Relations and Functions Test

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Relations and Functions Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x + y) = f(xy)$$ and $$f(1) = 5$$, then the value of $$\sum_{k\, =\, 0}^{6}\, f(k)$$.

A
$$25$$

B
$$35$$

C
$$36$$

D
$$24$$

Solution

$$f(1+0)=f(1.0)$$

$$f(0)=5$$

$$f(2)=f(1+1)=f(1.1)=f(1)=5$$

$$f(3)=f(2+1)=f(2.1)=f(2)=5$$

Similarly $$f(4),f(5) $$ and $$f(6) $$ all are 5.

$$\sum^{6}_{k=0}f(k)=7\times5=35$$
Question 2
1 / -0

R is a relation on set A then $$\displaystyle \left [ \left ( R^{-1} \right )^{-1} \right ]^{-1}$$ is_______

A
R

B
$$\displaystyle R^{-1}$$

C
$$\displaystyle A\times A$$

D
None of these

Solution

We know that Inverse of inverse of a Relation is the original Relation itself.

So, $$ \left[ \left( { R }^{ -1 } \right) ^{ -1 } \right] ^{ -1 } = \left[ R \right] ^{ -1 } $$
Question 3
1 / -0

A relation which satisfies reflexive symmetric and transitive is ________ relation

A
an identity

B
a constant

C
an equivalence

D
None of these

Solution

A relation which satisfies reflexive symmetric and transitive is _called an equivalence relation
Question 4
1 / -0

If $$f(x) + f( 1- x) = 10$$, then the value of $$f\left ( \frac{1}{10} \right )\, +\, f\left ( \frac{2}{10} \right )\, +\, ..........\, +\, f\left ( \frac{9}{10} \right )$$ is

A
$$45$$

B
$$50$$

C
$$90$$

D
Cannot be determined

Solution

Clearly required expression contains 10 terms which can be grouped into 5 pairs each having complementary arguments which add up to 1.

Answer $$=f\left(\dfrac1{10}\right)$$ +$$f\left(\dfrac9{10}\right)$$+$$f\left(\dfrac2{10}\right)$$ +$$f\left(\dfrac8{10}\right)$$+..............................

$$=f\left(\dfrac1{10}\right)$$+$$f\left(1-\dfrac1{10}\right)$$ +$$f\left(\dfrac2{10}\right)$$ +$$f\left(1-\dfrac2{10}\right)$$ +...............................

The first condition of the statement hence gives us the value to be $$5\times 10 =50$$
Question 5
1 / -0

If $$\displaystyle f:R\rightarrow R$$ defined by $$\displaystyle f\left ( x \right )=2x^{2}-3x+5$$, then find the value of $$\displaystyle \frac{f\left ( x+h \right )-f\left ( x \right )}{h}$$ at $$\displaystyle h\neq 0$$

A
4x+2h

B
4x+2h-3

C
4x-2h+3

D
4x-2h-3

Solution

Given, $$ f(x) = 2x^2 -3x + 5 $$
So, $$ \dfrac { f(x+h)-f(x) }{ h } =\dfrac { 2{ (x+h) }^{ 2 }-3(x+h)+5-(2{ x }^{ 2 }-3x+5) }{ h } $$
$$ =\dfrac { 2{ x }^{ 2 }+2{ h }^{ 2 }+4xh-3x-3h+5-2{ x }^{ 2 }+3x-5 }{ h } =\dfrac { 2{ h }^{ 2 }+4xh-3h }{ h } =\quad 2h+4x-3 $$
Question 6
1 / -0

If $$\displaystyle R_{n}=\left \{ x:\frac{-1}{n}< x< \frac{1}{n} \right \}$$ then $$\displaystyle R_{5}\cup R_{15}=$$ ________

A
$$\displaystyle R_{5}$$

B
$$\displaystyle R_{15}$$

C
$$\displaystyle R_{3}$$

D
$$\displaystyle R_{20}$$

Solution

$$ R_5 = $$ { $$ x:-\frac {1}{5} < x < \frac {1}{5} $$ }
And
$$ R_{15} = $$ {$$ x:-\frac {1}{15} < x < \frac {1}{15} $$ }

But, $$ \frac {1}{15} < x < \frac {1}{15} $$ is within the limits of $$ \frac {1}{5} < x < \frac {1}{5} $$

Hence, $$ R_5 \cup R_{15} = \frac {1}{5} < x < \frac {1}{5} = R_5 $$
Question 7
1 / -0

The domain of the function f(x) = $$\displaystyle \frac{\left | x \right |-2}{\left | x \right |-3}$$ is ________

A
R

B
R - {2, 3}

C
R - {2, -2}

D
R - {-3, 3}

Solution

The function has a real value only when denominator of the fraction is not equal to zero.

So, $$\left| x \right| - 3 \neq = 0 $$
$$ => \left| x \right| \neq = 3 $$
$$ => x \neq = 3 ; -3 $$

So, the domain of the function is $$ R - $$ { $$ -3,3 $$ }
Question 8
1 / -0

An equation that defines $$y$$ as a function of $$x$$ is given. Solve for $$y$$ in terms of $$x$$, and replace $$y$$ with the function notation $$f\left( x \right)$$.
$$x - 2y = 18$$

A
$$f\left( x \right) =\displaystyle\frac { 1 }{ 2 } x-18$$

B
$$f\left( x \right) =\displaystyle\frac { 1 }{ 2 } x-9$$

C
$$f\left( x \right) =-x+9$$

D
$$f\left( x \right) =-\displaystyle\frac { 1 }{ 2 } x+9$$

Solution

$$ x-2y = 18$$
$$-2y=18-x$$

$$y=\dfrac{18-x}{-2}$$

$$y=-(9-\dfrac{x}{2})$$
$$y=(\dfrac{x}{2}-9)$$
Question 9
1 / -0

$$A = \left \{1, 2, 3, 4\right \}$$ and $$B = \left \{a, b, c\right \}$$. The relations from $$A$$ to $$B$$ is

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

B
$$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$$

C
$$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

D
$$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$$

Solution

$$A=\{1,2,3,4\}$$
$$B=\{a,b,c\}$$

Possible Relation $$R:A\to B$$
$$=\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

Option C contains Relation $$R:A\to B$$
Question 10
1 / -0

Find $$\left( f+g \right) \left( 1 \right) $$ when $$f\left( x \right) =x+6$$ and $$g\left( x \right) =x-3$$.

A
$$5$$

B
$$-1$$

C
$$11$$

D
$$-7$$

Solution

Given, $$f(x)=x+6$$ and $$g(x)=x-3$$
We know $$(f+g)(1)$$ $$=f(1)+g(1)$$ .....(i)
So, $$f(1)=1+6=7$$
and $$g(1)=1-3=-2$$
On substituting this value in equation (i), we get
$$(f+g)(1)=7-2=5$$

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Relations and Functions Test - 29

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Relations and Functions Test - 29

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SHARING IS CARING

Question 1

$$25$$

$$35$$

$$36$$

$$24$$

Solution

Question 2

R

$$\displaystyle R^{-1}$$

$$\displaystyle A\times A$$

None of these

Solution

Question 3

an identity

a constant

an equivalence

None of these

Solution

Question 4

$$45$$

$$50$$

$$90$$

Cannot be determined

Solution

Question 5

4x+2h

4x+2h-3

4x-2h+3

4x-2h-3

Solution

Question 6

$$\displaystyle R_{5}$$

$$\displaystyle R_{15}$$

$$\displaystyle R_{3}$$

$$\displaystyle R_{20}$$

Solution

Question 7

R

R - {2, 3}

R - {2, -2}

R - {-3, 3}

Solution

Question 8

$$f\left( x \right) =\displaystyle\frac { 1 }{ 2 } x-18$$

$$f\left( x \right) =\displaystyle\frac { 1 }{ 2 } x-9$$

$$f\left( x \right) =-x+9$$

$$f\left( x \right) =-\displaystyle\frac { 1 }{ 2 } x+9$$

Solution

Question 9

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

$$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$$

$$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

$$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$$

Solution

Question 10

$$5$$

$$-1$$

$$11$$

$$-7$$

Solution

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