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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 11
If $$P(S)$$ denotes the set of all subsets of a given set S, then the number of one-to-one functions from the set $$S=\{1,2,3\}$$ to he set $$P(S)$$
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0%
$$24$$
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$$8$$
0%
$$336$$
0%
$$320$$
$$\begin{aligned} \text { If } A & = \{ x | x / 2 \in Z , 0 \leq x \leq 10 \} \\ B & = \{ x | x \text { is one digit prime } \} \\ C & = \{ x | x / 3 \in N , x \leq 12 \} \end{aligned}$$
,
Then $$A \cap ( B \cup C )$$ is equal to-
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$$\{ 2,6 \}$$
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$$\{3,6,12 \}$$
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$$\{ 2,6,12 \}$$
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$$\{ 6,8 \}$$
The function $$f:N\rightarrow N $$ defined by $$f\left( x \right) =x-5\left[ \dfrac { x }{ 5 } \right]$$, where $$N$$ is the set of natural numbers and $$[x]$$ denotes the greatest integer less then or equal to $$x$$ is
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One-one and onto
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One-one but not onto
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Onto but not one-one
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Neigher one-one nor onto
Explanation
undefined
The inverse of the function $$f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$ is
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$$\frac{1}{2}ln\frac{1+x}{1-x}$$
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$$\frac{1}{2}ln\frac{2+x}{2-x}$$
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$$\frac{1}{2}ln\frac{1-x}{1+x}$$
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2 ln (1+x)
Let $$f:R\rightarrow R$$ be defined as
$$f(x)={ x }^{ 3 }+{ 2x }^{ 2 }+4x+sin\left( \dfrac { \pi x }{ 2 } \right) $$ and $$g(x)$$be the inverse function of f(x), then $${ g }^{ ' }(8)$$is equal to :
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$$\dfrac { 1 }{ 2 } $$
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9
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$$\dfrac { 1 }{ 11 } $$
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11
The function $$f:R\rightarrow R$$ defined by $$f\left(x\right)=6^ {x}+6$$ is
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0%
one-one and onto
0%
many-one and onto
0%
one-one and into
0%
mauny-one and into
Explanation
$$f: R \rightarrow R \\$$
$$f(x)=6^{x}+6$$
$$\text { For one - one, } \\$$
$$\qquad f\left(x_{1}\right)=f\left(x_{2}\right) \\$$
$$\Rightarrow 6^{x_{1}}+6$$ &=$$6^{x_{2}}+$$
$$\Rightarrow \quad 6^{x_{1}}=6^{x_{2}} \\$$
$$\Rightarrow \quad x_{1}=x_{2}$$
Therefore f is one - one.
Now,
$$\begin{aligned} f(x) &=y=6^{x}+6 \\ & \Rightarrow(y-6)=6^{x} \end{aligned}$$
Taking log both side.
$$\Rightarrow \log _{6}(y-6)=x$$
Now,
$$f\left(\log _{6}(y-6)\right)=6^{\log _{6}(y-6)}+6$$
$$=y-6+6$$
$$=y$$
$$\therefore f \text { is onto }$$
Hence, f is one - one and onto.
Let n(A) = 4 and n(B) =Then the number of one - one functions from A to B is
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0%
120
0%
360
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24
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none of these
Explanation
$$n(A)=4$$
$$n(B)=6$$
$$\text { Number of one-one functions }(A \rightarrow B)$$
$$=6 c_{4} \times 4 !$$
$$=\frac{6 !}{2 ! 4 !} \times 4 !$$
$$=6 \times 5 \times 4 \times 3$$
$$=360$$
Let $$f:R\rightarrow R$$ be defined by $$f(x)=\dfrac {x|x|}{2}+\cos x+1$$ then $$f(x)$$ is
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One-one only
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Onto only
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Neither one-one nor onto
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Bijection
Let $$f : R \rightarrow (-1,1)$$ be defined as $$f(x)=\dfrac {e^{x}-e^{-x}}{e^{x}+e^{-x}}$$ then $$f$$ is
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0%
One-one onto
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One-one into
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Many-one onto
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Many-one-into
If $$A^2 - A + I = 0$$ then $$A^{-1}$$ =
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$$A^2$$
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$$A + I$$
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$$I - A$$
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$$A - I$$
Explanation
$$A^{2}-A+I=0$$
Pre- multiply with $$A^{-1}$$ on both sides,
$$A^{-1} A^{2}-A^{-1} A+A^{-1}=A^{-1} 0$$
$$\Rightarrow A-I+A^{-1}=0$$
$$\Rightarrow \quad A^{-1}=I-A$$.
option $$c$$ is correct.
A number of decimal system when represented in binary system then its first and last digits are same and the rest digits are also of another kind. Further this number appears to be palindrome of number. The number is
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$$57$$
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$$19$$
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$$8$$
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$$9$$
If $$f:Z\rightarrow Z, f(n)=\begin{cases} n+1;\quad n\quad is\quad even \\ n-3;\quad n\quad is\quad odd \end{cases}$$ is $$ f$$ is ...........
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only one one
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only Onto
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one one & Onto both
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Neither one one nor Onto
Which one of the following functions is not invertible?
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$$f:R\rightarrow R, f(x)=3x+1$$
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$$f:R\rightarrow [0, \infty), f(x)=x^{2}$$
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$$f:R^{+}\rightarrow R^{+}, f(x)=\dfrac{1}{x^{3}}$$
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None of these
If $$f:R\rightarrow R\quad defined\quad by\quad f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,\quad then\quad f\quad is$$
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one-one but not onto
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not one-one but onto
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one-one and onto
0%
neither one-one noronto
If 27 $$^{*}$$ 3 =243 and 5 $$^{*}$$ 4 = 80 .Then what is the value of 3 $$^{*}$$ 7 = ?
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84
0%
147
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63
0%
23
If $$f(x)=\frac { \alpha x }{ x+1 } $$, where $$x\neq -1$$ and (fof) (x) = x, then $$\alpha =$$
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$$\sqrt { 2 } $$
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$$-\sqrt { 2 } $$
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$$1$$
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$$-1$$
$$f(x)$$ is
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0%
One-one and onto
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One-one and into
0%
Many-one and onto
0%
Many-one and into
Let N be the set of natural numbers and two functions f and g be defined as
and g(n)=$$n-{ \left( -1 \right) }^{ n }$$ then fog is:
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one-one but not onto
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onto but not one-one
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neither one-one nor into
0%
both one-one and onto
The inverse of 19 mod 141 is :
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0%
50
0%
51
0%
52
0%
55
let $$f:R\rightarrow R$$ be a function defined by $$f(x)=\frac { { x }^{ 2 }-3x+4 }{ { x }^{ 2 }+3x+4 }$$ then f is
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0%
one-one but not onto
0%
onto but not one
0%
onto as well as one-one
0%
neither onto nor one-one
The function $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ \left| x \right| }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } $$ is
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0%
One-One and onto
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One-one but not onto
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Not one-one but onto
0%
Neither one-one nor onto
$$f:N\rightarrow N\quad where\quad f\left( x \right) =x-{ (-1) }^{ x }$$, then 'f' is
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0%
one-one and into
0%
many- one and into
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one-one and onto
0%
many-one and onto
Let f(x+y)=f f(x) f(y) and f(x) =1+x g(x) G(x), where $$\underset { x\rightarrow 0 }{ lim } g\left( x \right) =a and \underset { x\rightarrow 0 }{ lim } G\left( x \right) =b,$$ then f' (x) is equal to
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0%
1+ab
0%
ab
0%
f(x)
0%
ab f(x)
Consider the function $$f\left( x \right) ={ e }^{ x }$$ and $$g\left( x \right)=\sin ^{ -1 }{ x } $$, then which of the following is/are necessarily true.
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Domain of $$gof =$$ Domain of $$f$$
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Range of $$gof\ \subset$$ Range of $$g$$
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Domain of $$gof$$ is $$\left( -\infty ,0 \right) $$
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Range of $$gof$$ is $$\left( -\dfrac{\pi}{2} ,0 \right) $$
Explanation
$$\text { solution: } \quad f(x)=e^{x} \quad g(x)=\sin ^{-1}(x) \\$$
$$\text { gof }=\sin ^{-1}\left(e^{x}\right) \\$$
$$\therefore-1 \leqslant e^{x} \leqslant 1 \\$$
$$\text { so we need } x \in(-\infty \text { o] } \\$$
$$\text { domain of gof is }(-\infty, 0] \\$$
$$\text { Range of gof is same as range of } g \\$$
$$\text { Ansuer. : option (B) }$$
$$f:A\rightarrow B$$ will be an into function if
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$$f\left(A\right) \subset B$$
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$$f\left(A\right)=B$$
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$$B\subset f\left(A\right)$$
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$$f\left(B\right) \subset A$$
Suppose that $$g(x)=1+\sqrt { x } and\quad f(g(x))=3+2\sqrt { x } +x\quad then\quad f(x)\quad is$$
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$$\quad 1+2{ x }^{ 2 }$$
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$$\quad 2+{ x }^{ 2 }$$
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$$1+x$$
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$$2+x$$
Which one of the following is one-one?
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$$f:R\rightarrow R$$ given by $$f(x) =\left| x-1 \right| $$ for all $$x\in R$$
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$$g:\left[ -\pi /2,\pi /2 \right] \rightarrow given by$$ $$g(x)=\left| sinx \right| $$
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$$h:\left[ -\pi /2,\pi /2 \right] \in R$$ given by $$h(x) =sin x $$ for all $$x\in \left[ -\pi /2,\pi /2 \right] $$
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$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$x\in R$$
Let $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ -x^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,$$ then
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f(x) is one-one but not onto
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f(x) is neither one-one nor onto
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f(x) is many one but onto
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f(x) is one-one and onto
Explanation
$$\text { solution: } \quad f(x)=\frac{e^{x^{2}}-e^{-x^{2}}}{e^{x^{2}}+e^{-x^{2}}}=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1} \\$$
$$f^{\prime}(x)=\frac{\left\{\left(e^{x^{2}}\right)^{2}+1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}-\left\{\left(e^{x^{2}}\right)^{2}-1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}}{\left(\left(e^{x^{2}}\right)^{2}+1\right)^{2}} \\$$
$$\text { Since } f^{\prime}(x) \text { is nither positive, nor negative } \\$$
$$\text { for all } x \text { so } f(x) \text { is not one - one function } \\$$
$$\qquad f(x)=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1}$$
Range of $$f(x)$$ is $$[0,1 )$$
since Range of $$f(x)$$ is not equal to co-do
main
of $$f(x)$$ so $$f(x)$$ is not onto function.
Answer: option: (B)
Consider the binary operation $$*$$ on $$Q$$ defined by $$x*y=1+12x+xy,x,y\rightarrow Q$$ then $$2*3$$ equals
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0%
$$31$$
0%
$$41$$
0%
$$43$$
0%
$$51$$
The inverse of the function
$$f\left( x \right) =\frac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } +2$$ is given by
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$$\log _{ e }{ { \left( \frac { x-2 }{ x-1 } \right) }^{ 1/2 } } $$
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$$\log _{ e }{ { \left( \frac { x-1 }{ 3-x} \right) }^{ 1/2 } } $$
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$$\log _{ e }{ { \left( \frac { x }{ 2-x } \right) }^{ 1/2 } } $$
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$$\log _{ e }{ { \left( \frac { x-1 }{ x+1 } \right) }^{ -2 } } $$
Explanation
$$f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+2 \\$$
$$f(x)=y \quad \Rightarrow \quad x=f^{-1}(y) \\$$
$$\Rightarrow y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+2 \\$$
$$\Rightarrow y=\frac{e^{2 x}-1}{e^{2 x}+1}+2 \\$$
$$\Rightarrow \quad y-2=\frac{e^{2 x}-1}{e^{2 x}+1}\\$$
$$\Rightarrow(y-2)\left(e^{2 x}+1\right)=\left(e^{2 x}-1\right) \\$$
$$\Rightarrow(y-2) e^{2 x}+(y-2)=e^{2 x}-1\\$$
$$\Rightarrow(y-2) e^{2 x}-e^{2 x}=1-y$$
$$\Rightarrow e^{2 x}\{y-2-\}=1-y$$
$$\Rightarrow e^{2 x}=\frac{1-y}{y-3}$$
$$2 x \ln _{e} e=\ln _{e}\left(\frac{1-y}{y-3}\right)$$
$$\Rightarrow \quad 2 x=\ln \left(\frac{1-y}{y-3}\right)$$
$$\Rightarrow \quad x=\frac{1}{2} \ln \left(\frac{1-y}{y-3}\right)$$
$$\Rightarrow x=\ln \left(\frac{1-y}{y-3}\right)^{1 / 2}=f^{-1}(y)$$
so $$f^{-1}(x)=\ln \left(\frac{1-x}{x-3}\right)^{1 / 2}=\log _{e}\left(\frac{x-1}{3-x}\right)^{1 / 2}$$
Answer : option (B)
Let $$f:R\rightarrow R$$, be defined as $$f(x)={ e }^{ x^{ 2 } }+cosx$$ then f is
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0%
One-one and onto
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One-one and into
0%
Many-one and onto
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many-one and into
Let (X) be a function satisfying f' (X) = f (X) with f (0) = 1 and g (X) be a function that satisfies f (X) + g (x) = $${ x }^{ 2 },$$ Then the value of the integral $$\int _{ 0 }^{ 1 }{ f } (x)\quad g\quad (x)\quad dx,\quad is$$
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0%
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 5 }{ 2 } $$
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$$e+\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$
0%
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$
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$$e+\frac { { e }^{ 2 } }{ 2 } +\frac { 5 }{ 2 } $$
l qt f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g be the function satisfying f(x)+g(x)=$$X^{2}$$, the value of the integral $$\int _{ 0 }^{ 1 }{ f(x)g(x)\quad dx\quad is } $$
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$$\frac { 1 }{ 4 } (e-7)$$
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$$\frac { 1 }{ 4 } (e-2)$$
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$$\frac { 1 }{ 4 } (e-3)$$
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none of the above
the inverse of the function f(x)=$$\frac { { 10 }^{ x }-{ 10 }^{ -x } }{ { 10 }^{ x }+1{ 0 }^{ -x } } is$$
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$$\frac { 1 }{ 2 } { log }_{ 10 }\left( \frac { 1+x }{ 1-x } \right) $$
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$$\frac { 1 }{ 2 } { log }_{ 10 }\left( \frac { 1-x }{ 1+x } \right) $$
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$$\frac { 1 }{ 4 } { log }_{ 10 }\left( \frac { 2x }{ 2-x } \right) $$
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none of these
$$f : R \rightarrow R$$ defined by $$f ( x ) = \dfrac { x } { x ^ { 2 } + 1 } , \forall x \in R$$ is
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0%
one-one
0%
onto
0%
bijective
0%
neither one one nor onto
Explanation
$$\begin{aligned} f(x) &=\dfrac{x}{1+x^{2}} \\-f(4)=& \dfrac{4}{1+16}=\dfrac{4}{17} \\ f\left(\dfrac{1}{4}\right) &=\frac{\dfrac{1}{4}}{1+\dfrac{1}{16}}=\dfrac{4}{17} \end{aligned}$$
Hence two different input has Same output $$\cdot$$ so not one-one
$$f(x)=y$$
$$y=\dfrac{x}{1+x^{2}}=1 \quad y+y x^{2}-x=0$$
$$x=\dfrac{1 \pm \sqrt{1-4 y^{2}}}{2 y}$$
$$1-4 y^{2} \geq 0$$
$$(1+2 y)(1-x y) \geq 0$$
$$\dfrac{1}{-3} \leq y<\dfrac{1}{2}$$
Range of $$f(x)$$ is $$\left[\dfrac{-1}{2}, \dfrac{1}{2}\right]$$
Range of $$f(x) \neq$$ codomain $$f(x)$$ Hence f(x) is neither one-one nor onto option D is correct.
If f (x) = cosx and g (x) = x$$^2$$ then (gof) (x) is ....
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0%
cos$$^2$$ x
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cosx$$^2$$
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both (a)&(b)
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x$$^2$$ cosx
If the function $$f:[1,\infty ) \rightarrow [1, \infty) $$ is defind by $$f(x)=2^{x(x-1)}, $$ then $$f^{-1}(x)$$ is
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$$( \frac{1}{2})^{x(x-1)}$$
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$$\frac{1}{2}(1+\sqrt{1+4log_{2}x})$$
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$$\frac{1}{2}(1-\sqrt{1+4log_{2}x})$$
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None of these
If $$f ( x ) = \sqrt { x ^ { 2 } + 1 } , g ( x ) = \dfrac { x + 1 } { x ^ { 2 } + 1 }$$ and $$h ( x ) = 2 x - 3 ,$$ then $$f ^ { \prime } \left( h ^ { \prime } \left( g ^ { \prime } ( x ) \right) =\right.$$
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0%
$$0$$
0%
$$\dfrac { 1 } { \sqrt { x ^ { 2 } + 1 } }$$
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$$\dfrac { 2 } { \sqrt { 5 } }$$
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$$\dfrac { x } { \sqrt { x ^ { 2 } + 1 } }$$
$${ f }:{ R }\rightarrow { R }$$ where $$f ( x ) = \dfrac { x ^ { 2 } + a x + 1 } { x ^ { 2 } + x + 1 }.$$ Complete set of values of $$'a'$$ such that $$f ( x )$$ is onto to is :
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$$( - \infty , \infty )$$
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$$( - \infty , 0 )$$
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$$( 0 , \infty )$$
0%
None
Explanation
$$f(x)=\dfrac{x^{2}+a x+1}{x^{2}+x+1}$$
$$f: R \rightarrow R$$
domain co-domain
$$D\left(x^{2}+x+1\right)=-3<0$$
Hence $$x^{2}+x+1$$ is always positive
$$x^{2}+a x+1$$ coefficient of $$x^{2}=1>0$$
$$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow \infty} \dfrac{x^{2}+a x+1}{x^{2}+x+1} \dfrac{\infty}{\infty}$$ form
$$=\lim _{x \rightarrow 0} \dfrac{x^{2}\left(1+\dfrac{a}{x}+\dfrac{1}{x^{2}}\right)}{x^{2}\left(1+\frac{1}{x}+\dfrac{1}{x^{2}}\right)}=\dfrac{1+0+0}{1+0+0}=1$$
similarly
$$lim_{x\rightarrow-\infty} f(x)=1$$
Hence $$f(x)$$ is bounded
Range $$\neq R$$
$$\neq$$ co-domain
$$f(x)$$ cannot be onto for any value of $${a}$$
option $$D$$ is correct.
Let f be a real -valued function defined on the interval $$(-1,1)$$ such that
$${e^{ - x}}f(x) = 2 + \int\limits_{}^x {\sqrt {{t^2} + 1} } .dt,\forall x \in ( - 1,1)$$
and let $${f^{ - 1}}$$ be the inverse function of $$f$$ . Then $$\left[ {{f^{ - 1}}(2)} \right]$$ is equal to
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0%
$$1$$
0%
$$1/3\,$$
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$$1/2\,$$
0%
$$1/e\,$$
If the operation * is defined as $$\left( a\times b \right) ={ a }^{ 2 }+{ b }^{ 2 }$$ then $$\left( 3\times 4 \right) =5$$ is
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0%
650
0%
125
0%
625
0%
3125
Let f(x)= $$y=\begin{cases} { x }^{ 2 }-3x+4\quad \quad :\quad X<3 \\ \quad \quad \quad x+7\quad \quad :\quad X\ge 3 \end{cases}\quad and\quad g(x)=\begin{cases} \quad \quad x+6\quad \quad \quad :\quad X<4 \\ { x }^{ 2 }+x+2\quad \quad :\quad X\ge 4 \end{cases}$$
then which of the following is/ are true-
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$$(f+g) (1)=9$$
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$$(f-g)(3.5)= 1$$
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$$(f+g) (0)=24$$
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$$\left( \dfrac { f }{ g } \right) (5)=\dfrac { 8 }{ 3 } $$
If f(x) = x + 4, g(x) = 5x and h(x) = 12/x, find the value of $${ f }^{ -1 }$$(g(h(6)))-
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0%
10
0%
14
0%
6
0%
0
If a function $$f : R \rightarrow R$$ be such that $$f ( x + y ) = f ( x ) f ( y )$$ for all $$x , y \in R$$ where $$f ( x ) = 1 + x \phi ( x )$$ and $$\lim _ { x \rightarrow 0 } \phi ( x ) = 1 ,$$ then :
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$$f ^ { \prime } ( x )$$ does not exist
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$$f ^ { \prime } ( x ) = 2 f ( x )$$ for all $$x$$
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$$f ^ { \prime } ( x ) = f ( x )$$ for all $$x$$
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None of these
Choose correct answer (s) from given choice
If f(x) = x + 4, g (x) = 5x and h(x) = 12/x. Find the value of $${ f }^{ -1 }(g(h(6)))$$
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0%
10
0%
14
0%
6
0%
0
Find the inverse of $$f(x)=3x+2$$
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0%
$${f^{ - 3}}\left( z \right) = \frac{{z - 2}}{3}$$
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$${f^{ - 3}}\left( z \right) = \frac{{z + 2}}{3}$$
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$${f^{ - 3}}\left( z \right) = \frac{{z - 5}}{2}$$
0%
$${f^{ 3}}\left( z \right) = \frac{{z - 2}}{3}$$
If f(x)=x+tanx and g(x) is inverse of f(x) then g'(x) is equal to
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0%
$$\dfrac { 1 }{ 1+(g(x)-{ x) }^{ 2 } } $$
0%
$$\dfrac { 1 }{ 1-(g(x)-{ x) }^{ 2 } } $$
0%
$$\dfrac { 1 }{ 1+(g(x)-{ x) }^{ 2 } } $$
0%
$$\dfrac { 1 }{ 2-(g(x)-{ x) }^{ 2 } } $$
If $$f : N\rightarrow N$$ be defined by $$f(x) = 2x + 3$$ then $$f^{-1} (x) =$$
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0%
$$2x - 3$$
0%
$$\dfrac {x - 3}{2}$$
0%
$$\dfrac {x + 3}{2}$$
0%
Not defined
If f(x) = x + 4, g(x) = 5x and h(x) = 12/x. find the value of $$f^{ -1 }$$ (g(h(6))).
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0%
10
0%
14
0%
6
0%
0
Let $$f(x)=x+\cos x+2$$ and g(x) be the inverse function of f(x) then $$g^1(3)=$$
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$0$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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