Relations and Functions Test

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Relations and Functions Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\displaystyle f\left ( x \right ) = x^{3} + 2x^{2} + 3x + 4$$, then the equation $$\displaystyle \frac{1}{x - f\left ( 1 \right )} + \frac{2}{x - f\left ( 2 \right )} + \frac{3}{x - f\left ( 3 \right )} = 0$$, has

A
1 real root

B
2 real roots

C
all three roots are real

D
no real root exist

Solution

$$f\left( x \right) =x^{ 3 }+2x^{ 2 }+3x+4$$
Then
$$f\left( 1 \right) =1^{ 3 }+2.1^{ 2 }+3.1+4=1+2+3+4=10\\ f\left( 2 \right) =2^{ 3 }+2.2^{ 2 }+3.2+4=8+8+6+4=26\\ f\left( 3 \right) =3^{ 3 }+2.3^{ 2 }+3.3+4=27+18+9+4=56$$

$$\displaystyle \frac { 1 }{ x-f\left( 1 \right) } +\frac { 2 }{ x-f\left( 2 \right) } +\frac { 3 }{ x-f\left( 3 \right) } =0$$

$$\Rightarrow 1\left( x-f\left( 2 \right) \right) \left( x-f\left( 3 \right) \right) +2\left( x-f\left( 1 \right) \right) \left( x-f\left( 3 \right) \right) +3\left( x-f\left( 1 \right) \right) \left( x-f\left( 2 \right) \right) =0$$

$$ \Rightarrow \left( x-26 \right) \left( x-56 \right) +2\left( x-10 \right) \left( x-56 \right) +3\left( x-10 \right) \left( x-26 \right) =0$$

$$ \Rightarrow 6{ x }^{ 2 }-322x+3356=0$$

Comparing with General form of quadratic equation

Here, $$a=6, b=-322, c=3356$$

$$\therefore D=b^2 -4ac=322^2-4 \times 6 \times 3356 =23140> 0$$

Therefore, this quadratic equation have $$2$$ real roots
Question 2
1 / -0

If ,$$(x-1, y+2)= (7, 5)$$ then values of $$x$$ and $$y$$ are

A
$$5$$,$$8$$

B
$$8$$,$$3$$

C
$$-1$$,$$5$$

D
$$7$$,$$1$$

Solution

as the given ordered pairs are equal,
$$x-1=7$$ and $$y+2=5$$
$$\therefore x=8$$ and $$y=3$$
Question 3
1 / -0

Ordered pairs (a, 3) and (5, x) are equal ,the values of $$a$$ and $$x$$ are

A
$$2$$ and $$4$$

B
$$3$$ and $$6$$

C
$$5$$ and $$3$$

D
$$1$$ and $$-1$$

Solution

If two ordered pairs are equal, then both x - coordinates are equal and both y-coordinates are equal.

Since $$ (a, 3) = (5,x) $$ then ,$$ a = 5 , x = 6 $$
Question 4
1 / -0

If $$(x, y) = (3, 5)$$ ; then values of $$x$$ and $$y $$ are

A
3 and 5

B
4 and 7

C
-1 and 17

D
2 and 4

Solution
Question 5
1 / -0

Given $$M = (0, 1, 2)$$ and $$N = (1, 2, 3)$$. Find $$(N - M) \times (N \cap M)$$

A
{(3, 1), (3, 2)}

B
{(3, -1), (3, -2)}

C
{(-3, -1), (-3, 2)}

D
{(3, -1), (-3, -2)}

Solution

$$ N - M $$ means subtracting the common elements of $$M$$ and $$N$$ from $$N$$.
So, $$ N - M =\{3\}$$

Intersection of two sets has the elements which are common in both the sets.
So, $$ N \cap M =\{1,2\}$$
Cartesian product of two sets is found by forming ordered pairs, where $$x$$-coordinate is from first set and $$y$$-coordinate is from second set.
So, $$ (N - M) \times (N \cap M) =\{ (3,1), (3,2) \}$$.
Question 6
1 / -0

If $$A = \{5, 7\}, B= \{7, 9\}$$ and $$C = \{7, 9, 11\},$$ find $$(A \times B) \cup (A \times C)$$

A
$$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

B
$$\{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

C
$$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

D
none of these

Solution

Product of two sets is found by forming ordered pairs by multiplying every element of the first set with the second set.
So, $$ A \times B = $$ { $$ (5,7), (5,9), (7,7),(7,9) $$ }
And $$ A \times C = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Union of two sets has all the elements of both the sets.
So, $$ (A \times B) \cup ( A \times C) = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Question 7
1 / -0

If $$n(A) = 4$$ and $$n(B) = 5$$, then $$n(A \times B) = $$

A
$$20$$

B
$$25$$

C
$$4$$

D
$$15$$

Solution

Given, $$n\left( A \right) =4$$ and $$n\left( B \right) =5$$
$$ \Rightarrow n\left( A\times B \right) =n\left( A \right) \cdot n\left( B \right) $$
$$\Rightarrow n(A\times B)=4\cdot 5=20$$
So, answer is $$20$$.
Question 8
1 / -0

If $$R$$ be a relation defined from $$\displaystyle A=\left \{ 1,2,3,4 \right \}$$ to $$\displaystyle B=\left \{ 1,3,5 \right \},i.e.\left ( a,b \right )\in R$$ iff $$a<b$$ then $$\displaystyle R o R^{-1}$$ is

A
$$\displaystyle \left \{ \left ( 1,3 \right ),\left ( 1,5 \right ),\left ( 2,3 \right ),\left ( 2,5 \right ),\left ( 3,5 \right ),\left ( 4,5 \right ) \right \}$$

B
$$\displaystyle \left \{ \left ( 3,1 \right ),\left ( 5,1 \right )\left ( 3,2 \right ),\left ( 5,2 \right ) ,\left ( 5,3 \right ),\left ( 5,4 \right )\right \}$$

C
$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,5 \right ) ,\left ( 5,3 \right ),\left ( 5,5 \right )\right \}$$

D
$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,4 \right ) ,\left ( 4,5 \right )\right \}$$

Solution

$$\displaystyle R;A\in B$$ under given condition $$a<b$$ is given by,
$$\displaystyle R=\left \{ \left ( 1,3 \right ),\left ( 1,5 \right

),\left ( 2,3 \right ),\left ( 2,5 \right ),\left ( 3,5 \right ),\left (

4,5 \right ) \right \}$$
$$\displaystyle R^{-1}=\left \{ \left ( 3,1

\right ),\left ( 5,1 \right ),\left ( 3,2 \right ) ,\left ( 5,2 \right

),\left ( 5,3 \right )\left ( 5,4 \right )\right \}$$
$$\displaystyle

R o R^{-1}:$$ For composing $$\displaystyle R o R^{-1}$$ we will pick up an element of $$\displaystyle R^{-1}$$ first and then of $$R$$
$$\displaystyle

\left ( 3,1 \right )\in R^{-1}\left ( 1,3 \right )\in

R\rightarrow \left ( 3,3 \right )\in R o R^{-1}$$
$$\displaystyle

\therefore R o R^{-1}=\left \{ \left ( 3,3 \right ),\left ( 3,5 \right

),\left ( 5,3 \right ),\left ( 5,5 \right ) \right \}$$ only.
Question 9
1 / -0

A relation $$R$$ is defined on the set $$Z$$ of integers as follows: R=$$(x,y)$$ $$\displaystyle \in {R}:x^{2}+y^{2}= 25$$. Express $$R$$ and $$\displaystyle R^{-1}$$ as the sets of ordered pairs and hence find their respective domains.

A
$$0$$

B
Domain of $$\displaystyle R= \left \{ 0, \pm 3 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

C
Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

D
Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4, \pm 5 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Solution

$$R=\left\{ \left( 0,5 \right) ,\left( 0,-5 \right) ,\left( 3,4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( -3,-4 \right) ,\\ \left( 4,3 \right) ,\left( -4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( 5,0 \right) ,\left( -5,0 \right) \right\} \\ { R }^{- 1 }=\left\{ \left( 5,0 \right) ,\left( -5,0 \right) ,\left( 4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( -4,-3 \right) ,\\ \left( 3,4 \right) ,\left( 3,-4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( 0,5 \right) ,\left( 0,-5 \right) \right\} $$
Therefore domain of $$R\equiv \left\{ 0,3,-3,-4,4,-5,-5 \right\} =$$ domain of $${ R }^{ -1 }$$
Question 10
1 / -0

The value of $$b$$ and $$c$$ for which the identify $$f (x + 1) - f (x) = 8x + 3$$ is satisfied, where $$f (x)\, =\, bx^{2}\, +\, cx\, +\, d$$, are -

A
$$b = 2, c = 1$$

B
$$b = 4, c = - 1$$

C
$$b = - 1, c = 4$$

D
$$b = - 1, c = 1$$

Solution

Given $$f (x + 1) - f (x) = 8x + 3$$ and $$f(x) = bx^2+cx+d$$

Substitute $$x=0$$ we get $$f (0 + 1) - f (0) = 3$$

$$\Rightarrow (b + c + d) - d = 3\Rightarrow b+c=3 ...(i)$$

Again substitute $$x =-1$$

we get

$$ f (-1 + 1) - f (- 1) = - 8 + 3\Rightarrow f(0) - f(-1)= -5\Rightarrow d - (b - c + d) = -5\Rightarrow -b + c = -5 ......(ii)$$

Solving (i) and (ii) we get $$b = 4, c = - 1$$

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Relations and Functions Test - 26

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Relations and Functions Test - 26

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Question 1

1 real root

2 real roots

all three roots are real

no real root exist

Solution

Question 2

$$5$$,$$8$$

$$8$$,$$3$$

$$-1$$,$$5$$

$$7$$,$$1$$

Solution

Question 3

$$2$$ and $$4$$

$$3$$ and $$6$$

$$5$$ and $$3$$

$$1$$ and $$-1$$

Solution

Question 4

3 and 5

4 and 7

-1 and 17

2 and 4

Solution

Question 5

{(3, 1), (3, 2)}

{(3, -1), (3, -2)}

{(-3, -1), (-3, 2)}

{(3, -1), (-3, -2)}

Solution

Question 6

$$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

$$\{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

$$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

none of these

Solution

Question 7

$$20$$

$$25$$

$$4$$

$$15$$

Solution

Question 8

$$\displaystyle \left \{ \left ( 1,3 \right ),\left ( 1,5 \right ),\left ( 2,3 \right ),\left ( 2,5 \right ),\left ( 3,5 \right ),\left ( 4,5 \right ) \right \}$$

$$\displaystyle \left \{ \left ( 3,1 \right ),\left ( 5,1 \right )\left ( 3,2 \right ),\left ( 5,2 \right ) ,\left ( 5,3 \right ),\left ( 5,4 \right )\right \}$$

$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,5 \right ) ,\left ( 5,3 \right ),\left ( 5,5 \right )\right \}$$

$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,4 \right ) ,\left ( 4,5 \right )\right \}$$

Solution

Question 9

$$0$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4, \pm 5 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Solution

Question 10

$$b = 2, c = 1$$

$$b = 4, c = - 1$$

$$b = - 1, c = 4$$

$$b = - 1, c = 1$$

Solution

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