Relations and Functions Test

Result

Relations and Functions Test - 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\dfrac{1}{2}(3^{x}+3^{-x}), g(x)=\dfrac{1}{2}(3^{x}-3^{-x})$$ then $$ f(x)g(y)+f(y)g(x)$$ is equal to

A
$$f(x+y)$$

B
$$g(x+y)$$

C
$$2f(x)$$

D
$$2g(x)$$

Solution

$$\Rightarrow$$$$f(x)g(y)=\dfrac{1}{4}(3^{x}+3^{-x})(3^{y}-3^{-y})$$

$$\Rightarrow$$$$\dfrac{1}{4}(3^{x+y}+3^{y-x}-3^{x-y}-3^{-x-y})$$

$$\Rightarrow$$$$f(y)g(x)=\dfrac{1}{4}(3^{x+y}-3^{y-x}+3^{x-y}-3^{-x-y})$$

$$\Rightarrow$$$$f(x)g(y)+f(y)g(x)=\dfrac{1}{4}\times 2(3^{x+y}-3^{-x-y})$$

$$\Rightarrow$$$$\dfrac{1}{2}(3^{x+y}-3^{-(x+y)})$$

$$\Rightarrow$$$$g(x+y)$$
Question 2
1 / -0

If relation R$$=\left \{ (x, x+2) : x \in N, 1 \leq x <4 \right \}$$ then R is

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

B
$$\{(2, 3), (4,2), (5, 3)\}$$

C
$$\{(1, 3), (2, 4), (3, 5), (4, 6)\}$$

D
none of these

Solution

We have $$x\in N, 1\le x<4$$
$$\Rightarrow x =1,2,3$$
Hence $$R =\{(1,3),(2,4),(3,5)\}$$
Question 3
1 / -0

Directions For Questions

Consider the function $$y=\frac{x^{2}+x+d}{x^{2}+2x+d}$$

...view full instructions

If the range of function is $$\left [ \frac{5}{6},\frac{3}{2} \right ]$$, then the value of d is

A
-4

B
3

C
4

D
5
Question 4
1 / -0

If $$A=\{b,c,d\}$$ and $$B=\{x,y\}$$. Find which of the following are elements of $$A \times A$$.

A
$$\{b,b\}$$

B
$$\{b,c\}$$

C
$$\{b,d\}$$

D
All of the above

Solution

$$A\times A \Rightarrow$$ the first element will be from $$A$$ and the second element will also be from $$A$$.
$$A \times A = \left\{\{b,b\}, \{b,c\}, \{b,d\},\{c,b\},\{c,c\},\{c,d\},\{d,b\},\{d,c\},\{d,d\}\right\}$$
Thus, all the options $$A,B$$ and $$C$$ are the elements of $$A \times A$$
Question 5
1 / -0

The number of different solutions to the equation $$f (20x -11/ x) = 0$$ cannot be

A
0

B
3

C
2

D
2022

Solution

Let $$f (\alpha)=0 $$
$$\Rightarrow 20x -11/ x=\alpha$$
$$\Rightarrow 20x^2 - \alpha x -11 = 0$$
$$\Rightarrow x=\frac{\alpha \pm \sqrt{\alpha ^2+880}}{40}$$
so, $$x = x_1$$ & $$x_2$$ ; no. of solution will be 2 ; no. of solutions wil be 2 for each $$\alpha$$ . So they should be even
Question 6
1 / -0

Suppose $$S=\{1,2\}$$ and $$T=\{a,b\}$$ then $$T \times S$$

A
$$(a,1),(a,2),(b,1),(b,2)$$

B
$$(1,a),(2,b),(b,1),(b,2)$$

C
$$(a,1),(a,2),(1,b),(2,b)$$

D
None of the above

Solution

Given : $$S=\{ 1,2\} ,T=\{ a,b\} $$
$$T\times S$$ is the set of ordered pair of elements of $$T$$ and $$S$$ respectively
Then, $$T\times S=\{a,b\}\times \{1,2\}$$
$$T\times S=\{ (a,1),(a,2),(b,1),(b,2)\} $$
Question 7
1 / -0

Given $$A=\{b,c,d\}$$ and $$B=\{x,y\}$$ : find element of $$A\times B$$ .

A
$$\{b,x\}$$

B
$$\{b,y\}$$

C
$$\{c,x\}$$

D
All of the above

Solution

Given $$A=\{b,c,d\}$$ and $$B=\{x,y\}$$, then
$$A\times B=\{(b,x),(c,x),(d,x),(b,y),(c,y),(d,y)\}$$
Hence, all the elements given in options are elements of $$A\times B$$.
Question 8
1 / -0

Let $$f(x+y)=f(x) f(y) \forall x, y \in R, f(5) =2, f'(0)=3$$, then $$f'(5)$$ equals:

A
$$4$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$6$$

Solution

When $$f(x+y)=f(x)f(y)$$, then $$f(x)=a^{cx}$$
Now, $$f(5)=a^{5c}$$
$$\Rightarrow c=\dfrac{\log_a2}{5}$$
Now,$$f'(x)=ca^{cx}\ln a$$
$$\Rightarrow f'(0)=c\ln a$$
$$\Rightarrow 3=c\ln a$$
and $$f'(5)=\dfrac{3}{\ln a}.a^{\log_a2}.\log a$$
$$f'(5)=3\times 2$$
$$f'(5)=6$$
Question 9
1 / -0

If $$f(1)=1, f(n+1)=2f(n)+1, n \geq 1$$, then $$f(n)$$ is:

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

B
$$2^n$$

C
$$2^n -1$$

D
$$2^{n-1}-1$$

Solution

Given that $$f(n+1)=2f(n)+1,n\geq 1$$.
Therefore, $$f(2)=2f(1)+1$$
Since $$f(1)=1$$, we have
$$f(2)=2f(1)+1=2(1)+1=3=2^2-1$$.
Similarly $$f(3)=2f(2)+1=2(3)+1=7=2^3-1$$
and so on....
In general, $$ f(n)=2^n-1$$
Question 10
1 / -0

n(A)=m and n(B)=n ; then

A
n(A)+n(B)=n(A+B)

B
n(A)-n(B)=n(A+B)

C
A$$\times$$B=mn

D
n(A) $$\times$$n(B =n(A $$\times$$B)

Solution

C is not even logical while in the first 2 cases:

If there are elements common in A and B L.H.S of the first statement will be greater than R.H.S.

Clearly option B is also untrue as the combination of A and B cannot have lesser number of elements than A alone.

The last option has to be true as cartesian product of 2 sets gives a matrix having n(A) and n(B) as columns and rows.whose product will give the number of elements in the matrix.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 22

Result

Relations and Functions Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

$$f(x+y)$$

$$g(x+y)$$

$$2f(x)$$

$$2g(x)$$

Solution

Question 2

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

$$\{(2, 3), (4,2), (5, 3)\}$$

$$\{(1, 3), (2, 4), (3, 5), (4, 6)\}$$

none of these

Solution

Question 3

-4

3

4

5

Question 4

$$\{b,b\}$$

$$\{b,c\}$$

$$\{b,d\}$$

All of the above

Solution

Question 5

0

3

2

2022

Solution

Question 6

$$(a,1),(a,2),(b,1),(b,2)$$B×A={(a,1),(a,2),(b,1),(b,2)}

$$(1,a),(2,b),(b,1),(b,2)$$

$$(a,1),(a,2),(1,b),(2,b)$$

None of the above

Solution

Question 7

$$\{b,x\}$$

$$\{b,y\}$$

$$\{c,x\}$$

All of the above

Solution

Question 8

$$4$$

$$1$$

$$\dfrac{1}{2}$$

$$6$$

Solution

Question 9

$$2^{n+1}$$

$$2^n$$

$$2^n -1$$

$$2^{n-1}-1$$

Solution

Question 10

n(A)+n(B)=n(A+B)

n(A)-n(B)=n(A+B)

A$$\times$$B=mn

n(A) $$\times$$n(B =n(A $$\times$$B)

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$$(a,1),(a,2),(b,1),(b,2)$$