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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 2
If $$n (A) = 4$$ and $$n(B) = 6$$, then the number of surjections from $$A$$ to $$B$$ is
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$$4^{6}$$
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$$6^{4}$$
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$$0$$
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$$24$$
Explanation
$$n(A)<n(B)$$
$$\therefore$$ Co-domain can never be equal to range.
$$\therefore$$ No of surjections is zero
The number of injections that are possible from $$A$$ to itself is $$720,$$ then $$n (A) =$$
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$$5$$
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$$6$$
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$$7$$
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$$8$$
Explanation
first element of $$A$$ in domain can be mapped to
any of the $$n$$ elements of $$A$$ in range.
$$2^{nd}$$ element of $$A$$ in domain can be mapped to any of $$(n-1)$$ elements of $$A$$ in range.
$$\therefore$$ Total no of injections possible is
$$n\times (n-1)\times (n-2)...\times 3\times 2\times 1=n!$$
$$\therefore n!=720$$
$$\Rightarrow n=6$$
Let $$A=\{1,2,3\}, B =\{a, b, c\}$$ and If $$f=\{(1,a),(2,b),(3,c)\}, g=\{(1,b),(2,a),(3,b)\}, h=\{(1,b)(2,c),(3,a)\}$$ then
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$$g$$ and $$h$$ are injections
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$$f$$ and $$h$$ are injections
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$$f$$ and $$g$$ injections
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$$f,g$$ and $$h$$ are injections
Explanation
$$g(1)=g(3)=b \Rightarrow$$ g is not one-one.
domain of $$f =$$ domain of $$h = A.$$
range of $$f =$$ range of $$h = B$$
$$n(A)=n(B)=3$$
$$\therefore f$$ and $$h$$ are injections.
The number of one-one functions that can be defined from $$A = \left \{ 1,2,3 \right \} $$ to $$ B = \left \{ a,e,i,o,u \right \}$$ is
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$$3^{5}$$
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$$5^{3}$$
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$${_{}}^{5}P_{3}$$
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$$5!$$
Explanation
You can correlate this to permutation and combination problems.
We have to arrange 3 people on 5 places.
So total no of permutations
$$={_{}}^{5}P_{3}$$
The number of non-surjective mappings that can be defined from $$A = \left \{ 1,4,9,16 \right \} $$ to$$ B=\left \{ 2,8,16,32,64 \right \}$$ is
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$$1024$$
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$$20$$
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$$505$$
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$$625$$
Explanation
Here $$n(B)>n(A)$$, no function can be a surjective.
No of functions $$=$$ No of non-surjectives
Any elements of $$A$$ can be mapped to any of $$5$$ elements in $$B$$
So non surjective mapping from $$A$$ to $$B$$
$$=5\times5\times5\times5$$
$$=625$$
If $$ f:A\rightarrow B $$ is a constant function which is onto then $$B$$ is
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a singleton set
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a null set
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an infinite set
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a finite set
Explanation
$$f(x)$$ is a constant fucntion $$\Rightarrow$$ Range of $$f(x)$$ is a singleton set.
For $$f$$ to be an onto function, Co domain $$B$$ should be equal to Range.
$$\therefore B $$ is a singleton set.
If $$ f:A\rightarrow B $$ is a bijection then $$ f^{-1} of = $$
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$$fof^{-1}$$
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$$f$$
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$$f^{-1}$$
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an identity
Explanation
Since $$f$$ is bijective, its inverse will also be bijective.
$$f(x)=y$$
$$x=f^{-1}(y)$$
$$f(x)=f^{-1}(y)$$
$$y=f(f)^{-1}(y)=f(x)=y$$
Hence, it is an identity.
The number of injections possible from $$A=\{1,3,5,6\}$$ to $$B =\{2,8,11\}$$ is
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$$8$$
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$$64$$
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$$2^{12}$$
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$$0$$
Explanation
co-domain $$n(B) <$$ domain $$n(A)$$
There can't be any one-one function from $$A$$ to $$B.$$
The number of possible surjection from $$A=\{1,2,3,...n\}$$ to $$B = \{1,2\}$$ (where $$n \geq 2)$$ is $$62$$, then $$n=$$
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$$5$$
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$$6$$
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$$7$$
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$$8$$
Explanation
Each element in $$A$$ can be mapped onto any of two elements of $$B$$
$$\therefore$$ Total possible functions are $$2^{n}$$
For the $$f^{{n}'s}$$ to be surjections , they shouldn't be mapped alone to any of the two elements.
$$\therefore$$ Total no of surjections $$= 2^{n}-2$$
$$2^{n}-2=62$$
$$\Rightarrow n=6$$
If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=x^{2}, g(x)=\cos x$$ then $$(gof)(x)=$$
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$$\cos 2x$$
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$$x^{2}\cos x$$
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$$\cos x^{2}$$
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$$\cos^{2} x^{2}$$
Explanation
$$gof\left ( x \right )=g\left ( f\left ( x \right ) \right )=g\left ( x^{2} \right )$$
$$=\cos x^{2}$$
If $$f:R\rightarrow R $$ is defined by $$\displaystyle f(x)={\dfrac{2x+1}{3}}$$ then $$f^{-1}(x)=$$
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$$\dfrac{3x-1}{2}$$
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$$\dfrac{x-3}{2}$$
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$$\dfrac{2x-1}{3}$$
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$$\dfrac{x-4}{3}$$
Explanation
$$f\left ( x \right )=\dfrac{2x+1}{3}$$
Let $$f\left ( x \right )=y=\dfrac{2x+1}{3}$$
$$\Rightarrow 2x+1=3y$$
$$2x=3y-1$$
$$x=\dfrac{3y-1}{2}$$
$$\therefore f\left ( x \right )=y\Rightarrow f^{-1}\left ( y \right )=x$$
$$\Rightarrow f^{-1}\left ( y \right )=\dfrac{3y-1}{2}$$
$$\Rightarrow f^{-1}\left ( x \right )=\dfrac{3x-1}{2}$$
Let $$f(x)=\dfrac{Kx}{x+1}(x\neq -1)$$ then the value of $$K$$ for which $$(fof)(x)=x$$ is
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$$1$$
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$$-1$$
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$$2$$
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$$\sqrt{2}$$
Explanation
$$f\left ( f\left ( x \right ) \right )=f\left ( \dfrac{kx}{x+1} \right )$$
$$=\dfrac{k\left ( \dfrac{kx}{x+1} \right )}{\dfrac{kx}{x+1}=1}=\dfrac{k^{2}x}{\left ( k+1 \right )x+1}$$
$$\therefore$$ for $$f\left ( f\left ( x \right ) \right )=x$$
$$\Rightarrow k+1=0$$
$$k^{2}=1$$
$$\therefore k=-1$$
$$f:\left ( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right )\rightarrow \left ( -\infty ,\infty \right )$$ defined by $$f(x)=1+3x$$ 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 onto
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bijective
Explanation
$$f(x_{1})=f(x_{2}) \Rightarrow 1+3x_{1}=1+3x$$
$$x_{1}=x_{2}$$
$$\therefore f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$$
$$\therefore$$ f is one-one.
$$f$$ lies b/w $$(1-\dfrac{3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\therefore$$ co-domain $$(-\infty ,\infty )$$ is not equal to range $$(\dfrac{1-3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\Rightarrow$$ f is not onto.
$$\therefore$$ f is one-one but not onto.
If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=4x-1,g(x)=x^{3}+2,$$ then $$(gof)\left(\dfrac{a+1}{4}\right)=$$
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$$43$$
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$$4a^3-1$$
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$$a^{3}+2$$
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$$64a^3 - 8a^{2}-1$$
Explanation
$$g\left ( f\left ( \dfrac{a+1}{4} \right ) \right )=g\left ( 4\left ( \dfrac{a+1}{4} \right )-1 \right )$$
$$=g\left ( a \right )$$
$$=a^{3}+2$$
The function $$f:(0,\infty )\rightarrow (-\infty ,\infty )$$ is defined by $$ f(x)=\log_{3} x $$ then $$ f^{-1}(x)=$$
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$$3^{x}$$
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$$3^{-x}$$
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$$-3^{x}$$
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$$-3x^{-x}$$
Explanation
$$f\left ( x \right )=\log _{3}x$$
$$f^{-1}:\left ( -\infty ,\infty \right )\rightarrow \left ( 0, \infty \right )$$
Let $$\log _{3}x=y=f\left ( x \right )$$
$$\Rightarrow x=3^{y}$$
$$f\left ( x \right )=y\Rightarrow f^{-1}\left ( y \right )=x=3^{y}$$
$$\therefore f^{-1}\left ( x \right )=3^{x}$$
If $$f:R\rightarrow R,f(x)=3x-2$$ then $$ (fof)(x)+2=$$
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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
$$fof (x)=f(f(x))$$
$$=f(3x-2)$$
$$=3(3x-2)-2$$
$$=9x-8$$
$$fof(x)+2=9x-6=3(3x-2)$$
$$=3f(x)$$
If $$f(x)=2x+1$$ and $$g(x)=x^{2}+1$$ then $$ (go(fof))(2)=$$
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$$112$$
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$$122$$
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$$12$$
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$$124$$
Explanation
$$(go(fof))\left( x \right) =g\left( f\left( f\left( x \right) \right) \right) $$
$$=g\left ( f\left ( 2x+1 \right ) \right )$$
$$=g\left ( 2\left ( 2x+1 \right )+1 \right )$$
$$=g\left ( 4x+3 \right )$$
$$=\left ( 4x+3 \right )^{2}+1$$
$$(go(fof))\left ( 2 \right )=\left ( 11 \right )^{2}+1=122$$
If $$f(x)=\dfrac{1}{x}, g(x)=\sqrt{x}$$ and $$ (go\sqrt{f})(16)=$$
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$$2$$
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$$1$$
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$$\dfrac{1}{2}$$
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$$4$$
Explanation
$$\sqrt{f\left ( x \right )}=\dfrac{1}{\sqrt{x}}$$
$$g\left ( \sqrt{f\left ( x \right )} \right )=g\left ( \dfrac{1}{\sqrt{x}} \right )=\dfrac{1}{\sqrt{\sqrt{x}}}=\dfrac{1}{x^{\frac{1}{4}}}$$
$$\left ( go\sqrt{f} \right )(16)=\dfrac{1}{\left ( 16 \right )^{\dfrac{1}{4}}}=\dfrac{1}{2}$$
If $$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$ then $$(hogof)(x)$$ is equal to
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$$x^{2}+2$$
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$$2x^{2}+1$$
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$$x^{2}+1$$
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$$2(x^{2}+1)$$
Explanation
Given
$$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$
$$hogof(x)$$
$$=h(g(f(x)))$$
$$=h\left ( g\left ( x \right ) \right )\dots\dots \left[ \because f\left ( x \right )=x \right ]$$
$$=h\left ( 2x^{2}+1 \right )\dots\dots\left [ \because g\left ( x \right )=2x^{2}+1 \right ]$$
$$=2x^{2}+1+1$$
$$=2x^2+2$$
$$=2\left (x ^{2} +1\right )$$
$$\therefore hogof(x)=2(x^2+1)$$
If $$f(x)=\dfrac{e^{x}+e^{-x}}{2}$$, then the inverse of $$f(x)$$ is
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$$\log_{e}(x+\sqrt{x^{2}+1})$$
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$$\log_{e}\sqrt{x^{2}-1}$$
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$$\log_{e}\left(\dfrac {x+\sqrt{x^{2}-1}}{2}\right)$$
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$$\log_{e}(x+\sqrt{x^{2}-1})$$
Explanation
Let $$f\left ( x \right )=\dfrac{e^{x}+e^{-x}}{2}=y$$
$$\therefore x=f^{-1}\left ( y \right )$$
$$e^{x}+e^{-x}=2y$$
$$e^{2x}-2y e^{x}+1=0$$
$$\Rightarrow e^{x}=\dfrac{2y \pm \sqrt{4y^{2}-4}}{2}$$
$$e^{x}=\dfrac{y\pm \sqrt{y^{2}-1}}{1}$$
Range of $$f$$ is
$$\left ( -\infty ,\infty \right )\Rightarrow e^{x}=\dfrac{y+\sqrt{y^{2}-1}}{1}$$
As if we take $$e^{x}=\dfrac{y-\sqrt{y^{2}-1}}{1}$$, which is always small
$$\therefore x=\log _{e}\left ( \dfrac{y+\sqrt{y^{2}-1}}{1} \right )$$$$\therefore f^{-1}\left ( x \right )=\log_{e}\left ( \dfrac{x+\sqrt{x^{2}-1}}{1} \right )$$
If $$f:(-\infty ,\infty )\rightarrow (-\infty ,\infty )$$ is defined by $$f(x)=5x-6$$, then $$f^{-1}(x)=$$
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$$\dfrac{x+5}{6}$$
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$$\dfrac{x-5}{6}$$
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$$\dfrac{x-6}{5}$$
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$$\dfrac{x+6}{5}$$
Explanation
$$f(x)=5x-6=y$$
$$\Rightarrow 5x=y+6$$
$$x=\dfrac{y+6}{5}$$
$$\therefore f\left ( x \right )=y\Rightarrow x=f^{-1}\left ( y \right )$$
$$\therefore f^{-1}\left ( y \right )=\dfrac{y+6}{5}$$
If $$f(x)=\dfrac{5x+6}{7x+9}$$ then $$f^{-1}(x)=$$
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$$\dfrac{y+6}{7y+9}$$
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$$\dfrac{7y+9}{5y+6}$$
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$$\dfrac{9y-6}{-7y+9}$$
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$$\dfrac{9y-6}{-7y+5}$$
Explanation
Let $$f\left ( x \right )=y=\dfrac{5x+6}{7x+9}$$
$$7xy+9y=5x+6$$
$$\Rightarrow x\left ( 7y-5 \right )=6-9y$$
$$x=\dfrac{9y-6}{-7y+5}$$
$$x=f^{-1}\left ( y \right )=\dfrac{9y-6}{-7y+5}$$
If $$f$$ from $$R$$ into $$R$$ is defined by $$f(x)=x^{3}-1$$, then $$f^{-1}\left \{ -2,0,7 \right \}=$$
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$$\left \{ -1,1,2 \right \}$$
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$$\left \{ 0,1,2 \right \}$$
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$$\left \{ \pm 1,\pm 2 \right \}$$
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$$\left \{ 0,\pm 2 \right \}$$
Explanation
$$f$$ is invertible throughout $$R$$
$$\therefore f\left ( x \right )=x^{3}-1=y$$
$$\Rightarrow x=\sqrt[3]{y+1}=f^{-1}\left ( y \right )$$
$$\therefore f^{-1}\left ( -2 \right )=-1$$
$$f^{-1}\left ( 0 \right )=1$$
$$f^{-1}\left ( 7 \right )=2$$
$$\therefore f^{-1}\left \{ -2,0,7 \right \}=\left \{ -1,1,2 \right \}$$
If $$f(x)=3x-1$$ and $$g(x)=5x+6$$ then $$(g^{-1}of^{-1})(2)=$$
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$$10$$
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$$-1$$
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$$11$$
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$$12$$
Explanation
$$f\left ( x \right )=3x-1=y$$
$$\Rightarrow x=\displaystyle\frac{y+1}{3}$$
$$\therefore \displaystyle f^{-1}\left ( y \right )=\frac{y+1}{3}$$
$$\Rightarrow \displaystyle f^{-1}\left ( x \right )=\frac{x+1}{3}$$
$$g\left ( x \right )=5x+6=z$$
$$\Rightarrow x=\displaystyle \frac{z-6}{5}$$
$$g^{-1}\left ( z \right )=\displaystyle \frac{z-6}{5}$$
$$\Rightarrow g^{-1}\left ( x \right )=\displaystyle \frac{x-6}{5}$$
$$\displaystyle g^{-1}\left ( f^{-1} \left ( x \right )\right )=g^{-1}\left ( \frac{x+1}{3} \right )$$
$$=\displaystyle \frac{\dfrac{x+1}{3}-6}{5}$$
$$\displaystyle =\frac{x-17}{15}$$
$$\displaystyle \therefore g^{-1}\left ( f^{-1}\left ( 2 \right ) \right )=\frac{2-17}{15}=-1$$
If $$f(x)=e^{5x+13}$$ then $$f^{-1}(x)=$$
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$$\dfrac{13-\log y}{5}$$
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$$\dfrac{-13+\log y}{5}$$
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$$\dfrac{5+\log y}{13}$$
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$$\dfrac{5-\log y}{13}$$
Explanation
Let $$f\left ( x \right )=e^{5x+13}=y$$
$$f\left ( x \right )=y\Rightarrow x=f^{-1}\left ( y \right )$$
$$5x+13=\ln y$$
$$5 x=-13+\log y$$
$$ x=\dfrac{-13+\log y}{5}$$
$$\therefore f^{-1}\left ( y \right )=\dfrac{-13+\log y}{5}$$
If $$f:[1,\infty )\rightarrow [2,\infty) $$ is given by $$f(x)=x+\dfrac{1}{x}$$, then $$f^{-1}(x)=$$
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$$\dfrac{x+\sqrt{x^{2}-4}}{2}$$
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$$\dfrac{x}{1+x^{2}}$$
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$$\dfrac{x-\sqrt{x^{2}-4}}{2}$$
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$$x+\sqrt{x^{2}-4}$$
Explanation
Let $$f\left ( x \right )=x+\dfrac{1}{x}=y$$
$$\Rightarrow x=f^{-1}\left ( y \right )$$ & $$x^{2}-yx+1=0$$
Solving $$x^{2}-yx+1$$, we get
$$x^{2}-yx+1=0$$
$$x=\dfrac{y\pm \sqrt{y^{2}-4}}{2}$$
$$\therefore f^{-1}=\dfrac{x+\sqrt{x^{2}-4}}{2}$$
$$\because$$ $$f$$ is defined from $$\left ( 1,\infty \right )\rightarrow \left ( 2,\infty \right )$$ negative part is discarded.
If $$f:\left \{ 1,2,3,..... \right \}\rightarrow \left \{ 0,\pm 1,\pm 2,..... \right \}$$ is defined by $$f(n)=\begin{cases} \dfrac{n}{2} & \text{ if } n \space is \space even \\-\left (\dfrac{n-1}{2} \right ) & \text{ if } n \space is \space odd \end{cases}$$ then $$f^{-1}(-100)$$ is
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Function is not invertible.
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$$199$$
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$$201$$
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$$200$$
Explanation
$$f(n)$$ is positive if $$n$$ is even & negative if $$n$$ is odd.
$$\therefore f^{-1}\left ( -100 \right )=-2x+1$$
$$=-2\left ( -100 \right )+1$$
$$=200+1$$
$$=201$$
$$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}+4$$ then $$f^{-1}(13)=$$
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$$\left \{ -3,3 \right \}$$
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$$\left \{ -2,2 \right \}$$
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$$\left \{ -1,1 \right \}$$
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Not invertible
Explanation
$$f\left ( x \right )=x^{2}+4=y$$
$$\Rightarrow x^{2}=y-4$$
$$x=\pm \sqrt{y-4}$$
$$\therefore f^{-1}\left ( 13 \right )=\pm \sqrt{13-4}=\pm 3 $$
$$f(3)=13\,\&\,f(-3)=13$$
Thus image $$13$$ has two pre-images i.e, $$3$$ and $$-3$$
$$\therefore$$ $$f$$ is not invertible
If $$f(x)=2+x^{3}$$, then $$f^{-1}(x)$$ is equal to
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$$\sqrt[3]{x}+2$$
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$$\sqrt[3]{x}-2$$
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$$\sqrt[3]{x-2}$$
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$$\sqrt[3]{x+2}$$
Explanation
Let $$f\left ( x \right )=y$$
$$\Rightarrow 2+x^{3}=y$$
$$\Rightarrow x^{3}=y-2$$
$$\Rightarrow x=\left ( y-2 \right )^{1/3}$$
$$\therefore f^{-1}\left ( y \right )=x$$$$=\left ( y-2 \right )^{{1}/{3}}$$
$$\therefore f^{-1}\left ( x \right )=\left ( x-2 \right )^{{1}/{3}}$$
The solution of $$8x\equiv 6(mod \ 14) $$ is
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$$\{8, 6\}$$
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$$\{6,14\}$$
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$$\{6,13\}$$
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$$\{8,14,6\}$$
Explanation
Since,$$8\mathrm{x}\equiv 6(mod \ 14)$$
i.e., $$8\mathrm{x}-6=14\mathrm{k}$$ for $$\mathrm{k}\in \mathrm{I}$$.
$$ \Rightarrow 8x = 14k+6$$
$$ \Rightarrow 4x = 7k+3$$
The values $$6$$ and $$13$$ satisfy this equation (when $$k =3$$ and $$k=7$$),
while
$$8, 14$$ and $$16$$ do not.
If $$f(x)=(1-x)^{1/2}$$ and $$g(x)= \ln(x)$$ then the domain of $$(gof)(x)$$ is
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$$(-\infty ,2)$$
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$$(-1,1)$$
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$$(-\infty ,1]$$
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$$(-\infty ,1)$$
Explanation
Given $$f(x)=(1-x)^{\frac{1}{2}}$$ and $$g(x)=ln(x)$$
$$gof(x)$$
$$=g(f(x))$$
$$=\ln\left ( 1-x \right )^{1/2}$$
$$=\dfrac{1}{2}\ln\left ( 1-x \right )$$
$$\therefore$$ For the composite function to be defined $$1-x>0$$
$$x<1$$
$$\therefore$$ Domain is $$\left ( -\infty ,1 \right )$$
If $$f:R^{+}\rightarrow R$$ such that $$f(x)=\log_{5} x$$ then $$f^{-1}(x)=$$
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$$\log_{x}10$$
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$$5^{x}$$
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$$3^{-x}$$
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$$3^{1/x}$$
Explanation
$$f\left ( x \right )=y$$
$$\Rightarrow \log _{5}x=y$$
$$\Rightarrow \dfrac{\log x}{\log 5}=y$$
$$\Rightarrow \log x=y\log 5$$
$$\Rightarrow e^{\log x}=e^{y\log 5}$$
$$\Rightarrow e^{\log x} =e^{\log 5^{y}}$$
$$[\because a \log x = \log x^{a}$$]
$$\Rightarrow x=5^{y}=f^{-1}\left ( y \right )$$
$$\therefore f^{-1}\left ( x \right )=5^{x}$$
If $$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$$ then $$fofofof(x)=$$
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$$f(x)$$
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$$2\left ( \dfrac{x+1}{x-1} \right )$$
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$$\dfrac{x-1}{x+1}$$
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$$x$$
Explanation
$$fofofof\left(x\right)=fofof\left(\dfrac{x+1}{x-1}\right)$$
$$=fof\left(\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\right)=fof\left(\dfrac{2x}{2}\right)$$
$$=fof\left(x\right)$$
$$=f\left(\dfrac{x+1}{x-1}\right)=x$$
If $$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$$ then $$ (GoG)(n)=$$ (where $$k$$ is odd)
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$$1$$
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$$n$$
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$$2$$
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$$n-1$$
Explanation
$$G(n)=n-(-1)^{k-1}(n-1)$$
$$GoG(n)=G(n-(-1)^{k-1}(n-1))$$
$$=n-(-1)^{k-1}(n-1)-(-1)^{k-1}((n-1)-(-1)^{k-1}(n-1))$$
$$=n-(n-1)$$
$$=1$$
If $$f:[1,\infty )\rightarrow B$$ defined by the function $$ f(x)=x^{2}-2x+6$$ is a surjection, then $$B$$ is equals to
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$$[1,\infty )$$
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$$[5,\infty )$$
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$$[6,\infty )$$
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$$[2,\infty )$$
Explanation
$$f(x)=x^2-2x+6$$ is a surjection.
So the range of $$f(x)$$ will be equal to its codomain.
$$f(x)=x^2-2x+6$$
$$f^1(x)=2x-2$$
$$=2(x-1)$$
$$f(x)$$ will be increasing when $$x\geqslant 1$$.
$$\therefore f(1)=1-2+6$$
$$=5$$
$$\therefore B=[5, \infty)$$
If $$f:R\rightarrow R^{+}$$ then $$\displaystyle f(x)=\left(\dfrac{1}{3}\right)^{x}$$, then $$f^{-1}(x)=$$
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$$\displaystyle \left(\dfrac{1}{3}\right)^{-x}$$
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$$3^{x}$$
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$$\displaystyle \log_{1/3}$$$$ x$$
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$$\displaystyle \log_{x}\left(\dfrac{1}{3}\right)$$
Explanation
Let $$\displaystyle f(x)=\left(\frac{1}{3}\right)^{x}=y$$
Taking logarithm on both sides,
$$\displaystyle x\log \frac{1}{3}=\log y$$
$$\displaystyle \Rightarrow x=\frac{\log y}{\log \frac{1}{3}}$$
$$\displaystyle \Rightarrow x=\log _ { 1/3 } y\quad \quad \quad \left[ \because \frac { \log b }{ \log a } =\log _{ a }{ b } \right] $$
$$\therefore f^{-1}\left ( x \right )=\log _{1/3}x$$
If $$ X =\{1, 2,3,4,5\} $$ and $$Y =\{1,3,5,7,9\}$$, determine which of the following sets represent a relation and also a mapping?
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$$R_{1}= \{(x,y)$$:$$ y=x+2, x \in Y,y \in Y\}$$
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$$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$$
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$$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$$
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$$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$$
Explanation
Here we will check all options one by one:
Option A:
$$R_1 = \{(x,y): y= x+2, x\in X, y\in Y\}$$
$$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$$
Since $$4$$ amd $$6$$ are the images of $$2$$ and $$4$$ respectively but $$4$$ and $$6$$ do not belong to $$Y$$
$$\therefore (2, 4),(4,6)\not\in X \times Y$$
Hence $$R_{1}$$ is not a relation as well as not a mapping.
Option B:
$$R_{2}$$: It is a relation but not a mapping because the element $$1$$ has two different images.
Option C:
$$R_{2}$$: It is a relation but not a mapping because the element $$3$$ has two different images.
Option D:
$$R_{4}$$: It is both a relation and a mapping because every element in $$X$$ is mapped to the elements in $$Y$$
. Also, every element of $$X$$ has a one and only one image in $$Y$$ and every element in $$Y$$ has its pre-image in $$X$$. Hence, it is also one-one and onto mapping and hence it is a bijection.
If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$
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$$\dfrac{x}{\sqrt{1+3x^{2}}}$$
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$$\dfrac{x}{\sqrt{1-x^{2}}}$$
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$$\dfrac{2x}{\sqrt{1+2x^{2}}}$$
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$$\dfrac{x}{\sqrt{1+x^{2}}}$$
Explanation
$$fofof\left(x\right)=fof\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$=f\left(\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+x^{2}}}}\right)$$
$$=f\left(\dfrac{x}{\sqrt{1+2x^{2}}}\right)$$
$$=\left(\dfrac{\dfrac{x}{\sqrt{1+2x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+2x^{2}}}}\right)$$
$$=\dfrac{x}{\sqrt{1+3x^{2}}}$$
If A $$=$${$$x : x^{2}-3x+2= 0$$}, and $$R$$ is a universal relation on $$A$$, then $$R$$ is
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$$\{(1,1),(2, 2)\}$$
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$$\{(1,1)\}$$
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$$\phi $$
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$$\{(1,1),(1, 2)(2,1),(2,2)\}$$
Explanation
Consider, $$x^2-3x+2=0$$
$$\implies x^2-2x-x+2=0$$
$$\implies (x-2)(x-1)=0$$
$$\implies x=1,2$$
$$\therefore A=\{1,2\}$$
Also R is universal relation on set A, then every element of set A is related every other element of A
So $$R= \{(1,1), (1,2), (2,1), (2,2)\}$$.
Assertion(A):
If $$X=\left \{ x:-1\leq x\leq 1 \right \}$$ and $$f:X\rightarrow X$$ defined by $$f(x)=\sin \pi x; \forall x\in A$$ is not invertible function
Reason (R):
For a function $$f$$ to have inverse, it should be a bijection
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Both A and R are true and R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
$$f(-\pi )=f(0)=f(\pi )=0$$
$$\therefore f$$ is not a bijection
$$\therefore \sin\pi x$$ is not invertible.
If $$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}} $$, then $$(fog)(x)=$$
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$$x$$
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$$\dfrac{x}{\sqrt{1+x^{2}}}$$
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$$\sqrt{1+x^{2}}$$
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$$2x$$
Explanation
$$fog\left(x\right)=f\left(g\left(x\right)\right)$$
$$\displaystyle =f\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$\displaystyle =\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1-\dfrac{x^{2}}{1+x^{2}}}}=x$$
If $$f(x)=1+x+x^{2}+x^{3}+\ldots\ldots $$ for $$\left | x \right |<1$$ then $$f^{-1}(x)=$$
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$$\dfrac{x-1}{x+1}$$
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$$\dfrac{x+1}{x}$$
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$$\dfrac{x}{x-1}$$
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$$\dfrac{x-1}{x}$$
Explanation
$$f(x)=1+x+x^{2}+\ldots\ldots$$
$$=\dfrac{1}{1-x}$$
Let $$y=\dfrac{1}{1-x}=f(x)$$
$$1-x=\dfrac{1}{y}$$
$$x=1-\dfrac{1}{y}=\dfrac{y-1}{y}$$
But $$x=f^{-1}(y)=\dfrac{y-1}{y}$$
If the function is $$f:R\rightarrow R, g:R\rightarrow R$$ are defined as $$f(x)=2x+3, g(x)=x^{2}+7$$ and $$f[g(x)]=25$$ then $$x=$$
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$$f(x)$$
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$$\pm 2$$
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$$\pm 3$$
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$$\pm 4$$
Explanation
$$f(g(x))=f(x^{2}+7) =25$$
$$=2(x^{2}+7)+3$$
$$=2x^{2}+17$$
$$\Rightarrow 2x^{2}=8$$
$$x^{2}=4 \Rightarrow x=\pm 2$$
If $$f(x)=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$, then $$f^{-1}(x)=$$
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$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-1}{x+1} \right )$$
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$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-1} \right )$$
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$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x+1}{x-2} \right )$$
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$$\displaystyle\frac{1}{2}\log_{2}\left ( \frac{x-2}{x-1} \right )$$
Explanation
Let $$f\left(x\right)=y=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$
$$\Rightarrow 2^{x}\left(y-1\right)=2^{-x}\left(1+y\right)$$
$$2^{2x}=\displaystyle\frac{y+1}{y-1}$$
$$\displaystyle 2x=\log_{2}\left(\frac{y+1}{y-1}\right)$$
$$\Rightarrow \displaystyle x=f^{-1}\left(y\right)=\frac{1}{2}\log_{2}\left(\frac{y+1}{y-1}\right)$$
So, $$\displaystyle f^{-1}x=\frac{1}{2}\log_{2}\left(\frac{x+1}{x-1}\right)$$
If $$f(x)=\dfrac{x}{\sqrt{1-x^{2}}}$$, then $$ (fof)(x)=$$
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$$\dfrac{x}{\sqrt{1-x^{2}}}$$
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$$\dfrac{x}{\sqrt{1-2x^{2}}}$$
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$$\dfrac{x}{\sqrt{1-3x^{2}}}$$
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$$x$$
Explanation
$$fof\left ( x \right )=f\left ( f\left ( x \right ) \right )=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-\left ( x^{2}/\left ( 1-x^{2} \right ) \right )} \right )$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\sqrt{\left ( 1-x^{2}-x^{2} \right )/\left ( 1-x^{2} \right )}$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-2 x^{2}} /\sqrt{1-x^{2}}\right)$$
$$=x/\sqrt{1-2x^{2}}$$
If $$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}-10x+21 $$ then $$ f^{-1}(-3)$$ is
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$$\left \{ -4,6 \right \}$$
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$$\left \{ 4,6 \right \}$$
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$$\left \{ -4, 4, 6 \right \}$$
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Not Invertible
Explanation
Let $$f^{-1}(-3)=t$$
$$\Rightarrow f(t)=-3$$
$$t^{2}-10t+21=-3$$
$$t^{2}-10t+24=0$$
$$t^{2}-6t-4t+24=0$$
$$t(t-6)-4(t-6)=0$$
$$\Rightarrow t=6,4 = f^{-1}(-3)$$
I: If $$f:A\rightarrow B$$ is a bijection only then does $$f$$ have an inverse function
II: The inverse function $$f:R^{+}\rightarrow R^{+}$$ defined by $$f(x)=x^{2}$$ is $$f^{-1}(x)=\sqrt{x}$$
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only I is true
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only II is true
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both I and II are true
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neither I nor II true
Explanation
$$f$$ is a bijection $$\Rightarrow$$ $$f$$ is one one and onto.
$$\therefore$$ there exists on inverse value for every value in co-domain.
$$\therefore$$ $$f$$ is invertible if and only if $$f$$ is a bijection
$$f:R^{+}\rightarrow R^{+} ; f(x)=x^{2}=y$$
$$\Rightarrow x=\sqrt{y}=f^{-1}(y)$$
If $$f(x)=\sin^{-1}\left \{ 3-(x-6)^{4} \right \}^{1/3}$$ then $$ f^{-1}(x)=$$
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$$6+\sqrt[4]{3+\sin^{3}x}$$
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$$6+\sqrt[4]{3-\sin^{3}x}$$
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$$6+\sqrt[4]{3+\sin x}$$
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$$6+\sqrt[4]{3-\sin x}$$
Explanation
Let $$f(x)=y=\sin^{-1}(3-(x-6)^{4})^{\frac{1}{3}}$$
$$\sin y=(3-(x-6)^{4})^{\frac{1}{3}}$$
$$\sin^{3}y=3-(x-6)^{4}$$
$$(x-6)^{4}=3-\sin^{3}y$$
$$x-6=(3-\sin^{3}y)^{\frac{-1}{4}}$$
$$x=(3-\sin^{3}y)^{\frac{1}{4}}+6$$
$$x=f^{-1}(y)=(3-\sin^{3}y)^{\frac{1}{4}}+6.$$
Which of the following functions defined from $$(-\infty ,\infty )$$ to $$ (-\infty ,\infty )$$ is invertible ?
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$$f(x) = \sin (2x+3)$$
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$$f(x) = x^{2} + 4$$
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$$ f (x) =x^{3}$$
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$$f (x) = \cos x$$
Explanation
$$f(x)=\sin(2x+3)$$ lies only from $$[-1,1]$$ and hence is not onto.
So, $$f(x)$$ is not bijective
$$\therefore$$ it is not invertible
$$f(x)=x^{2}+4$$ is always positive and so not onto.
So, $$f(x)$$ is not bijective
$$\therefore$$ It is also not invertible.
$$f(x)=x^{3}$$ varies from $$(-\infty ,\infty )$$ as $$x$$ varies from $$(-\infty ,\infty ) $$
So, $$f(x)$$ is bijective
$$\therefore$$ It is invertible.
$$f(x)=\cos x$$ always lies between $$[-1,1]$$ and so is into function.
So, $$f(x)$$ is not bijective
Hence, not invertible.
lf $${f}\left({x}\right)=\sin^{2}{x}+\sin^{2}\left({x}+\displaystyle \dfrac{\pi}{3}\right)+ \cos x \cos \left({x}+\displaystyle \dfrac{\pi}{3}\right)$$ and $${g}\left(\displaystyle\dfrac{5}{4}\right)=1$$, $$g\left(1\right) = 0 $$ then $$\left({g}{o}{f}\right)\left({x}\right)=$$
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$$1$$
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$$0$$
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$$\sin x$$
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Data is insufficient
Explanation
$$\displaystyle {f}\left({x}\right)=\sin^{2}{x}+\left(\sin{x}\cos\frac{\pi}{3}+ \cos{x} \displaystyle \sin\frac{\pi}{3}\right)^{2}+ {c}{o}{s} {x}\left(\displaystyle \cos {x}\cos\frac{\pi}{3}-\sin {x}\sin\frac{\pi}{3}\right)$$
$$\displaystyle =\sin ^{ 2 }{ x } +{ \left[ \frac { \sin{ x } }{ 2 } +\frac { \sqrt { 3 } \cos{ x } }{ 2 } \right] }^{ 2 }+\frac { \cos^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \cos x \sin x$$
$$\displaystyle =\sin ^{ 2 }{ x } +\frac { \sin ^{ 2 }{ x } }{ 4 } +\frac { 3 }{ 4 } \cos ^{ 2 }{ x } +\frac { \sqrt { 3 } }{ 2 } \sin x\cos{ x }+\frac { \cos ^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \sin { x\cos { x } } $$
$$=\displaystyle \frac{5}{4}\left(\sin^{2}{x}+\cos^{2}{x}\right)=\frac{5}{4}$$
$$\displaystyle \therefore $$ $$[{g}{o}{f}]\left({x}\right)={g}[{f}\left({x}\right)]={g}\left(\displaystyle \frac{5}{4}\right)=1$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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