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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 8
Let $$f\left( x \right) = {x^2}$$ and $$g\left( x \right) = {2^x}$$. Then the solution of the equation $$fog\left( x \right) = gof\left( x \right)$$ is
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$$R$$
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$$\left\{ {0} \right\}$$
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$$\left\{ {0,\,2} \right\}$$
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none
Explanation
$$f\left(x\right)={x}^{2}$$
$$f{g\left(x\right)}=f{\left({2}^{x}\right)}={\left({2}^{x}\right)}^{2}={2}^{2x}$$
$$g\left(x\right)={2}^{x}$$
$$g{f\left(x\right)}=g{\left({x}^{2}\right)}={2}^{{x}^{2}}$$
Given:$$f{g\left(x\right)}=g{f\left(x\right)}$$
$$\Rightarrow\,{2}^{2x}={2}^{{x}^{2}}$$
Since bases are same we can equate the powers
$$\Rightarrow\,2x={x}^{2}$$
$$\Rightarrow\,{x}^{2}-2x=0$$
$$\Rightarrow\,x\left(x-2\right)=0$$
$$\Rightarrow\,x=0,2$$
When $$x=0,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{0}={2}^{0}=1$$
When $$x=2,\,f{g\left(x\right)}=g{f\left(x\right)}\Rightarrow\,{2}^{2\times 2}={2}^{{2}^{2}}\Rightarrow\,16=16$$
Hence $$x=\left\{0,2\right\}$$
Let $$g(x)=1+x-[x]\quad $$ and $$f(x)=\begin{cases} -1\quad if\quad x<0 \\ 0\quad \quad if\quad x=0 \\ 1\quad \quad if\quad x>0 \end{cases}$$ , then $$\forall \:x,fog(x)$$ equals
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$$x$$
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$$1$$
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$$f(x)$$
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$$g(x)$$
Explanation
given g(x)=1+x=[x] 1+{x} {x} donates the fractional part of x so,
g(x) is always positive
so, input to f(g(x)) is always positive and
hence, fog(x)=1
If $$f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$$ where $$a > 0$$ and $$n$$ is a positive integer then $$( f o f ) ( x )$$ is
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$$f ( x )$$
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$$x$$
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$$0$$
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$$1$$
Explanation
$$f(n)= (a-x^{n})^{1/n}$$
$$(fof)^{(n)}=(a-((a-x^{n})^{1/n})^{n})^{1/n}$$
$$=(a-a+x^{n})^{1/n}$$
$$fof(n)=x$$
The inverse of the function $$f(x)=\dfrac {e^{x}-e^{-x}}{e^{x}+e^{-x}}+2$$ is given by
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$$\log { e{ \left( \dfrac { x-1 }{ x+1 } \right) }^{ -2 } }$$
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$$\log { e{ \left( \dfrac { x-2 }{ x+1 } \right) }^{ 1/2 } }$$
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$$\log { e{ \left( \dfrac { x }{ 2-x } \right) }^{ 1/2 } }$$
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$$\log { e{ \left( \dfrac { x-1 }{ 3-x } \right) }^{ 1/2 } }$$
$$f:R \rightarrow R$$ such that $$f(x)=\ell n(x+\sqrt {x^{2}+1})$$. Another function $$g(x)$$ is defined such that $$gof(x)=x\ \forall\ x \in\ R$$. Then $$g(2)$$ is -
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$$\dfrac {e^{2}+e^{-2}}{2}$$
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$$e^{2}$$
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$$\dfrac {e^{2}-e^{-2}}{2}$$
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$$e^{-2}$$
Explanation
$$f(x)= ln (x+ \sqrt{x^{2}+1})$$
$$g(x)= g(f(x))$$ $$ g(2)=?$$
ln $$(x+ \sqrt{x^{2}+1})=y$$
$$\sqrt{x^2+1}= e^y-{x}$$
$$x^{2}+1= e^{2y}-2e^{y}x+x^{2}$$
$$e^{2y}-2e^{y}x-1=o \Rightarrow$$
$$x =\dfrac{e^{2y}-1}{2e^{y}}$$
$$=\dfrac{(e^{y}- e^{-y})}{2}$$
$$\therefore g(x) = \dfrac{e{^y}- e^{-y}}{2} \Rightarrow g (2)= \dfrac{e^{2}-e^{-2}}{2}$$
Let $$f:R\rightarrow R$$ is a function satisfying $$f(2-x)=f(2+x)$$ and $$f(20-x)=f(x)\forall x\in R$$
If $$f(0)=5$$ then the minimum possible no. of values of $$x$$ satisfying $$f(x)=5$$ for $$x=[0.,70]$$, is
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$$21$$
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$$12$$
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$$11$$
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$$22$$
Explanation
Given, $$f(2 - x) = f(2 + x) .. (i)$$
$$\therefore f(x)$$ is symmetric about $$x = 2$$
$$f(20 - x) = f(x) ..(ii)$$
$$x \rightarrow x + 10$$
$$f(10 - x) = f(10 + x)$$
$$f(x)$$ is symmetric about $$x = 10$$
$$x \rightarrow x + 2$$
$$f(18 - x) = f(x + 2) .. (ii)$$
$$f(2 - x) = f(18 - x)$$
$$x \rightarrow -x$$
$$f(2 + x) = f(18 + x)$$
$$\therefore x + 2 \rightarrow x$$
$$\therefore f(x) = f(x + 16)$$
$$f(x)$$ has period = $$16$$
$$f(x) = 5 , \, x \in [0, 170]$$
put $$x = 2$$ in (i) $$f(0) = f(4)$$
put $$x = 4$$ in (ii) $$f(4) = f(16)$$
when $$x \in [0, 16] \rightarrow f(x) $$ has two solution
when $$x \in [0, 160]$$ has $$20$$ solution.
When $$x in [0, 170]$$ has $$'21'$$ solution
All values of a for which f : R $$ \to R$$ defined by f(x)= $${x^3} + a{x^2} + 3x + 100$$ is a one one functions, are
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$$( - \infty , - 2)$$
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$$( - \infty ,4)$$
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$$(4, - 4)$$
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$$( - 3,3)$$
Explanation
$$f ^ { \prime } ( x ) = 3 x ^ { 2 } + 2 a x + 3$$
$$f ^ { \prime } ( x ) = 0$$ and f ( x ) > 0
$$f ^ { \prime } ( x ) < 0$$
$$ 3 x ^ { 2 } + 2 a x + 3 = 0$$
$$ x = \dfrac { - 2 a \pm \sqrt { 4 a ^ { 2 } - 36 } } { 6 }$$
$$ 4 a ^ { 2 } < 36$$
$$ a \in ( - 3,3 ) $$
Let $$A = \{ 1,2,3 \}$$ . Which of the following functions on
A is invertible?
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$$f = \{ ( 1,1 ) , ( 2,1 ) , ( 3,1 ) \}$$
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$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,1 ) \}$$
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$$f = \{ ( 1,2 ) , ( 2,3 ) , ( 3,2 ) \}$$
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$$f = \{ ( 1,1 ) , ( 2,2 ) , ( 3,1 ) \}$$
Explanation
$$A=\left\{ 1,2,3 \right\} $$
$$\therefore$$ $$A$$ is invertible when obviously
$$f=\left\{ \left( 1,1 \right) ,\left( 2,1 \right) ,\left( 3,1 \right) \right\} $$
If $$f ( x ) = \sin ^ { - 1 } ( \sin x ) + \cos ^ { - 1 } ( \sin x ) \text { and } \phi ( x ) = f ( f ( f ( x ) ) )$$ then $$\phi ^ { \prime } ( x )$$
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1
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$$\sin x$$
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0
0%
none of these
Explanation
$$f(x)=\sin ^{ -1 }{ \left( \sin { x } \right) } +\cos ^{ -1 }{ \left( \sin { x } \right) } =\cfrac { \pi }{ 2 } \left[ \because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\cfrac { \pi }{ 2 } \right] $$
$$f(f(x))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=f(f(f(x)))=f\left(\cfrac { \pi }{ 2 }\right )=\cfrac { \pi }{ 2 } $$
$$\phi (x)=\cfrac { \pi }{ 2 } $$
$$\phi '(x)=0$$
if $$f\left( x \right) = 3x + 2$$ , $$g\left( x \right) = {x^2} + 1$$,then the values of $$\left( {f_og} \right)\left( {{x^2} - 1} \right)$$
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$$3{x^4} - 6{x^2} + 8$$
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$$3{x^4} + 3x + 4$$
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$$6{x^4} + 3{x^2} - 2$$
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$$6{x^4} + 3{x^2} + 2$$
Explanation
$$f\left(x\right)=3x+2$$
$$g\left(x\right)={x}^{2}+1$$
$$f.g\left(x\right)=f\left({x}^{2}+1\right)$$
$$f.g\left(x\right)=3\left({x}^{2}+1\right)+2=3{x}^{2}+5$$
$$f.g\left(x\right)=3{x}^{2}+5$$
$$f.g\left({x}^{2}-1\right)=3{\left({x}^{2}-1\right)}^{2}+5$$
$$=3\left({x}^{4}-2{x}^{2}+1\right)+5$$
$$=3{x}^{4}-6{x}^{2}+3+5$$
$$=3{x}^{4}-6{x}^{2}+8$$
Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If $$f:A\rightarrow B$$ is an one-one function and $$f(x)=x$$ holds only for one value of $$x\epsilon \{ 1,2,3,4,5\} ,$$ then the number of such possible function is
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$$120$$
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$$36$$
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$$45$$
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$$44$$
Difference between the greatest and the least values of the function
$$f(x) = x(ln x - 2)$$ on $$[1, e^{2}]$$ is
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$$2$$
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$$e$$
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$$e^{2}$$
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$$1$$
The function $$f :\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\rightarrow \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ defined by $$f(x)=\sin^{-1}(3x-4x^{3})$$ is
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both one-one and onto
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onto but not one-one
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one-one but not onto
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neither one-one nor onto
Explanation
$$f:[-\cfrac{1}{2},\cfrac{1}{2}]\rightarrow[-\cfrac{\pi}{2},\cfrac{\pi}{2}]\\f(x)=\sin^{-1}(3x-4x^3)$$
Range of $$\sin^{-1}(3x-4x^3)$$ is $$
[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$
And $$-\cfrac{\pi}{2}\le\sin^{-1}(3x-4x^3)\le\cfrac{\pi}{2}$$
$$\Rightarrow \sin(\cfrac{-\pi}{2})\le3x-4x^3\le\sin\cfrac{\pi}{2}$$
$$\Rightarrow-1\le3x-4x^3\le1$$
$$3x-4x^3\le-1$$
$$\Rightarrow 4x^3-3x-1\le0$$
$$\Rightarrow x\le\cfrac{1}{2}$$
and$$3x-4x^3\le1\Rightarrow 4x^3-3x+1\le0\\ \Rightarrow x\le-\cfrac{1}{2}\\ \therefore x\epsilon [-\cfrac{1}{2},\cfrac{1}{2}]\\f{x}=\sin^{-1}(3x-4x^3)\\ \Rightarrow f^1(x)=\cfrac{1}{\sqrt{1-(3x-4x^3)^2}}\times(3-12x^2)\\ \Rightarrow f^1(x)=3-12x^2\\ \quad=-(12x^2-3)\\x^2\le0\Rightarrow 12x^2\le0\Rightarrow 12x^2-3\le-3\\-3(12x^2-3)\le3\\ \therefore f^1(x)\le0$$
$$\therefore f^1(x)$$ is a decreasing function.
Hence $$f(x)$$ is one-one and range of function $$=$$ its co-domain.
Hence it is both one-one and onto.
If $$f\left( x \right) = \frac{{x - 1}}{{x + 1}}$$, then $$f^{-1}\left( x \right)$$ is
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$$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$
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$$\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$
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$$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$$
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$$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$$
Let g be the inverse function of differentiable function f and $$G\left( x \right) =\frac { 1 }{ g\left( x \right) } if\quad f\left( 4=2 \right) $$ and $$f'\left( 4 \right) =\frac { 1 }{ 16 } $$, then the value of $${ \left( G'\left( 2 \right) \right) }^{ 2 }$$ equals to:
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1
0%
4
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16
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64
If $$f:( - 1,1) \to B$$ , is a function defined by $$f(x) = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$$, then find $$B$$ when $$f(x)$$ is both one-one and onto function.
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$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
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$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
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$$\left( {0,\frac{\pi }{2}} \right)$$
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$$\left[ {0,\frac{\pi }{2}} \right)$$
Explanation
For $$x \epsilon (-1,1)$$, we have
$$f(x)=\tan^{-1}\left[\dfrac {2x}{1-x^2}\right]$$
Substituting $$x=\tan\theta$$ in above equation.
Therefore, $$f(\tan \theta)=\tan^{-1}\left[\dfrac {2\tan \theta}{1-\tan^2\theta}\right]$$
$$=\tan^{-1}\tan (2\theta)=2\theta$$
$$=2\tan ^{-1}x$$
Thus $$-\dfrac {\pi}{2}<\tan ^{-1}\left[\dfrac {2x}{1-x^2}\right]<\dfrac {\pi}{2}$$
Thus option B is correct.
If $$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)\ \forall x\ \epsilon \ R$$, then $$f(x)$$ is
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one-one and onto
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one-one and into
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many-one and onto
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non-invertible
The multiplicative inverse of the product of the additive inverse of x+1 is ________________.
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$$x-1$$
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$$\dfrac { 1 }{ 1-x } $$
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$${ x }^{ 2 }-1$$
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$$\dfrac { 1 }{ 1-{ x }^{ 2 } } $$
Let S be a non-empty set and P(S) be the power set of set S. Find the identity element for the union $$(\cup)$$ as a binary operation on $$P(S)$$.
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$$\phi$$
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$$1$$
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$$0$$
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None of these
Explanation
We observe that
$$A \cup \phi = A = \phi \cup A$$ for every subset A of set S.
$$A \cup \phi = A =\phi \cup A$$ for all $$A \in P(S)$$
$$\phi$$ is the identity element for union $$(\cup)$$ on $$P(S)$$.
If $$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$$ is not invertible, then $$k=$$
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$$2$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$3$$
Explanation
We have,
$$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$$
Since, $$|A| = 0$$ if it is not invertible,
$$1(2k) - ((\cos 60) ( 2 \tan 45 )) = 0$$
$$2k - 1 = 0$$
$$k = \dfrac{1}{2}$$
Hence, this is the answer.
Let $$N$$ be the set of natural numbers and two functions $$f$$ and $$g$$ be defined as $$f,g : N\to N$$ such that :
$$f (n)= \begin{cases}\dfrac{n+1}{2}& \text{if n is odd}\\ \dfrac{n}{2} & \text{in n is even} \end{cases}$$
and $$g(n) = n - (-1)^n$$. The fog is:
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Both one-one and onto
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One-one but not onto
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Neither one-one nor onto
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onto but not one-one
Explanation
$$fx = \begin{cases} \dfrac{n+1}{2}& \text{n is odd} \\\dfrac{n}{2}& \text{n is even}\end{cases}$$
$$g(x) = n -(-1)^n \begin{cases}n+1; \text{n is odd} \\n-1; \text{n is even} \end{cases}$$
$$f(g(n)) = \begin{cases}\dfrac{n}{2}; & \text{n is even}\\ \dfrac{n+1}{2}; &\text{n is odd} \end{cases}$$
$$\therefore$$ onto but not one-one
The numbers system which uses alphabets as well as numbers is-
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Binary numbers system
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Octal numbers system
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Decimal numbers system
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Hexadecimal numbers system
Explanation
HEXADECIMAL NUMBERS SYSTEM
$$f : R \rightarrow R , f ( x ) = e ^ { | x | } - e ^ { - x }$$ is many-one into function.
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True
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False
Explanation
$$f(x)=e^{(x)}-e^{-x}$$
For every $$x$$, there will be different $$f(x) $$
$$\therefore $$ It is a one - one function
$$\therefore False $$
Number of one-one functions from A to B where $$n(A)=4, n(B)=5$$.
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$$4$$
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$$5$$
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$$120$$
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$$90$$
Explanation
$$n(A)=4$$ and $$n(B)=5$$
For one-one mapping
4 elements can be selected out of 5 elements of set B in $${}^5C_4$$ ways
and then those 4 selected elements can be mapped with 4 elements of set A in $$4!$$ ways.
Number of one-one mapping from $$A$$ to $$B$$ $$={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120$$
Let $$f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1$$. If $$f(x)$$ divided by $$x^ {3}-x$$, then the remainder is some function of $$x$$ say $$g(x)$$. Then $$g(x)$$ is an:-
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one-one function
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many one function
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into function
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onto function
Explanation
$$f : R \rightarrow R , f ( x ) = 2 x + | \sin x |$$ is one-one onto.
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True
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False
Explanation
$$ f(x) = 2x + \left | sin x \right |$$
for query x,there will be a defferent f(x)
$$ \therefore one-one $$ & domain $$\rightarrow$$ R
$$ \therefore onto $$
$$ \therefore True. $$
If $$ f : R \rightarrow R $$ be given by $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}}, $$ then $$fof(x)$$ is
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0%
$$
x^{\dfrac{1}{3}}
$$
0%
$$
1^{3}
$$
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x
0%
$$
\left(3-x^{3}\right)
$$
Explanation
Given,
$$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}} $$.
Now,
$$fof(x)$$
$$=f[f(x)]$$
$$ =\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}}, $$
$$ =\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}}, $$
$$ =\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}}, $$
$$=[x^3]^{\dfrac{1}{3}}$$
$$=x$$.
Let : $$R\rightarrow R$$ defined as $$f\left( x \right) =\dfrac { x\left( x+1 \right) \left( { x }^{ 4 }+1 \right) +{ 2x }^{ 4 }+{ x }^{ 2 }+2 }{ { x }^{ 2 }+x+1 } $$
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odd and one-one
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even and one-one
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many to one and even
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many to one and neither even nor odd
If a binary operation is defined $$a\star b=a^b$$ then 2$$\star 2$$ is equal to:
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$$4$$
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$$2$$
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$$9$$
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$$8$$
Explanation
$$a \star b = {a}^{b} \quad \left( \text{Given} \right)$$
Therefore,
$$2 \star 2 = {2}^{2} = 4$$
Hence the correct answer is $$4$$.
If is a binary operation such that $$a * b = a^2 + b^2$$ then $$3 * 5$$ is
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34
0%
9
0%
8
0%
25
Explanation
$$a \star b = {a}^{2} + {b}^{2} \quad \left( \text{Given} \right)$$
Therefore,
$$3 \star 5 = {3}^{2} + {5}^{2} = 9 + 25 = 34$$
Hence the correct answer is $$34$$.
Consider $$f(x) = \dfrac{x^2}{1 + x^3}$$ ; $$g(t) = \displaystyle \int f(t) dt$$ . If $$g(1) = 0$$ then $$g(x)$$ equals
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$$\dfrac{1}{3} ln(1 + x^3)$$
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$$\dfrac{1}{3} ln\left ( \dfrac{1 + x^3}{2} \right )$$
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$$\dfrac{1}{2} ln\left ( \dfrac{1 + x^3}{3} \right )$$
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$$\dfrac{1}{3}l n\left ( \dfrac{1 + x^3}{3} \right )$$
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
$$f(x)=x^{3}+x^{2}+3 x+\sin x$$
$$Suppose$$
$$f(x_{1})=f(x_{2})$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$$
$$Equating\;above\;equations\;to \;0$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$$
$$Solving\;them$$
$$It\; will\; not\; prove\; that$$
$$x_{1}=x_{2}$$
$$So,\;f(x)\;is\;one-one$$
$$Range\;f(x)=Real$$
$$Co-domain\;f(x)=Real$$
$$Range=Co-domain$$
$$Hence,\; f(x) \;is \;onto$$
$$So,\;option\; C \;is\;correct.$$
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
0%
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.
Let $$f:[2,\infty )\rightarrow X$$ be defined by $$f(x)=4x-{x}^{2}$$. Then, $$f$$ is invertible, if $$X=$$
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$$[2,\infty)$$
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$$(-\infty,2]$$
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$$(-\infty,4]$$
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$$[4,\infty)$$
Explanation
Since, $$f$$ is invertible, range of $$f=$$ Co-domain of $$f=X$$
So, we need to find the range of $$f$$ to find $$X.$$
Foe finding the range, let
$$f(x)=y$$
$$\Rightarrow$$ $$4x-x^2=y$$
$$\Rightarrow$$ $$x^2-4x=-y$$
$$\Rightarrow$$ $$x^2-4x+4=4-y$$ [ Adding $$4$$ on both sides ]
$$\Rightarrow$$ $$(x-2)^2=4-y$$
$$\Rightarrow$$ $$x-2=\pm\sqrt{4-y}$$
$$\Rightarrow$$ $$x=2\pm\sqrt{4-y}$$
This is defined only when,
$$4-y\ge 0$$
$$\Rightarrow$$ $$y\le 4$$
$$X=$$ Range of $$f=(-\infty , 4]$$
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 $$A.$$
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
$$g(x)=x^2+x-1$$
$$g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}$$
$$g\left(f\left(\dfrac{5}{4}\right)\right)=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$
$$-\dfrac{5}{4}=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$
$$f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0$$
$$\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0$$
$$\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}$$
The inverse function of
$$f(x) = \dfrac{8^{2x}-{8^{-2x}}} {8^{2x}+8^{-2x}}\in (-1, 1)$$, is ________.
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$$\dfrac{1}{4} \log_e\left(\dfrac{1-x}{1+x}\right)$$
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$$\dfrac{1}{4} \log_e\left(\dfrac{1+x}{1-x}\right)$$
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$$\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1-x}{1+x}\right)$$
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$$\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1+x}{1-x}\right)$$
Explanation
$$f(x) = \dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}$$
$$y =\dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}$$
$$\dfrac{1+y}{1-y} = \dfrac{8^{2x}}{8^{-2x}}$$
$$8^{4x} = \dfrac{1+y}{1-y}$$
$$4x = \log_8\left(\dfrac{1+y}{1-y}\right)$$
$$x = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)$$
$$f^{-1}(y) = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)$$
$$\therefore \boxed {f^{-1}(x) = \dfrac{1}{4} \log_8 \left(\dfrac{1+x}{1-x}\right)}....Answer$$
Hence option $$'B'$$ is the answer.
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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$$x$$
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$$(2x-3)$$
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$$\dfrac{4x-6}{3x+4}$$
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None of these
If $$f(x)=\sqrt[3]{3-x^3}$$ then $$(f o f)(x)=?$$
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$$x^{1/3}$$
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$$x$$
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$$(1-x^{1/3})$$
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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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$$3x -2$$
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$$4x -5$$
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$$3x -4$$
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$$2x -3$$
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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$$ 2 $$
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$$ 1 $$
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$$ 3 $$
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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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$$ 16 $$
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$$ 32 $$
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$$ 64 $$
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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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2
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1/3
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2/3
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-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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$$\phi$$
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$$R$$
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$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$
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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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One-one and onto
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One-one and into
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Many-one and onto
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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 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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2log 3
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$$ (\log_{e}3)^{2} $$
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$$ \log_{e} 6 $$
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None of these
Explanation
Given f(x) is continuous at $$x=0$$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0) $$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$$
$$ \Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$$ $$ (\log_e3)^{2} $$
Let f(x)= max { 1+sinx, 1, 1 -cosx}, $$x \epsilon [0, 2 \pi]$$ and g(x)= max {1, |x-1|} $$x \epsilon R$$, then
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g(f(0))=1
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g(f(1))=1
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f(f(1))=1
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f(g(0))=1+sin1
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 = $$2^{2n + 1}$$
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| A - B| = 1`
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$$lim_{n \to \infty} \dfrac{A}{B} = -1$$
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None of these
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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