Relations Test 29

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Relations Test 29

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Weekly Quiz Competition

Question 1
1 / -0

Let $$A = \left\{ {x,\,y,\,z} \right\}$$ and $$B = \left\{ {1,\,2} \right\}$$. The number relations from A to B is

A
64

B
32

C
16

D
28

Solution

$$\Rightarrow$$ $$A=\{X,\,Y,\,Z\}$$ and $$B=\{1,\,2\}$$ [ Given ]
$$\Rightarrow$$ Number of relation from $$A$$ to $$B$$ $$=2^{Number\,of\,elements\,in\,A\times B}$$
$$=2^{n(A)\times n(B)}$$
$$\Rightarrow$$ Number of elements in set $$A=3$$
$$\Rightarrow$$ Number of elements in set $$B=2$$
$$\Rightarrow$$ Number of relation from $$A$$ to $$B$$ $$=2^{n(A)\times n(B)}$$
$$=2^{3\times 2}$$
$$=2^6$$
$$=2\times 2\times 2\times 2\times 2\times 2$$
$$=64$$
Question 2
1 / -0

If $$A = \left \{1, 2, 3, 4\right \}$$ and $$I_{A}$$ be the identify relation on $$A$$, then

A
$$(1, 2) \epsilon I_{A}$$

B
$$(2, 2) \epsilon I_{A}$$

C
$$(2, 1) \epsilon I_{A}$$

D
$$(3, 4) \epsilon I_{A}$$

Solution
Question 3
1 / -0

If x is real, then $$\dfrac{x^2+2x+c}{x^2+4x+3c}$$ can take all real values if?

A
$$0 < c < 2$$

B
$$0 < c < 1$$

C
$$-1 < c < 1$$

D
None of the above

Solution

Let us assume that $$\dfrac{x^2+2x+c}{x^2+4x+3c}=y$$
$$\Rightarrow yx^2+4xy+3yc=x^2+2x+c$$
$$\Rightarrow x^2(y-1)+2x(2y-1)+3yc-c=0$$
For x to be real, Discrimination of this equation should be $$\geq 0$$
$$\therefore 4(2y-1)^2-4\times (y-1)\times (3yc-c)\geq 0$$
$$4(4y^2-4y+1)-4(3cy^2-4cy+c)\geq 0$$
Dividing both sides by $$4$$, we get
$$(4-3c)y^2+4y(c-1)+1-c\geq 0$$
For this equation to hold true, coefficient of $$y^2$$ should be greater than $$0$$ and D $$< 0$$
$$\Rightarrow 4-3c > 0$$ i.e. $$c < \dfrac{4}{3}$$ ..$$(1)$$
Also,
$$16(c^2-2c+1)-4\times (1-c)\times (4-3c) < 0$$
$$\Rightarrow 16c^2-32c+16-4(4-7a+3a^2) < 0$$
$$\Rightarrow 16c^2-32c+16-12c^2+28c-16 < 0$$
$$\Rightarrow 4c(c-1) < 0$$
$$\therefore 0 < c < 1$$.. $$(2)$$
From $$(1)$$ and $$(2)$$, we can say that $$0 < c < 1$$ is the required condition.
Question 4
1 / -0

Let $$R = \left \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 2)\right \}$$ be a relation on the set $$A = \left \{1, 2, 3, 4\right \}$$. The relation $$R$$ is

A
Reflexive

B
Transitive

C
Not symmetric

D
none of these

Solution
Question 5
1 / -0

Let $$A$$ and $$B$$ be two sets containing four and two elements respectively.Then the number of subset of the set $$A \times B$$, each having at least three elements is

A
$$256$$

B
$$275$$

C
$$510$$

D
$$219$$

Solution

$$A$$ and $$B$$ containing $$4$$ and $$2$$ respectively
set A has 4 elements
set B has 2 elements
total number of element in $$(A\times B)=4\times 2=8$$
total number of subsets of $$(A\times B)={2}^{8}=256$$
Number of subsets having 0 element $$={ _{ }^{ 8 }{ C } }_{ 0 }=1$$
Number of subsets having 1 element $$={ _{ }^{ 8 }{ C } }_{ 1 }=8$$
number of subsets having 2 element $$={ _{ }^{ 8 }{ C } }_{ 2 }=\cfrac{8\times 7}{2}=28$$
Number of subsets having atleast 3 elements
$$=256-28-8-1=219$$
Question 6
1 / -0

If $$a\ne R$$ and the equation $$-3(x-[x])^2+2(x-[x])+a^2=0$$ ( where $$[x]$$ denotes the greatest integer $$\le x$$) has no integral solution, then all possible values of a lie in the interval :

A
$$(-2,-1)$$

B
$$(-\infty ,-2)\cup(2,\infty)$$

C
$$(-1 ,0)\cup(0,1)$$

D
$$(1,2)$$

Solution

REF.Image
As $$x-[x]=\left \{ x \right \}$$, let $$\left \{ x \right \}=y$$

As x is non-integral, $$0<y<1$$ $$(y\neq 0)$$
equation becomes, $$-3y^{2}+2y+a^{2}=0$$

$$\Rightarrow a^{2}=3y^{2}-2y=y(3y-2)$$

Drawing graph of $$3y^{2}-2y=0$$ (parabola)

Minima $$\Rightarrow \frac{d}{dy}(3y^{2}-2y)=0\Rightarrow 6y-2=0\Rightarrow y=1/3$$

Taking positive solutions $$(a^{2}\geq 0)$$

we get $$0 < a^{2} < 1$$

$$\Rightarrow a\epsilon (-1,0)\cup (0,1)$$

Option (C)
Question 7
1 / -0

$${x\epsilon R:\frac{14x}{x+1}-\frac{9x-30}{x-4}<0}$$ is equal to

A
(-1,4)

B
(1,4) $$\cup $$(5,7)

C
(1,7)

D
(-1,1) $$\cup $$ (4,6)

Solution

$$ x\epsilon R :\frac{14x}{x+1}-\frac{9x-30}{x-4}< 0 $$
$$ x+1\neq 0 $$ $$ x-4 \neq 0 $$
$$ \Rightarrow x \neq -1 $$ $$ x \neq 4 $$
$$ 14(x-4)-(x+1)(7x-30)< 0 $$
$$ 14x^{2}-56x-(9x^{2}-30x+9x-30)< 0 $$
$$ 14x^{2}-56x-9x^{2}+30x-9x+30< 0 $$

$$ 5x^{2}-35x+30< 0 $$
$$ x^{2}-7x+6< 0 $$
$$ x^{2}-6x+6< 0 $$
$$ x(x-6)-1(x-6)< 0 $$
$$ (x-1)(x-6)< 0 $$
$$ \Rightarrow x> 1 $$ & $$ x< 6 $$
$$ \Rightarrow x \epsilon ; (1,4)\cup (4,6) $$
Question 8
1 / -0

If $$\left| { z }_{ 1 }-a \right| <a,\left| { z }_{ 2 }-a \right| <b,\left| { z }_{ 3 }-a \right| <c$$, $$(a,b,c\in R)$$ then $$\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| $$ is

A
less than $$(a+b+c)$$

B
more than $$(a+b+c)$$

C
less than $$2(a+b+c)$$

D
more than 2$$(a+b+c)$$

Solution

Given$$,$$
$$\left| {{z_1} - a} \right| < a$$
$$\left| {{z_2} - b} \right| < b$$
$$\left| {{z_3} - c} \right| < c$$
Then$$,$$
$$\left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| < a + b + c$$
$$\left| {{z_1} + {z_2} + {z_3}} \right| < \left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right|$$
and$$,$$
$$\left| {\left( {{z_1} - a} \right) + \left( {{z_2} - b} \right) + \left( {{z_3} - c} \right) + a + b + c} \right|$$ $$ < \left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| + \left| {a + b + c} \right|$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ $$ < a + b + c + a + b + c$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ $$ <2 \left( {a + b + c} \right)$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ less then $$2\left( {a + b + c} \right)$$
Hence$$,$$ Option $$(C)$$ is correct$$.$$
Question 9
1 / -0

The area of the region $$R = \{(x, y) : |x| \le |y| \, \text{and} \, x^2 + y^2 \le 1 \}$$ is

A
$$\dfrac{3 \pi}{8}$$ sq. units

B
$$\dfrac{5 \pi}{8} $$ sq. units

C
$$\dfrac{\pi}{2}$$ sq. units

D
$$\dfrac{\pi}{8} $$ sq. units

Solution
Question 10
1 / -0

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets $$A \times B$$ having 3 or more elements is

A
$$219$$

B
$$211$$

C
$$256$$

D
$$220$$

Solution

Let $$A=\left\{x,y\right\}$$
$$B=\left\{a,bc,d\right\}$$
$$A\times B$$ has $$2\times 4=8$$ elements
Total substance of $$A\times B={2}^{8}=256$$
$$\therefore\,$$Total number of subsets of $$A\times B$$ having $$3$$ or more elements
$$=256-\left(1\,null \,set+8\,single\,ton\,set-^{8}C_{2}\,having\,2\,elements\right)$$
$$=256-1-8-\dfrac{8!}{6!2!}$$
$$=256-1-8-\dfrac{8\times 7\times 6!}{6!2!}$$
$$=256-1-8-28=219$$