Relations and Functions Test

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Relations and Functions Test - 48

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Question 1
1 / -0

If $$f(x)=\begin{cases} 2x+3\quad \quad x\le 1 \\ a^{ 2 }x+1\quad x>1 \end{cases}$$, then the values of $$a$$ for which $$f(x)$$ is injective.

A
$$-3$$

B
$$1$$

C
$$0$$

D
none of these

Solution

A function is called injective (one-one) if it is monontonic.
Clearly for $$x<1$$ f is increasing and it's maximum values is $$2(1)+3=5$$
Hence for f to be monotonic $$(x>1)$$, $$f(1) \geq 5$$
$$\Rightarrow a^2(1)+1\geq 5 \Rightarrow a^2\geq 2 \Rightarrow a\in R-(-2,2)$$
Question 2
1 / -0

The relation $$R$$ defined in $$A=\left\{ 1,2,3, \right\} $$ by $$a\ R$$ $$b$$ if $$\left| { a }^{ 2 }-{ b }^{ 2 } \right| \le 5$$. Which of the following is false?

A
$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$

B
$${R}^{-1}=R$$

C
Domain of $$R=\left\{ 1,2,3 \right\} $$

D
Range of $$R=\left\{ 5 \right\} $$

Solution

$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$

Clearly $$R^{-1} = R$$

Domain $$=\{1,2,3\}$$ Range $$=\{1,2,3\}$$

Hence option 'D' is a false statement with regard to $$R$$.
Question 3
1 / -0

If the function $$f:[1,\infty )\rightarrow [1,\infty )$$ is defined by $$f\left( x \right) ={ 2 }^{ x\left( x-1 \right) },$$ then $${ f }^{ -1 }\left( x \right) $$ is

A
$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }^{ x\left( x-1 \right) }$$

B
$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }\left( 1+\sqrt { 1+4\log _{ 2 }{ x } } \right) $$

C
$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }\left( 1-\sqrt { 1+4\log _{ 2 }{ x } } \right) $$

D
Not defined

Solution

$$\displaystyle f\left( x \right)={ 2 }^{ x\left( x+1 \right) },$$ i.e., $$y={ 2 }^{ x\left( x+1 \right) },$$

$$\displaystyle \Rightarrow \log _{ 2 }{ y } =x\left( x-1 \right) ={ x }^{ 2 }-x$$

$$\Rightarrow { x }^{ 2 }-x-\log _{ 2 }{ y } =0$$

$$\displaystyle \Rightarrow x=\frac { 1\pm \sqrt { 1+4\log _{ 2 }{ y } } }{ 2 } $$

Since $$x\in [1,\infty )$$ $$\therefore$$ -ve sign is ruled out.

$$\displaystyle \therefore \Rightarrow x=\frac { 1 }{ 2 } \left( 1+\sqrt { 1+4\log _{ 2 }{ y } } \right) $$

$$\displaystyle \Rightarrow f^{ -1 }\left( x \right) =\frac { 1 }{ 2 } \left( 1+\sqrt { 1+4\log _{ 2 }{ y } } \right) $$

$$\displaystyle \Rightarrow f^{ -1 }\left( x \right) =\frac { 1 }{ 2 } \left( 1+\sqrt { 1+4\log _{ 2 }{ x } } \right) $$
Question 4
1 / -0

If $$g(x)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$, then $$f(x)=$$

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

B
$$2+{ x }^{ 2 }$$

C
$$1+x$$

D
$$2+x$$

Solution

We have $$g\left( x \right)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$ ...(1)

Also, $$f\left( g\left( x \right) \right) =f\left( 1+\sqrt { x } \right) $$ ...(2)

From (1) and (2), we get

$$f\left( 1+\sqrt { x } \right) =3+2\sqrt { x } +x$$

Let $$1+\sqrt { x } =y\Rightarrow x={ \left( y-1 \right) }^{ 2 }$$

$$\therefore f\left( y \right) =3+2\left( y-1 \right) +{ \left( y-1 \right) }^{ 2 }\\ =3+3y-2+{ y }^{ 2 }-2y+1=2+{ y }^{ 2 }\\ \therefore f\left( x \right) =2+{ x }^{ 2 }$$
Question 5
1 / -0

Let $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$. Which of the following is a relation from $$X$$ to $$Y$$.

A
$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$

B
$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

C
$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

D
$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Solution

Given $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$

$$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$$
$$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\} $$

Now, $$R_1=\{(1,3),(2,2)\}$$

Since, $$R_1\subset X\times Y$$

Hence, $$R_1$$ is a relation from $$X$$ to $$Y$$

$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

Since, $$(5,5) \in R_2$$ but $$\notin X\times Y$$

So, $$R_2$$ is not a relation from $$X$$ to$$Y$$

$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

Since, $$(3,5),(3,7),(5,7) \in R_3$$ but $$\notin X\times Y$$

So, $$R_3$$ is not a relation from $$X$$ to $$Y$$

$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Since, $$(2,5),(7,9) \in R_4$$ but $$\notin X\times Y$$

So, $$R_4$$ is not a relation from $$X$$ to $$Y$$
Question 6
1 / -0

Let $$f:\left( 4,6 \right) \rightarrow \left( 6,8 \right) $$ be a function defined by $$f(x)=x+2$$ (where $$[.]$$ denote the greatest integer function), then $$f^{ -1 }\left( x \right) $$ is equal to

A
$$\displaystyle x-\left[ \frac { x }{ 2 } \right] $$

B
$$-x-2$$

C
$$x-2$$

D
$$\displaystyle \frac { 1 }{ x+\left[ \frac { x }{ 2 } \right] } $$

Solution

Since $$f:\left( 4,6 \right) \rightarrow \left( 6,8 \right) $$

$$f\left( x \right) =x+2$$

Let $$y=x+2\Rightarrow x=y-2$$

$$f^{ -1 }\left(y \right) =y-2$$

Replacing $$x$$ by $$y$$ and $$y$$ by $$x$$, we get

$$f^{ -1 }\left( x \right) =x-2$$
Question 7
1 / -0

Which of the following is an onto function

A
$$f\, :\, [0,\, \pi]\, \rightarrow\, [-\, 1\, 1], f(x)\, =\, \sin\, x$$

B
$$f\, :\, [0, \pi]\, \rightarrow\, [-1, 1],\, f(x)\, =\, \cos\, x$$

C
$$f\, :\, R\, \rightarrow\, R,\, f(x)\, =\, e^{x}$$

D
$$f\, :\, Q\, \rightarrow\, R, f(x)\, =\, x^{3}$$

Solution

A function is called onto iff, Range$$=$$ co-domain.
A. In $$[0,\pi]$$, Range of $$\sin x$$ is $$[0,1]\Rightarrow$$ Not onto.
B. In $$[0,\pi]$$, Range of $$\cos x$$ is $$[-1,1] =$$co-domain $$\Rightarrow $$onto function.
C. Range of $$e^x$$ is $$(0, \infty)\neq$$ co-domain.
D. Range of $$x^3$$ is not R, since $$x\in Q$$
Question 8
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Let f : {x,y,z} $$\rightarrow$$ {a,b,c} be a one-one function. It is known that only one of the following statment is true, and only one such function exists :

find the function f (as ordered pair).(i) f(x) $$\neq$$ b
(i) f(y) = b

(ii) f(z) $$\neq$$ a

A
{(x,b), (y,a), (z,c)}

B
{(x,a), (y,b), (z,c)}

C
{(x,b), (y,c), (z,a)}

D
{(x,c), (y,a), (z,b)}

Solution

When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a,b,c $$

$$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = b$$ or $$f(x) = c, f(y) = a, f(z) = b$$

But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then $$f(y) = b, f(z) =c , f(x) = a\Rightarrow f(x) = a, f(y) = b, f(z) = c$$

Clearly if (ii) is true then it is satisfying every condition.

Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
Question 9
1 / -0

Suppose f and g both are linear function with $$\displaystyle f(x)=-2x+1$$ and $$\displaystyle f \left ( g\left ( x \right ) \right )=6x-7$$ then slope of line $$y=g(x)$$ is

A
$$3$$

B
$$-3$$

C
$$6$$

D
$$-2$$

Solution

Given, $$f(x)=-2x+1$$

$$\therefore f(g(x))=-2g(x)+1=6x-7$$

$$\Rightarrow g(x)=-3x+4$$

$$\Rightarrow g'(x)=-3$$

Hence slope of $$g(x)$$ is $$-3$$
Question 10
1 / -0

If $$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$

and $$x\, \in\, (1, 2)$$, then $$g(f(x))$$ is equal to

A
$$x^{2}\, +\, 3$$

B
$$x^{2}\, -\, 3$$

C
$$5\, -\, x^{2}$$

D
$$1 - x$$

Solution

$$x\, \epsilon\, (1, 2)$$

Hence for x >1, $$f(x) = 5-x^2$$and $$f(x) \epsilon (1,4) $$

$$f(x) >1$$

$$g(f(x)) = 2 - f(x)= 2-5+x^2 = x^2-3$$

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Relations and Functions Test - 48

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Relations and Functions Test - 48

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SHARING IS CARING

Question 1

$$-3$$

$$1$$

$$0$$

none of these

Solution

Question 2

$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$

$${R}^{-1}=R$$

Domain of $$R=\left\{ 1,2,3 \right\} $$

Range of $$R=\left\{ 5 \right\} $$

Solution

Question 3

$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }^{ x\left( x-1 \right) }$$

$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }\left( 1+\sqrt { 1+4\log _{ 2 }{ x } } \right) $$

$$\displaystyle { \left( \frac { 1 }{ 2 } \right) }\left( 1-\sqrt { 1+4\log _{ 2 }{ x } } \right) $$

Not defined

Solution

Question 4

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

$$2+{ x }^{ 2 }$$

$$1+x$$

$$2+x$$

Solution

Question 5

$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$

$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Solution

Question 6

$$\displaystyle x-\left[ \frac { x }{ 2 } \right] $$

$$-x-2$$

$$x-2$$

$$\displaystyle \frac { 1 }{ x+\left[ \frac { x }{ 2 } \right] } $$

Solution

Question 7

$$f\, :\, [0,\, \pi]\, \rightarrow\, [-\, 1\, 1], f(x)\, =\, \sin\, x$$

$$f\, :\, [0, \pi]\, \rightarrow\, [-1, 1],\, f(x)\, =\, \cos\, x$$

$$f\, :\, R\, \rightarrow\, R,\, f(x)\, =\, e^{x}$$

$$f\, :\, Q\, \rightarrow\, R, f(x)\, =\, x^{3}$$

Solution

Question 8

{(x,b), (y,a), (z,c)}

{(x,a), (y,b), (z,c)}

{(x,b), (y,c), (z,a)}

{(x,c), (y,a), (z,b)}

Solution

Question 9

$$3$$

$$-3$$

$$6$$

$$-2$$

Solution

Question 10

$$x^{2}\, +\, 3$$

$$x^{2}\, -\, 3$$

$$5\, -\, x^{2}$$

$$1 - x$$

Solution

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