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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 1 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 1
If $$f(x) =x+\tan x$$ and $$f$$ is inverse of $$g$$, then $$g'(x)$$ is equal to
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$$\dfrac{1}{1+[g(x)-x]^2}$$
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$$\dfrac{1}{2-[g(x)+x]^2}$$
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$$\dfrac{1}{2+(x-g(x))^2}$$
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$$None\ of\ these$$
Explanation
Since,
$$f=g^{-1}$$
$$\implies f(g(x))=x$$
$$\Rightarrow f(g(x))=g(x)+\tan(g(x))=x$$
$$\tan (g(x))=x-g(x)$$
Differentiating both sides with respect to $$x$$
$$g'(x)+\sec ^2(g(x))g'(x)=1$$
$$g'(x)=\dfrac{1}{1+\sec ^2(g(x))}=\dfrac{1}{2+\tan ^2(g(x))}$$
$$g'(x)=\dfrac{1}{2+(x-g(x))^2}$$
Hence, option $$C$$ is correct answer.
$$f:R\rightarrow R$$ is a function defined by $$f(x)=10x-7$$. If $$g=f^{-1}$$ then $$g(x)=$$
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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
Given $$f(x)=10x-7$$ and $$g=f^{-1}$$
Let $$f\left ( x \right )=10x-7=y$$
$$\Rightarrow x=\dfrac{y+7}{10}=f^{-1}\left ( y \right )=g\left ( y \right )$$
$$\therefore g\left ( x \right )=\dfrac{x+7}{10}$$
A constant function $$f:A\rightarrow B$$ will be one-one if
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$$n (A) = n(B)$$
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$$n(A) = 1$$
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$$n (B) = 1$$
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$$n (A) < n (B)$$
Explanation
Given f is a constant functions.
$$\Rightarrow$$ range of f is $$\left \{ c \right \}(say)$$
Since f is one-one $$\Rightarrow$$ domain of A should also contain
one element.
$$\therefore n(A)=1.$$
If $$f:\mathbb{N} \rightarrow \mathbb{N}$$ and $$f(x) = x^{2}$$ then the function is
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not one to one function
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one to one function
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into function
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none of these
Explanation
Given : $$f(x)=x^2$$
$$\because x^2>0 \implies f(x)>0$$ for every $$x\in \mathbb{N}$$
Let' find domain and range of $$f(x)$$
$$ x =1\longrightarrow f(x)=1$$
$$x =2\longrightarrow f(x)=4$$
$$x =3\longrightarrow f(x)=9$$
$$ x=4\longrightarrow f(x)=16$$
$$x =5\longrightarrow f(x)=25$$
Here we get that, no two elements of the domain has the same image and no element of co-domain is the image of more than one element in the domain.
$$\therefore$$ $$f$$ is one-one.
Function $$f:X\rightarrow Y$$ is onto if, for any $$y\in {Y}$$ there exist $$x\in {X}$$ such that $$f(x)=y$$
Let's prove that $$f(x)=x^2$$ is not onto.
Let's take an example $$y=3\in \mathbb{N}$$
$$\implies f(x)=y$$ ............ by definition of onto function
$$\implies x^2=3$$
$$\implies x=\sqrt{3}\ or\ -\sqrt3$$ which does not belong to $$\mathbb{N}$$
Hence, for $$y=3\in \mathbb{N}$$ there does not exist any $$x\in \mathbb{N}$$ such that $$f(x)=y$$.
$$\therefore$$ $$f$$ is not onto.
Hence, $$f$$ in only one-one.
$$f(x)=1$$, if $$x$$ is rational and $$f(x)=0$$, if $$x$$ is irrational
then $$(fof) (\sqrt{5})=$$
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$$0$$
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$$1$$
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$$\sqrt{5}$$
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$$\dfrac{1}{\sqrt{5}}$$
Explanation
Given,
$$f(x)=\begin{cases}1 & \text {; if x is rational} \\ 0 & \text{; if x is irrational}\end{cases}$$
Consider $$x=\sqrt{5}$$,
$$\text{(f o f)}(\sqrt{5})=f(f(\sqrt{5}))$$
$$=f(0)$$ .......... $$[\because \sqrt{5}$$ is irrational$$]$$
$$=1$$ .......... $$[\because 0$$ is rational$$]$$
$$\therefore \text{(f o f)}(\sqrt{5})=1$$
Hence, option B is correct.
If $$f(x) = 3x + 2, g(x) = x^2 + 1$$, then the value of $$(fog) (x^2 +1)$$ is
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$$3x^4 + 6x^2 + 8$$
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$$3x^4 + 3x + 4$$
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$$6x^4 + 3x^2 + 2$$
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$$3x^2 + 6x + 2$$
Explanation
We have $$f(x) = 3x + 2, g(x) = x^2 + 1$$
$$fog(x) = f[g(x)] =3(x^2+1)+2= 3x^2 + 5$$
so $$(fog) (x^2 +1) = 3(x^2 + 1)^2 + 5 =3(x^4+2x^2+1)+5= 3x^4+ 6x^2 + 8$$
If $$f:A\rightarrow B $$ is surjective then
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no two elements of $$A$$ have the same image in $$B$$
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every element of $$A$$ has an image in $$B$$
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every element of $$B$$ has at least one pre-image in $$A$$
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$$A$$ and $$B$$ are finite non empty sets
Explanation
Surjective means onto function.
co domain $$=$$ Range
So every element of $$B$$ has at least one pre-image in $$A$$.
If $$f:(0,\infty )\rightarrow (0,\infty )$$ is defined by $$f(x)=x^{2}$$, then $$f^{-1}(x)=$$
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$$\sqrt{x}$$
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$$\dfrac{1}{\sqrt{x}}$$
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Not invertible
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$$\dfrac{2}{\sqrt{x}}$$
Explanation
As $$f$$ is defined from $$\left ( 0,\infty \right )\rightarrow \left ( 0,\infty \right )$$
$$\therefore f$$ is onto and one - one.
Let $$f\left ( x \right )=x^{2}=y$$
$$\Rightarrow x=f^{-1}\left ( y \right )$$
Also, $$x^2=y\Rightarrow x=\sqrt{y}$$ as $$x> 0$$
$$\therefore f^{-1}\left ( y \right )=\sqrt{y}$$
Hence, $$f^{-1}\left ( x \right )=\sqrt{x}$$
$$f:R\rightarrow R , g:R\rightarrow R$$ and $$f(x)= \sin x$$, $$g(x)=x^{2}$$ then $$fog(x)=$$
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$$x^{2}+\sin x$$
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$$x^{2}\sin x$$
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$$\sin^{2}x$$
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$$\sin x^{2}$$
Explanation
$$fog\left ( x \right )=f\left(g ( x \right ))=f (x^{2})$$
$$=\sin x^{2}$$
Find the value of $$\displaystyle \left( g\circ f \right) \left( 6 \right) $$ if $$\displaystyle g\left( x \right) ={ x }^{ 2 }+\frac { 5 }{ 2 } $$ and $$\displaystyle f\left( x \right) =\frac { x }{ 4 } -1$$.
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2.75
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3
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3.5
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8.625
Explanation
$$f(x)=\dfrac {x}{4}-1$$ and $$g(x)=x^2+\dfrac {5}{2}$$
$$\therefore f(6)=\dfrac{6}{4}-1=\dfrac{3}{2}-1=\dfrac{1}{2}$$
$$\therefore (gof)(6)=g(\dfrac{1}{2})=\left(\dfrac{1}{2}\right)^2+\dfrac{5}{2}=\dfrac{11}{4}=2.75$$
The first component of all ordered pairs is called
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Range
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Domain
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Function
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None of these
Explanation
The first components of all order pair is called Domain.
The second components of all ordered pair is called Range.
The second component of all ordered pairs of a relation is
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Range
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Domain
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mapping
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none of these
Explanation
The second components of all ordered pairs of a relation is Range.
A ______ maps elements of one set to another set.
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order
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set
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relation
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function
Explanation
A relation map elements of one set to another set
$$R:A\to B$$
Elements in A is mapped to elements in set B.
Suppose y is equal to the sum of two quantities of which one varies directly as x and the other inversely as x If y = 6 when x = 4 and $$\displaystyle y=\frac{10}{3}$$ when x = 3 then what is the relation between x and y?
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$$\displaystyle y=x+\frac{4}{x}$$
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$$\displaystyle y+2x=\frac{4}{x}$$
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$$\displaystyle y=2x+\frac{8}{x}$$
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$$\displaystyle y=2x-\frac{8}{z}$$
Explanation
(A)If$$y=X+\frac{4}{x}\Rightarrow 4+\frac{4}{4}=3 \neq 6=6$$
(B)$$y+2x=\frac{4}{x}\Rightarrow 6+2(4)=\frac{4}{4}\Rightarrow 6+8\neq 1$$
(C)$$y=2x+\frac{8}{x}\Rightarrow 6=2(4)+\frac{8}{4}\Rightarrow 6\neq 8+2$$
(D)$$y=2x-\frac{8}{x}\Rightarrow 6=2(4)-\frac{8}{4}=8-2=6$$ If x=4 and y=6
$$y=2x-\frac{8}{x}\Rightarrow \frac{10}{3}=2\times 3-\frac{8}{3}\Rightarrow \frac{10}{3}=\frac{10}{3}$$ If x=3 and y=$$y=\frac{10}{3}$$
Then option (D) $$Y=2x-\frac{8}{x}$$
If X is brother of the son of Y's son. How is X related to Y?
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Son
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Brother
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Cousin
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Grandson
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Uncle
Explanation
Son of Y's Son- Grandson, Brother of Y's Grandson- Y's Grandson
Option D is correct.
In the group $$G = \left \{1, 5, 7, 11\right \}$$ under $$\otimes_{12}$$ the value of $$7\otimes_{12} 11^{-1}$$ is equal to
($$\otimes_{12}$$: under multiplication modulo $$12$$)
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$$5$$
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$$7$$
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$$11$$
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$$1$$
Explanation
Clearly $$11^{-1} = 11(\because 11\otimes_{12} 11 = 1)$$
$$\therefore 7 \otimes_{12} 11^{-1} = 7\otimes_{12} 11 = 5$$
What is the relation for the statement "A is taller than B"?
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is taller than
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A is taller
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B is taller
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is less than
Explanation
A is taller than B.
A is related to B by 'taller than'.
x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x =If y = 16 and z = 7 what is x?
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180
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160
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280
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200
Explanation
Given
x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x = 10. If y = 16 and z = 7
Then
$$x=k\frac{y}{z^{2}}$$
$$\Rightarrow 10=k\times \frac{4}{14\times14}$$
$$\Rightarrow k=490$$
Thus
$$x=490\times \frac{16}{7\times 7}=160$$
If $$f: R \rightarrow R$$ and $$g: R \rightarrow R$$ are defined by $$f(x) =2x +3, g(x)=x^2 + 7$$, what are the values of $$x$$ such that $$g(f(x))=8$$?
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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
Given, $$f(x)=2x+3$$
Also given $$ g(x)={ x }^{ 2 }+7$$
Therefore, $$ g[f(x)]={ \left( f(x) \right) }^{ 2 }+7$$
$$ \Rightarrow g[f(x)]={ \left( 2x+3 \right) }^{ 2 }+7$$
Thus $$ { \left( 2x+3 \right) }^{ 2 }+7=8\\ \Rightarrow { \left( 2x+3 \right) }^{ 2 }=1\\ \Rightarrow 2x+3=\pm 1\\ \Rightarrow 2x=\pm 1-3\\ \Rightarrow 2x=-2,-4\\ \Rightarrow x=-1,-2$$
Option C is correct.
Find the correct
expression for $$\displaystyle f\left( g\left( x \right) \right) $$ given that
$$\displaystyle f\left( x \right) =4x+1$$ and $$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$$
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$$\displaystyle -{ x }^{ 2 }+4x+1$$
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$$\displaystyle { x }^{ 2 }+4x-1$$
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$$\displaystyle 4{ x }^{ 2 }-7$$
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$$\displaystyle 4{ x }^{ 2 }-1$$
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$$\displaystyle 16{ x }^{ 2 }+8x-1$$
Explanation
Given, $$g(x)=x^2-2$$
$$f(g(x))=f(x^2-2)$$
Also given, $$f(x)=4x+1$$
$$\therefore f(x^2-2)=4(x^2-2)+1$$
$$\Rightarrow 4x^2-8+1$$
$$\Rightarrow 4x^2-7$$
If $$a,b\in A, a*b\in A$$ then
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$$*$$ is a unary operation in $$A$$
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$$a * b = b * a$$
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$$*$$ is a binary operation in $$A$$
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$$a * b \neq b * a$$
Explanation
If $$a,b\in A,a*b\in A,$$ then $$*$$ is a binary operation in A.
If $$f(x) = \sqrt {x^{2} - 3x + 6}$$ and $$g(x) = \dfrac {156}{x +17}$$, find the value of the composite function $$g(f(4))$$.
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$$5.8$$
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$$7.4$$
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$$7.7$$
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$$8.2$$
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$$10.3$$
Explanation
$$f(x)=\sqrt { x^{ 2 }-3x+6 } ,\quad g(x)=\frac{ 156 }{ x+17 }\\ g(f(4))=\quad ?\\ f(4)=\sqrt { 10 } \\ g(f(4))=g(\sqrt { 10 } )=\quad 3.16\\ g(f(4))=7.7$$
Squaring a given number is a
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relation in some set
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relation
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unary operation
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binary operation
Explanation
Squarring a given number is a Binary operation beacuse squarring two numbers is multiplying the same number twice.
If $$*$$ is a binary operation in $$A$$ then
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$$A$$ is closed under $$*$$
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$$A$$ is not closed under $$*$$
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$$A$$ is not closed under $$+$$
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$$A$$ is closed under $$-$$
Explanation
If $$ *$$ is a binary in $$A$$,then A is closed under $$*$$
$$+$$ is
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binary operation on $$R$$
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not a binary operation on $$R$$
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a binary operation in $$Q^c$$
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not a binary operation in $$ E$$
Explanation
Addition is a binary operation on R since it satisfies the following properties:
It is closed in R, i.e; sum of two real numbers is also real.
It follows Commutativity and Associativity.
Existence of Identity element 0, which when added to a number gives back the same number.
Existence of Additive Inverse. The negative of a number added to that number gives the identity element 0.
$$*$$ is said to be commutative in $$A$$ for all $$a,b\in A$$
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$$a + b = b + a$$
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$$a * b = b * a$$
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$$a - b = b - a$$
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$$a * b \neq b * a$$
Explanation
$$(B) a*b = b*a$$
here $$*$$ is commutative.
If $$f: R \rightarrow R$$ and $$g: R \rightarrow R$$ are defined by $$f(x) =3x -4$$, and $$g(x)=2 + 3x$$, find $$(g^{-1}\, of^{-1})(5)$$.
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$$1$$
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$$\dfrac {1}{2}$$
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$$\dfrac {1}{3}$$
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$$\dfrac {1}{5}$$
Explanation
Given, $$f(x)=3x-4$$
Let $$y=3x-4$$
$$ \Rightarrow x=\dfrac { y+4 }{ 3 } \\ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x+4 }{ 3 } $$
Also given $$ g(x)=2+3x$$
$$\Rightarrow y=2+3x\\ \Rightarrow x=\dfrac { y-2 }{ 3 } \\ \Rightarrow g^{ -1 }\left( x \right) =\dfrac { x-2 }{ 3 } $$
Thus $$g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { \left( \dfrac { x+4 }{ 3 } \right) -2 }{ 3 } $$
$$ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { x-2 }{ 9 } \\ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( 5 \right) =\dfrac { 5-2 }{ 9 } =\dfrac { 1 }{ 3 } $$
So, option C is correct.
For what value of x is $$fog = gof$$ if $$f(x)=x - 2$$ and $$g(x)=x^3+3$$?
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$$\dfrac{-2}{3}$$
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$$-1$$
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$$\dfrac{3}{2}$$
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$$\dfrac{-3}{2}$$
Explanation
$$f(x)=x-2\\ g(x)={ x }^{ 3 }+3\\ f[g(x)]=g(x)-2\\ \Rightarrow fog={ x }^{ 3 }+3-2={ x }^{ 3 }+1\\ g[f(x)]={ \{ f(x)\} }^{ 3 }+3\\ \Rightarrow gof={ (x-2) }^{ 3 }+3\\ fog=gof\\ { x }^{ 3 }+1={ (x-2) }^{ 3 }+3\\ { x }^{ 3 }-2={ (x-2) }^{ 3 }\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6x(x-2)\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6{ x }^{ 2 }+12x\\ \Rightarrow { x }^{ 2 }+1+2x=0\\ \Rightarrow { (x+1) }^{ 2 }=0\\ \Rightarrow x=-1$$
So none of the above options are correct
Find number of all such functions $$y = f(x)$$ which are one-one?
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$$0$$
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$$3^{5}$$
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$$^{5}P_{3}$$
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$$5^{3}$$
Explanation
One-one functions are those in which each element in the domain has a unique image in the range of the function.
Here $$B$$ is the co-domain of the function which has lesser number of elements than the domain itself. This shows that it is not possible for the function to be one-one.
The inverse of the function $$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } } $$ is
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$$x=\log _{ 2 }{ \cfrac { 1 }{ 1-{ 2 }^{ y } } } $$
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$$x=\log _{ 2 }{ \left( 1-\cfrac { 1 }{ y } \right) } $$
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$$x=\log _{ 2 }{ \left( \cfrac { 1 }{ 1-y } \right) } $$
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$$x=\log _{ 2 }{ \left( \cfrac { y }{ 1-y } \right) } $$
Explanation
Given,
$$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } } \Rightarrow y+{ 2 }^{ x }y={ 2 }^{ x }$$
$$\Rightarrow y={ 2 }^{ x }(1-y)\Rightarrow { 2 }^{ x }=\cfrac { y }{ 1-y } $$
Taking log on both sides at base $$2$$, we get
$$\log _{ 2 }{ \left( { 2 }^{ x } \right) } =\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
$$\Rightarrow x=\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
$$\therefore$$ Inverse is $$\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
If $$f : R - \left \{\dfrac {3}{5}\right \}\rightarrow R - \left \{\dfrac {3}{5}\right \}; f(x) = \dfrac {3x + 1}{5x - 3}$$, then ___________.
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$$f^{-1} (x) = 2f(x)$$
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$$f^{-1} (x) = f(x)$$
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$$f^{-1} (x) = -f(x)$$
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$$f^{-1} (x)$$ does not exists
Explanation
$$f(x) = \dfrac {3x + 1}{5x - 3}$$ .... $$(i)$$
We know that if $$f(x) = \dfrac {ax + b}{cx + d}$$ and $$a + d = 0$$, then
$$f(x) = f^{-1}(x)$$
From $$(i)$$, we get
$$a=3,b=1,c=5,d=-3$$
$$a+d=3-3=0$$
Hence, $$f^{-1}(x)=f(x)$$
Suppose that $$g\left( x \right) =1+\sqrt { x }$$ and $$f\left( g\left( x \right) \right) =3+2\sqrt { x } +x$$, then $$f\left( x \right)$$ is
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$$1+2{ x }^{ 2 }$$
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$$2+{ x }^{ 2 }$$
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$$1+x$$
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$$2+x$$
Explanation
$$g(x)=1+\sqrt{x}$$
$$(g(x))^{2}=1+2\sqrt{x}+x$$
$$2+(g(x))^{2}=3+2\sqrt{x}+x$$
$$\therefore f(x)=2+x^{2}$$
The number of real linear functions $$f(x)$$ satisfying $$f\left\{ f(x) \right\} =x+f(x)$$
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$$0$$
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$$4$$
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$$5$$
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$$2$$
Explanation
Let $$f(x) = ax+b $$
$$f(f(x))=a(ax+b)+b = a^2x+ab+b$$
Given, $$f(f(x))=x+f(x)$$
$$\implies a^2x+ab+b = x+ ax+b$$
$$\implies (a^2-a-1)x+ab=0$$
Since above equation is valid for all values of x, we have
$$a^2-a-1 =0$$ and $$ ab=0$$
$$a^2-a-1 =0$$
$$a=\dfrac{ 1\pm \sqrt{1+4}}{2}$$
$$\implies a=\dfrac{ 1\pm \sqrt{5}}{2}$$
$$ab=0$$
Since $$a$$ is non-zero,
$$\implies b=0$$
$$f(x) =\dfrac{ 1+ \sqrt{5}}{2}x$$ and
$$f(x) =\dfrac{ 1- \sqrt{5}}{2}x$$
satisfy the condition.
Hence the answer is option (D)
Let A = {0, 1} and N the set of all natural numbers. Then the mapping $$f : N \rightarrow A$$ defined by
$$f(2n - 1) = 0, f (2n) = 1 \forall n \epsilon N$$
is many-one onto.
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True
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False
Explanation
Let $$A={0,1}$$ and N the set of all natural numbers. Then the mapping $$f:N \longrightarrow A$$ defines by $$f(2n-1)=0,$$ $$f(2x)=1$$ $$\forall$$ $$n$$ $$\epsilon$$ $$N$$ is many-one onto. The above statement is absolutely true.
If $$D$$ be subset of the set of all rational numbers which can be expressed as terminating decimals, then $$D$$ is closed under the binary operations of:
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addition, subtraction and division
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addition, multiplication and division
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addition, subtraction and multiplication
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subtraction, multiplication and division
Explanation
Division of rational number that can be expressed as terminating decimal may not be rational. $$\therefore$$ $$D$$ is closed under addition, subtraction, and multiplication.
If $$a\ast b={ a }^{ 3 }+{ b }^{ 3 }$$ on $$z$$, then $$\left( 1\ast 2 \right) \ast 0=........$$
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$$0$$
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$$729$$
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$$81$$
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$$27$$
Explanation
$$a*b=a^{3}+b{3}$$
So
$$1*2=1^{3}+2{3}=1+8=9$$
$$(1*2)*0=9*0=9^{3}+0^{3}=729+0=729$$
State True or False.
Let $$f : R \rightarrow R$$ be defined by $$f (x) = cos (5x + 2)$$. Then $$f$$ is invertible.
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True
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False
Explanation
$$f\left( x \right) =\cos\left( 5x+2 \right) $$
$$x=\left[ { \cos }^{ -1 }\left( f\left( x \right) \right) -2 \right] \dfrac { 1 }{ 5 } $$
As, $$-1\le f\left( x \right)\le 1$$
So, we can't put other values which makes it INTO function.
So, $$f\left( x \right) =cos\left( 5x+2 \right) $$ is not invertible.
If a language of natural numbers has a binary regularly of $$0$$ and $$1$$, then which one of the following strings represents the natural number $$7$$?
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$$1$$
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$$101$$
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$$110$$
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$$111$$
Explanation
Converting decimal 7 to binary number
The number of binary operations on $$\left\{ 1,2,3,4 \right\} $$ is ______.
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$${ 4 }^{ 2 }$$
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$${ 4 }^{ 8 }$$
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$${ 4 }^{ 3 }$$
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$${ 4 }^{ 16 }$$
Explanation
In $$\{1,2,3,4\}$$ there are 4 elements.
The number of binary operations in a set with n elements$$=n^{(n\times n)}$$
Here $$n=4$$
So t
he number of binary operations in a set with n elements$$=4^{(4\times 4)}=4^{16}$$
If $$ f: R->R$$ is defined by $$f(x) = |x|$$, then
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$$f^{-1}_{}(x) = -x$$
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$$f^{-1}_{}(x) = \dfrac{1}{|x|}$$
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The function $$f^{-1}_{}(x)$$ does not exist
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$$f^{-1}_{}(x) = \dfrac{1}{x}$$
Explanation
For a function $$f(x)$$ to be invertible, the function must be one-one and onto.
The range of $$f(x) = |x|$$ is $$[0, \infty)$$, while the co-domain of $$f(x)$$ is given as $$\mathbb{R}$$. Hence $$f(x)$$ is not onto.
Also, since $$f(x) = f(-x)$$, $$f(x)$$ is also not one-one in its domain.
Hence, f(x) is not invertible, ie, the function $$f^{-1}(x)$$ does not exist.
Option C is the right answer.
If $$a \times b =2 a - 3b + ab$$, then $$3 \times 5+5\times 3$$ is equal to
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$$22$$
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$$24$$
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$$26$$
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$$28$$
Explanation
$$ a\times b = 2a-3b+ab $$
find $$ 3\times 5+5\times 3$$
$$ 3\times 5[a = 3, b = 5]$$
$$ \therefore 3\times 5 = 2(3)-3(5)+3(5)$$
$$ 3\times 5= 6 $$
$$ 5\times 3 [a = 5, b = 3]$$
$$ \therefore 5\times 3 = 2(5)-3(3)+(5)(3)$$
$$ = 10-9+15$$
$$ 5\times 3= 16$$
$$ \therefore 3\times 5+5\times 3 = 6+16$$
$$ 3\times 5+5\times 3= 22 $$
Let $$R$$ be the relation on $$Z$$ defined by $$R = \{(a, b): a, b \in z, a - b$$ is an integer$$\}$$. Find the domain and Range of $$R$$.
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$$z, z$$
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$$z^+, z$$
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$$z, z^-$$
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None of these
Explanation
Given:
$$R=\{ (a, b) : a, b, \in z, a-b \text{ is an integer}\}$$
As difference of integers are also integers so,
Domain of $$R = z$$
Range of $$R = z$$, as
$$a-b$$ spans the whole integer values.
If $$x \times y = x^{2}+y^{2}-xy$$ then the value of $$9 \times 11$$ is :
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$$93$$
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$$103$$
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$$113$$
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$$121$$
Explanation
$$\textbf{Hint: Put}$$ $$\boldsymbol{x=9, y=11}$$ $$\textbf{in the given expression and then simplify to get the result.}$$
Correct Option: $$\textbf{B}$$
Solution:
$$\textbf{Step 1: Put}$$ $$\boldsymbol{x=9, y=11}$$ $$\textbf{in the given expression}$$
We have given, $$ x \times y = x^{2}+y^{2}-xy $$
By putting $$x=9, y=11$$, we get
$$ 9\times 11 = 9^{2}+11^{2}-9(11) $$
$$\textbf{Step 2: Simplify the expression in R.H.S}$$
$$\Rightarrow9 \times 11 = 81+121-99 $$
$$\Rightarrow 9\times 11 = 202-99 $$
$$\Rightarrow 9\times 11 = 103 $$
$$\textbf{Hence, the correct option is B}$$
Let $$f(x)={x}^{3}-6{x}^{2}+15x+3$$. Then,
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$$f(x)> 0$$ for all $$x\in R$$
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$$f(x)> f(x+1)$$ for all $$x\in R$$
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$$f(x)$$ is invertible
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$$f(x)< 0$$ for all $$x\in R$$
Explanation
$$f(x)=x^3-6x^2+15x+3$$
$$f'(x)=3x^2-12x+15$$
$$=3(x^2-4x+5)$$
$$=3(x^2-4x+4+1)$$
$$f'(x)=3(x-2)^2+{3} > 0$$
Therefore $$f(x)$$ is strictly increasing function
$$\Rightarrow f^{-1}(x)$$ exists
Hence $$f(x)$$ is a invertible function.
The number of binary operation on {1, 2, 3... n} is..
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$$2^n$$
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$$n^2$$
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$$n^3$$
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$$n^{2n}$$
Explanation
we have general formula,
for the $$n$$ number of series, the number of binary operation is given by,
$$2^n$$
Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
Consider the following statements:
$$f[g(x)]$$ is a polynomial of degree 3.
$$g[g(x)]$$ is a polynomial of degree 2.
Which of the above statements is/are correct ?
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1 only
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2 only
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Both 1 and 2
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Neither 1 nor 2
Explanation
Neither 1 nor 2
Given,
$$f(x)=x^2+2x-5$$
$$g(x)=5x+30$$
(i)
$$f[g(x)]$$
$$=f[5x+30]$$
$$=(5x+30)^2+2(5x+30)-5$$
upon solving the above equation, we get,
$$f[g(x)]=25x^2+310x+955$$
degree $$=2$$
(ii)
$$g[g(x)]$$
$$=g[5x+30]$$
$$=5(5x+30)+30$$
$$=25x+150+30$$
$$=25x+180$$
$$g[g(x)]=25x+180$$
degree $$=1$$
Let $$f(x)=\cfrac { 1 }{ 1-x } $$. Then $$\left\{ f\circ \left( f\circ f \right) \right\} (x)$$
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$$x$$ for all $$x\in R$$
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$$x$$ for all $$x\in R-\left\{ 1 \right\} $$
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$$x$$ for all $$x\in R-\left\{ 0,1 \right\} $$
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none of these
Explanation
$$f(x)=\dfrac{1}{1-x}$$ for $$x\in R-\{1\}$$
$$\{fof\}(x)=\dfrac{1}{1-\dfrac{1}{1-x}}=\dfrac{1}{\dfrac{1-x-1}{1-x}}=\dfrac{x-1}{x}=1-\dfrac{1}{x}$$ for $$x\in R - \{0,1\}$$
$$\{fofof\}(x)=\dfrac{1}{1-fof(x)}=\dfrac{1}{1-\dfrac{x-1}{x}}=\dfrac{1}{\dfrac{x-x+1}{x}}=x$$ for $$x\in R-\{0,1\}$$
Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
What are the roots of the equation $$g[f(x)] = 0$$ ?
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0%
$$1, -1$$
0%
$$-1, -1$$
0%
$$1, 1$$
0%
$$0, 1$$
Explanation
Given,
$$f(x)=x^2+2x-5$$
$$g(x)=5x+30$$
$$g[f(x)]=0$$
$$g[x^2+2x-5]=0$$
$$5(x^2+2x-5)+30=0$$
$$5x^2+10x-25+30=0$$
$$5x^2+10x+5=0$$
$$x^2+2x+1=0$$
$$(x+1)^2=0$$
$$\therefore x=-1,-1$$
Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
If $$h(x) = 5f(x) - xg (x)$$, then what is the derivative of $$h(x)$$ ?
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$$-40$$
0%
$$-20$$
0%
$$-10$$
0%
$$0$$
Explanation
Given,
$$f(x)=x^2+2x-5$$
$$g(x)=5x+30$$
$$h(x)=5f(x)-xg(x)$$
$$=5(x^2+2x-5)-x(5x+30)$$
$$=5x^2+10x-25-5x^2-30x$$
$$h(x)=-20x-25$$
$$\dfrac{d}{dx}[h(x)]=\dfrac{d}{dx}(-20x-25)$$
$$=-20$$
The number of one-one functions that can be defined from $$A=\{4,8,12,16\}$$ to $$B$$ is $$5040,$$ then $$n(B)=$$
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0%
$$7$$
0%
$$8$$
0%
$$9$$
0%
$$10$$
Explanation
The first element $$4$$ of $$A$$ can be mapped to any of the $$n$$ elements in $$B$$
Similarly second element $$8$$ of $$A$$ can be mapped to any of $$(n-1)$$ elements in $$B$$
$$\Rightarrow$$ Total no of one - one functions is
$$n(n-1)(n-2)(n-3)=5040$$
$$\Rightarrow n=10$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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