Relations and Functions Test

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Relations and Functions Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The inverse of the function $$y=\large{\frac{e^x-e^{-x}}{e^x+e^{-x}}}$$ is

A
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1+x}{1-x}}$$

B
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{2+x}{2-x}}$$

C
$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1-x}{1+x}}$$

D
$$2 \log (1+x)$$

Solution

Given $$y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$

or, $$\dfrac{1+y}{1-y}=\dfrac{2e^x}{2e^{-x}}$$

or, $$e^{2x}=\dfrac{1+y}{1-y} $$

or, $$ x=\dfrac{1}{2}\log \dfrac{1+y}{1-y}$$

So the inverse is $$\dfrac{1}{2}\log \dfrac{1+x}{1-x}$$.
Question 2
1 / -0

Let $$A = \{ 1,2,3,4,5,6\} .$$ The number of onto functions from $$A$$ to$$A$$ such that.$$f\left( x \right) \ne x$$ for all $$x \in A$$ is

A
$$720$$

B
$$240$$

C
$$245$$

D
$$265$$

Solution

$$A=\{1,2,3,4,5,6\}$$

$$f:A\rightarrow A$$

$$\therefore \ $$Number of onto function$$=n!$$

$$ =6!$$

$$ =6\times 5\times 4\times 3\times 2\times 1$$

$$\therefore \ $$Number of onto function$$ =720$$
Question 3
1 / -0

If $$f: A\rightarrow A$$ defined by $$f(x) =\dfrac{4x +3}{6x-4}$$ where $$A= R-\dfrac{2}{3}$$. Find $$f^{-1}$$

A
$$2x$$

B
$$\dfrac {4x+3}{6x-4}$$

C
$$\dfrac x2$$

D
None of these

Solution

$$f\left( x \right) =\cfrac { 4x+3 }{ 6x-4 } \\ let\quad y=\cfrac { 4x+3 }{ 6x-4 } \\ \Rightarrow 6xy-4y=4x+3\\ \Rightarrow 6xy-4x=3+4y\\ \Rightarrow x(6y-4)=3+4y\\ \Rightarrow x=\cfrac { 3+4y }{ 6y-4 } \\ Replace\quad x\quad by\quad y\quad \& \quad y\quad by\quad x.\\ y=\cfrac { 3+4x }{ 6x-4 } \\ f^{ ' }\left( x \right)=\cfrac { 4x+3 }{ 6x-4 } $$
Question 4
1 / -0

If the binary operation $$*$$ is defined on a set of integers as $$a * b = a + 3 b ^{2} $$ , then the value of $$2 * 3$$ is

A
$$27$$

B
$$29$$

C
$$2$$

D
None of these

Solution

$$a\ast b =a+3b^{2}$$
$$\Rightarrow 2\ast 3=2+3(3)^{2}$$
$$=2+27$$
$$=29$$
Question 5
1 / -0

The solution of $$( 3 * 4) * 3$$, when $$*$$ is a binary operation on Z such that: $$a * b = a + b$$, is.

A
$$10$$

B
$$-10$$

C
$$-\dfrac16$$

D
$$-6$$

Solution

Given $$( 3 * 4) * 3$$

$$(3*4)*3=(3+4)+3$$ (as$$(a \ast b)=(a+b)$$)

$$=7+3$$

$$=10$$
Question 6
1 / -0

If $$g\left( x \right) = {x^2} + x - 2$$ and $$\frac{1}{2}gof\left( x \right) = 2{x^2} + 5x + 2$$, then $$f\left( x \right)$$ is

A
$$2x-3$$

B
$$2x+3$$

C
$$2{x^2} + 3x + 1$$

D
$$2{x^2} - 3x -1$$

Solution

$$g(x)={ x }^{ 2 }+x-2\\ \cfrac { 1 }{ 2 } g(f(x))=2{ x }^{ 2 }+5x+2\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-2=4{ x }^{ 2 }-10x+4\\ \Rightarrow f^{ 2 }\left( x \right) +f(x)-(4{ x }^{ 2 }-10x+6)=0\\ f(x)=\cfrac { -1\pm \sqrt { 1+4(4{ x }^{ 2 }-10x+6) } }{ 2 } \\ f(x)=\cfrac { -1\pm \sqrt { 1+16{ x }^{ 2 }-40x+24 } }{ 2 } \\ f(x)=\cfrac { -1\pm (4x-5) }{ 2 } =(2x-3)$$
Question 7
1 / -0

If $$f\left( x \right)$$ is an invertible function, and $$g\left( x \right) = 2$$ $$f\left( x \right) + 5$$, then the value of $${g^{ - 1}}\left( x \right)$$ is

A
$$2{f^{ - 1}}\left( x \right) - 5$$

B
$${\dfrac{1} {2{f^{ - 1}}\left( x \right) + 5}}$$

C
$${\dfrac{1} {2}}{f^{ - 1}}\left( x \right) + 5$$

D
$${f^{ - 1}(x)}\left( \dfrac {x - 5} {2} \right)$$

Solution

Given:$$g\left(x\right)=2f\left(x\right)+5$$
Replace $$x\rightarrow\,{g}^{-1}{\left(x\right)}$$
$$\Rightarrow\,g{g}^{-1}{\left(x\right)}=2f\left({g}^{-1}{\left(x\right)}\right)+5$$
$$\Rightarrow\,x=2f\left({g}^{-1}{\left(x\right)}\right)+5$$
$$\Rightarrow\,x-5=2f\left({g}^{-1}{\left(x\right)}\right)$$
$$\Rightarrow\,f\left({g}^{-1}{\left(x\right)}\right)=\dfrac{x-5}{2}$$
$$\Rightarrow\,{g}^{-1}{\left(x\right)}={f}^{-1}\left(\dfrac{x-5}{2}\right)$$
Question 8
1 / -0

Let f : $$R \to R$$ and g : $$R \to R$$ be two one-one and onto functions such that they are the mirror images of each other about the line y = 2. If h(x) = f(x) + g(x), then h(0) equal to

A
2

B
4

C
0

D
1

Solution

Given $$g(x)$$ and $$f(x)$$ are mirror images about $$y=2$$
$$\implies g(x)-2=2-f(x)\implies f(x)+g(x)=4\implies h(x)=4$$
$$h(0)=4$$
Question 9
1 / -0

Let $$A$$ be a set of $$4$$ elements and $$B$$ has $$3$$ elements . From the set of all functions from $$A$$ to $$B$$, the probability that it is an onto function is

A
$$\dfrac{4}{9}$$

B
$$0$$

C
$$\dfrac{29}{32}$$

D
$$1$$

Solution

No .of functions from $$A$$ to $$B$$ is $$3^4=81$$ elements
No.of onto functions $$4\times 3^2=36$$
Probability $$\dfrac{36}{81}=\dfrac 49$$
Question 10
1 / -0

If $$f : A \rightarrow B $$ defined as $$f(x) = x^2+2x+\frac{1}{1+(x+1)^2}$$ is onto function, then set B is equal to

A
$$[0, \infty)$$

B
$$(-\infty, 0)$$

C
$$[2, \infty)$$

D
$$(-\infty, \infty)$$

Solution

$$f:A\rightarrow B,\quad f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ \because \cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } >0\\ \therefore f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ a=1,\quad b=2\\ D=4-4\times 1\times (\cfrac { 1 }{ +ve } )\quad \ge 0\\ \because x\epsilon R$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 63

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Relations and Functions Test - 63

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SHARING IS CARING

Question 1

$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1+x}{1-x}}$$

$$\large{\frac{1}{2}}$$ $$\log \large{\frac{2+x}{2-x}}$$

$$\large{\frac{1}{2}}$$ $$\log \large{\frac{1-x}{1+x}}$$

$$2 \log (1+x)$$

Solution

Question 2

$$720$$

$$240$$

$$245$$

$$265$$

Solution

Question 3

$$2x$$

$$\dfrac {4x+3}{6x-4}$$

$$\dfrac x2$$

None of these

Solution

Question 4

$$27$$

$$29$$

$$2$$

None of these

Solution

Question 5

$$10$$

$$-10$$

$$-\dfrac16$$

$$-6$$

Solution

Question 6

$$2x-3$$

$$2x+3$$

$$2{x^2} + 3x + 1$$

$$2{x^2} - 3x -1$$

Solution

Question 7

$$2{f^{ - 1}}\left( x \right) - 5$$

$${\dfrac{1} {2{f^{ - 1}}\left( x \right) + 5}}$$

$${\dfrac{1} {2}}{f^{ - 1}}\left( x \right) + 5$$

$${f^{ - 1}(x)}\left( \dfrac {x - 5} {2} \right)$$

Solution

Question 8

2

4

0

1

Solution

Question 9

$$\dfrac{4}{9}$$

$$0$$

$$\dfrac{29}{32}$$

$$1$$

Solution

Question 10

$$[0, \infty)$$

$$(-\infty, 0)$$

$$[2, \infty)$$

$$(-\infty, \infty)$$

Solution

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