Relations and Functions Test

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Relations and Functions Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=2x-1, g(x)=x^{2} $$ then $$ (3f-2g)(x)=$$

A
$$5x-x^{2}+9$$

B
$$6x-5x^{2}-4$$

C
$$2x-x^{2}-3$$

D
$$6x-2x^{2}-3$$

Solution

$$3f\left ( x \right )-2g\left ( x \right )$$
$$=3\left ( 2x-1 \right )-2\left ( x^{2} \right )$$
$$=6x-2x^{2}-3$$
Question 2
1 / -0

If $$f(x)$$ is a polynomial in $$x(>0)$$ satisfying the equation $$ f(x)+f\left (\dfrac1x\right)=f(x).f\left (\dfrac1x\right)$$ and $$f(2)=-7$$, then $$f(3)$$ is equal to

A
$$-26$$

B
$$-27$$

C
$$-28$$

D
$$-29$$

Solution

Whenever, $$f(x)+f\left (\dfrac{1}{x}\right)=f(x).f\left (\dfrac{1}{x}\right)$$, then $$f(x)=1+x^n$$ or $$f(x)=1-x^n$$ keeping in mind that $$n$$ must be greater than $$0$$.
So here, $$f(2)=-7$$.
So we choose $$f(x)=1-x^n$$.
So, $$f(2)=1-2^n$$
$$\Rightarrow 2^n=8$$
$$\Rightarrow n=3$$
So, the function is $$f(x)=1-x^3$$.
Now,$$f(3)=1-3^3$$
$$\Rightarrow f(3)=-26$$
Question 3
1 / -0

If $$f(x+y)=f(x)f(y)$$ and $$f(5)=32$$ then $$f(7)=$$

A
$$35$$

B
$$36$$

C
$$ \displaystyle \frac{7}{5}$$

D
$$128$$

Solution

$$f(x+y)=f(x).f(y)$$
Put $$x=1, y=1$$
$$f(x+y)=f(x).f(y)$$
$$\Rightarrow f(2)=f(1)f(1)$$
$$\Rightarrow f(3)=f(2)f(1)=f(1)^3$$
$$\Rightarrow f(4)=f(3)f(1)=f(1)^4$$
$$\Rightarrow f(5)=f(4)f(1)=f(1)^5$$
Given$$ f(5)=32$$
$$\therefore f(1)=2$$
And $$f(2)=4$$
Now,
$$f(7)=f(5)f(2)=32\times4=128$$
Question 4
1 / -0

If a non-zero function $$f$$ satisfies the relation $$f(x+y)+f(x-y)=2f(x).f(y)$$ for all $$x, y$$ in $$R$$ and $$f(0) \neq 0$$; then $$f(10)-f(-10)=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Given, $$f(x+y)+f(x-y)=2f(x)\cdot f(y)$$

Put $$y=0, x=x$$

$$f(x)+f(x)=2f(x)f(0)$$

$$2f(x)=2f(x)f(0)$$

$$f(x)=0$$ that is not possible

$$f(0)=1$$

Take $$x = 0$$ and $$y = 10$$

$$f(10) + f(-10) = 2f(10)f(0)$$

$$f(10) + f(-10) = 2f(10)$$

$$f(-10) = f(10)$$

$$\therefore f(10)-f(-10)=0$$
Question 5
1 / -0

If relation $$R$$ is defined by $$\mathrm{R}=\{(\mathrm{x},\ \mathrm{y}):2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6\}$$, then the domain of $$\mathrm{R}$$ is

A
$$[-3,3]$$

B
$$[-\sqrt{3},\sqrt{3}]$$

C
$$[-\sqrt{2},\sqrt{2}]$$

D
$$[-2,2]$$

Solution

Domain $$=$$ value of $$\mathrm{x}$$ in $$f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ then
$$2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ = $$\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1$$
Since it is an equation of ellipse then value of $$x$$ $$ \Rightarrow \mathrm{x}\in[-\sqrt{3},\sqrt{3}]$$
Question 6
1 / -0

The domain and range of relation $$R=\{(x,y) | x, y \in N$$, $$x+2y=5\} $$ is?

A
$$\{1,3\}, \{2,1\}$$

B
$$\{2,1\}, \{3,2\}$$

C
$$\{1,3\}, \{1,1\}$$

D
$$\{1,2\}, \{1,3\}$$

Solution

The possible values of $$x$$ which satisfies the given relation is the domain of the relation
The possible values of $$y$$ which satisfies the given relation is the range of the relation
Given that $$x,y$$ are natural numbers and the relation is $$x+2y=5$$
For $$x=1$$ , the value of $$y$$ is $$2$$
For $$x=3$$ , the value of $$y$$ is $$1$$
If $$x\geq 5$$, then $$y$$ is negative which does not belong to naturals.
If $$x=2,4$$ then $$y$$ becomes rational number which is not in naturals.
Thus, only $$1$$ and $$3$$ gives $$y$$ as natural numbers $$2$$ and $$1$$.
Therefore, the domain is $$\{1,3\}$$ and the range is $$\{2,1\}$$
Question 7
1 / -0

If $$R$$ is the relation 'less than' from $$A=\{1, 2, 3, 4, 5\}$$ to $$B=\{1, 4\}$$, the set of ordered pairs corresponding to $$R$$, then the inverse of $$R$$ is

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

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

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

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

Solution

$$\because R$$ is the relation 'less than' from $$A$$ to $$B$$.

So $$R=\{(1,4), (2,4), (3,4)\}$$

$$R^{-1}=\{(4,1), (4,2), (4, 3)\}$$
Question 8
1 / -0

If $$A$$ is the set of even natural numbers less than $$8$$ and $$B$$ is the set of prime numbers less than $$7$$, then the number of relations from $$A$$ to $$B$$ is

A
$$2^{9}$$

B
$$9^{2}$$

C
$$3^{2}$$

D
$$2^{9}-1$$

Solution

$$A = \{2, 4, 6\}, B = \{2,3,5\}$$
$$n (A\times B) = 3\times 3 = 9$$
Number of relations from $$A$$ to $$B $$ is $$2^9$$
Question 9
1 / -0

The range of the function $$f(x)=\dfrac{\sin(\pi [x])}{x^{2}+1}$$ (where $$[.]$$ denotes greatest integer function) is

A
$$\{0\}$$

B
$$R$$

C
$$(0,1)$$

D
$$(0,\infty )$$

Solution

$$f(x)=\dfrac {\sin(\pi[x])}{x^2+1}$$
$$\because$$ $$[x]$$ always gives integer value and $$\sin n\pi =0$$
$$\therefore f(x)=0$$
Question 10
1 / -0

If $$A =\{1,2,3\}$$, $$B=\{1,4,6,9\}$$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by '$$x$$ is greater than $$y$$'. The range of $$R$$ is

A
$$\{1, 4, 6, 9\}$$

B
$$\{4, 6, 9\}$$

C
$$\{1\}$$

D
$$\{4, 6\}$$

Solution

$$R=\left \{ ( x,y \right ): x\in A, y\in B$$ and $$x> y\}$$
$$R= \left \{ \left ( 2,1 \right ),\left ( 3,1 \right ) \right \}$$
Therefore, $$R=\left \{ 1 \right \}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 21

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Relations and Functions Test - 21

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SHARING IS CARING

Question 1

$$5x-x^{2}+9$$

$$6x-5x^{2}-4$$

$$2x-x^{2}-3$$

$$6x-2x^{2}-3$$

Solution

Question 2

$$-26$$

$$-27$$

$$-28$$

$$-29$$

Solution

Question 3

$$35$$

$$36$$

$$ \displaystyle \frac{7}{5}$$

$$128$$

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 5

$$[-3,3]$$

$$[-\sqrt{3},\sqrt{3}]$$

$$[-\sqrt{2},\sqrt{2}]$$

$$[-2,2]$$

Solution

Question 6

$$\{1,3\}, \{2,1\}$$

$$\{2,1\}, \{3,2\}$$

$$\{1,3\}, \{1,1\}$$

$$\{1,2\}, \{1,3\}$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$2^{9}$$

$$9^{2}$$

$$3^{2}$$

$$2^{9}-1$$

Solution

Question 9

$$\{0\}$$

$$R$$

$$(0,1)$$

$$(0,\infty )$$

Solution

Question 10

$$\{1, 4, 6, 9\}$$

$$\{4, 6, 9\}$$

$$\{1\}$$

$$\{4, 6\}$$

Solution

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