = 1`
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$$\displaystyle \lim_{n \to \infty} \dfrac{A}{B} = -1$$
0%
None of these
Explanation
Since $$a_1 = g(x) = 3 + 4x$$
$$\therefore a_2 = g\{g^2(x)\} = g(3+4x) = 3 + 4(3+4x) = (4^2 - 1) + 4^x$$
$$a_3 = g\{g^(x)\} = g(15 + 4^2x) = 3 + 4 (15 + 4^2x) = 63 + 4^3x = (4^3 - 1) + 4^3 x$$
Similarly, we get $$a_n = (4^n - 1) + 4^n x$$
$$\Rightarrow A = 4^n - 1 \space and B = 4^n$$
$$\Rightarrow A + B + 1 = 2^{2n + 1}, |a - b| = 1\space and\space lim_{n \to \infty}\dfrac{4^n - 1}{4^n}$$
$$= lim_{n \to \infty}\left(1 - \dfrac{1}{4^n}\right) = 1$$
Let $$ f(x) + f(y) = f(x\sqrt{1 - y^2} + y\sqrt{1 - x^2})$$ (f(x) is not identically zero) the.
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$$f(4x^3 - 3x) + 3f(x) = 0$$
0%
$$f(4x^3 - 3x) = 3f(x)$$
0%
$$f(2x\sqrt{1-x^2}) + 2f(x) = 0$$
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$$f(2x\sqrt{1-x^2}) = 2f(x)$$
Consider two functions $$f(x) = \begin{cases} [x], -2 \leq x \leq -1\\ |x| + 1, -1 < x \leq 2 \end{cases}$$ and g(x) = \begin{cases} [x], -\pi \leq x < 0 \\ sin x, 0 \leq x \leq \pi \end{cases} where [.] denotes thegreatest integer function
The exhaustive domain of g(f(x)) is
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0%
[0,2]
0%
[-2, 0]
0%
[-2,2]
0%
[-1,2]
The graph of y = g(x) in its domain is broken at
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0%
1 point
0%
2 point
0%
3 point
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None of these
Let $$g(x) = f(x) - 1$$. If $$f(x) + f(1 - x) = 2 \space \forall \space x \space \epsilon \space R$$, then $$g(x)$$ is symmetrical about
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0%
Origin
0%
The line $$x = \frac{1} {2}$$
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The point $$\left (1, 0 \right )$$
0%
The point $$\left (\frac {1} {2}, 0 \right )$$
Domain (D) and range (R) of $$f(x) = \sin^{-1}\left (\cos^{-1} [x] \right )$$ where [.] denotes the greatest integer function is
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0%
$$D \equiv x \epsilon \left [ 1, 2 \right ), R \epsilon \left \{0 \right \}$$
0%
$$D \equiv x \epsilon \left [ 0, 1 \right ], R \equiv \left \{-1, 0, 1 \right \}$$
0%
$$D \equiv x \epsilon \left [ -1, 1 \right ), R \epsilon \left \{0, \sin^{-1} \left (\frac{\pi} {2}\right), \sin^{-1} (\pi) \right \}$$
0%
$$D \equiv x \epsilon \left [-1, 1 \right ), R \epsilon \left \{-\frac{\pi} {2}, 0 , \frac{\pi} {2} \right \}$$
g(f(x)) is not defined if
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0%
$$a\space \epsilon ( 10, \infty) b\space \epsilon (5, \infty)$$
0%
$$a\space \epsilon ( 4, 10) b\space \epsilon (5, \infty)$$
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$$a\space \epsilon ( 10, \infty) b\space \epsilon (0, 1)$$
0%
$$a\space \epsilon ( 4, 10) b\space \epsilon (1, 5)$$
Let $$A=\left\{0,1\right\}$$ and $$N$$ be the set of natural numbers. Then, the mapping $$f:N\to A$$ defined by $$f(2n-1)=0, f(2n)=1, \forall n \in R$$, is onto.
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0%
True
0%
False
Explanation
True
Given
A
=
{
0
,
1
}
A={0,1}
is onto.
For set $$A, B$$ and $$C$$, let $$f:A\to B, g:B\to C$$ be functions such that $$gof $$ is surjective.
Then $$g$$ is surjective function.
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0%
True
0%
False
Explanation
Suppose that $$g\circ f$$ is surjective.
Let $$z∈C$$.
Then since $$g\circ f$$ is surjective, there exists $$x\in A$$ such that $$(g\circ f)(x) =g(f(x)) =z$$.
Therefore if we assume $$y=f(x)\in B$$,
then $$g(y) =z$$.
Thus $$g$$ is surjective
Let $$A$$ be a finite set. Then, each injective function from $$A$$ into itself is not surjective.
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0%
True
0%
False
Explanation
False
Let $$f:A\to B$$ and $$g:B\to C$$ be the bijective function. Then $$(gof)^{-1}$$ is
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0%
$$f^{-1}og^{-1}$$
0%
$$fog$$
0%
$$g^{-1} of^{-1}$$
0%
$$gof$$
Explanation
Given that, $$f:A\to B$$ and $$g:B\to C$$ be the bijective functions.
$$(gof)^{-1}=f^{-1}og^{-1}$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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