Relations Test 31

Result

Relations Test 31

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Question 1
1 / -0

The domain of $$\dfrac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}}$$ is

A
$$(\dfrac{1}{2},11)$$

B
$$(\dfrac{1}{2},31)$$

C
$$(\dfrac{1}{2},17)$$

D
$$(\dfrac{1}{2},41)$$

Solution

$$+(x) = \dfrac {1}{\sqrt {x (1 - x)}} + \sqrt {3x - 1 - 2x^{2}}$$

Domain of $$f(x)$$ : -
$$x (1 - x) > 0$$ and $$3x - 1 - 2x^{2} \geq 0$$

$$2x^{2} - 3x + 1 \leq 0$$

$$2x (x - 1) - 1 (x - 1) \leq 0$$

$$(2x - 1) (x - 1)\leq 0$$

Thus taking the intersection of above $$2$$.

$$x \leftarrow [\dfrac {1}{2}, 1)$$.
Question 2
1 / -0

The set of values of '$$b$$' for which the origin and the point $$(1,1)$$ lie on the same side of the straight line $${ a }^{ 2 } x + ab y + 1 = 0 \;\;\forall \;\; a\in R,\; b > 0$$ are:

A
$$b\in (2,4)$$

B
$$b\in (0,2)\quad $$

C
$$b\in (0,4)\quad $$

D
$$b\in (2,6)$$

Solution
Question 3
1 / -0

Total number of equivalence relation defined in the set $$s=\{a, b, c\}$$ is

A
$$5$$

B
$$31$$

C
$$2^3$$

D
$$3^3$$

Solution

By pascal's triangle
$$1\\ 1\quad \quad 1\\ 1\quad \quad 3\quad \quad 1\\ 1\quad \quad 7\quad \quad 6\quad \quad 1\\ 1\quad \quad 15\quad 25\quad 10\quad \quad 1\\ 1\quad \quad 31\quad 90\quad 65\quad \quad 15\quad \quad 1$$
Here no. of elements is $$3$$.
$$\therefore$$ No. of equivalent relations $$=1+3+1=5$$.
Question 4
1 / -0

The domain of $$f(x)=\frac{1}{\sqrt{(x-1)(x-2)(x-3)}}$$ is

A
$$(-\infty ,1)\cup (3,\infty )$$

B
$$(1,2)\cup (3,\infty )$$

C
$$(-\infty ,2)$$

D
R

Solution

$${\textbf{Step -1: Put x=0 and change the sign as negative multiply by negative is positive}}$$ $${\textbf{and positive multiply by negative is negative }} $$
$$ x = 0 \Rightarrow - \times - \times - = - $$

$${\textbf{Step -2:Put x=1.5 and change the sign as positive multiply by negative is negative}}$$ $${\textbf{and negative multiply by negative is positive }}$$
$$x = 1.5 \Rightarrow + \times - \times - = + $$

$${\textbf{Step -3:Put x=2.5 and change the sign as positive multiply by positive is positive}}$$ $${\textbf{and positive multiply by negative is negative}}$$
$$x = 2.5 \Rightarrow + \times + \times - = - $$

$${\textbf{Step -4: Put x=4 and change the sign as positive multiply by positive is always positive }}$$
$$x = 4 \Rightarrow + \times + \times + = + $$

$${\text{x is positive between 1 and 2 and between 3 and infinity so }} x \in (1,2) \cup (3,\infty )$$
$$\textbf{Hence option B is correct}$$
Question 5
1 / -0

Let A be a finite set and n(A) = 5, then the number of relations which are not symmetric is _____________________.

A
$$2^{25}$$

B
$$2^{15}$$

C
$$2^{25} - 2^{10}$$

D
$$2^{25} - 2^{15}$$

Solution
Question 6
1 / -0

Let $$A=\left\{ 2,4,6,8 \right\} $$. A relation $$R$$ on $$A$$ is defined by $$R={(2,4),(4,2),(4,6),(6,4)}$$. Then $$R$$ is:

A
$$anti-symmetric$$

B
$$reflexive$$

C
$$symmetric$$

D
$$transitive$$

Solution
Question 7
1 / -0

If $$x\ \epsilon \ R$$, the number of solutions of $$\sqrt{2x+1}-\sqrt{2x-1}-1$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$infinite$$

Solution

We have,

$$x\in R$$

Then,

$$\sqrt{2x+1}-\sqrt{2x-1}-1$$

Now,

Put $$x=1,2,3,4.........$$

So,

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

$$ \sqrt{2\left( 2 \right)+1}-\sqrt{2\left( 2 \right)-1}-1=\sqrt{5}-\sqrt{3}-1 $$

And so on….

Then,

Infinite number of solution.

Hence, this is the answer.
Question 8
1 / -0

If $$\sqrt { \log _ { 4 } \left( \log _ { 2 } \left( \log _ { 2 } \left( x ^ { 2 } - 2 x + a \right) \right) \right. } )$$ is defined $$\forall x \in R$$, then

A
$$[ 10 , \infty )$$

B
$$[ 9 , \infty )$$

C
$$[ 2 , \infty )$$

D
$$[ 5 , \infty )$$

Solution
Question 9
1 / -0

Let $$T$$ be the set of all triangle in the Euclidean plane, and let a relation $$R$$ on $$T$$ be defined as $$aRb$$ if a is congruent to $$b\ \forall \ a, b\ \epsilon \ T$$. Then $$R$$ is

A
reflective but not transitive

B
transitive but not symmetric

C
equivalence

D
none of these

Solution
Question 10
1 / -0

Let S be the set of all real nos, then the relation $$R = \{ ( a , b ) : | a - b |$$ is a multiple of 4$$\}$$ then the relation.

A
Reflexive but not symmetric

B
Symmetric but not reflexive

C
Reflexive only

D
Equivalence

Solution