Relations and Functions Test

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Relations and Functions Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Let $$R$$ be a relation on $$N$$ defined by $$x+2y=8$$. The domain of $$R$$ is

A
$$\left\{ 2,4,8 \right\} $$

B
$$\left\{ 2,4,6,8 \right\} $$

C
$$\left\{ 2,4,6 \right\} $$

D
$$\left\{ 1,2,3,4 \right\} $$

Solution

$$x+2y=8$$
$$x=8-2y$$
Since y and x both must be natural,
Possible values of y are 1,2,3 which corresponds to possible value of x to be 6,4,2 .
Thus range of R = $$\{2,4,6\}$$
Question 2
1 / -0

If y is a function of x and $$\log (x + y) = 2xy$$, then the value of y'(0) = .....

A
2

B
0

C
-1

D
1

Solution

$$\log (x + y) = 2xy$$.....(1)
$$\therefore \frac{1}{x + y} \cdot (1 + \frac{dy}{dx}) = 2x \frac{dy}{dx} + 2y$$
$$\therefore ( \frac{1}{x + y} - 2x) \frac{dy}{dx} = 2y - \frac{1}{x + y}$$
$$\therefore \frac{dy}{dx} = \frac{2y (x + y) - 1}{1 - 2x (x + y)}$$
If x = 0, then from (1)
log y = 0 = log 1
$$\therefore$$ y = 1
$$\therefore$$ y'(0) $$= \frac{(2(1)(0 + 1)-1)}{1 - 2(0)(0 + 1)} = 1$$
Question 3
1 / -0

If $$f(x+y)=f(x)f(y)$$ and $$ \sum^{\infty}_{x=1}f(x)=2\,x\,y\epsilon N$$, where $$N$$ is the set of all natural number, then the value of $$f(4)/f(2)$$ is:

A
$$\frac{2}{3}$$

B
$$\frac{1}{9}$$

C
$$\frac{1}{3}$$

D
$$\frac{4}{9}$$

Solution

$$f(x+y)=f(x)f(y)$$

Put $$x=1,y=1$$

$$f(2)=(f(1))^2$$

Put $$x=2,y=1$$

$$f(3)=f(2).f(1)=f(1)^3$$

Put $$x=2,y=2$$

$$f(4)=f((2))^2=f((1))^4$$

$$f(n)=(f(1))^n$$

$$\sum^{\infty}_{x=1} f(x)=f(1)+f(2)+f(3)+....=2$$

$$\Rightarrow f(1)+f((1))^2+f(1))^3+.....=2$$

$$f(1)/(1-f(1))=2$$

$$f(1)=\frac{2}{3}$$

$$f(2)=[\frac{2}{3}]^2$$

$$f(4)=[\frac{2}{3}]^4$$

$$\frac{f(4)}{f(2)}=\frac{[\frac{2}{3}]^4}{{[\frac{2}{3}]^2}}=\frac{4}{9}$$
Question 4
1 / -0

If $$\displaystyle f(x)= \cos{ \left[ \pi ^{ 2 } \right] x}+ \cos{( \left[ -\pi ^{ 2 } \right]x)}$$, where $$[.]$$ denotes the step function, then

A
$$f(0)=1$$

B
$$f\left (\dfrac{\pi} 4\right )=2$$

C
$$f\left (\dfrac{\pi} 2\right )=-1$$

D
$$f(\pi )=1$$

Solution

$$\pi^2=9.87$$

$$\Rightarrow$$$$9<\pi^2<10$$ and $$-10<-\pi^2<-9$$

$$\therefore [\pi^2]=9$$ and $$[-\pi^2]=-10$$

$$\therefore f(x)=\cos9x+\cos10x$$

$$\therefore f\left ( 0 \right )=1+1=2$$

$$\Rightarrow$$$$f\left ( \dfrac{\pi }{4} \right )=\cos9\times\dfrac{\pi}{4}+\cos10\times\dfrac{\pi}{4} $$

$$\Rightarrow$$$$f\left ( \dfrac{\pi }{4} \right )$$$$=-\dfrac{1}{\sqrt{2}}$$

$$\Rightarrow$$$$f\left ( \dfrac{\pi }{2} \right )=\cos9\times\dfrac{\pi}{2}+\cos10\times\dfrac{\pi}{2} $$

$$\Rightarrow$$$$f\left ( \dfrac{\pi }{2} \right )=$$$$-1$$

$$\Rightarrow$$$$f(\pi)=0$$

Hence, option 'C' is correct.
Question 5
1 / -0

Let $$A$$ be a non-empty set such that $$A \times A$$ has $$9 $$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$, then

A
$$A=\left\{ -1,0 \right\}$$

B
$$A=\left\{ 0,1 \right\}$$

C
$$A=\left\{ -1,0,1 \right\}$$

D
$$A=\left\{ -1,1 \right\}$$

Solution

Given $$A\times A$$ has $$9$$ elements $$\Rightarrow n(A)=3$$
Elements of $$A\times A $$ are $$(-1,0)$$ & $$(0,1)$$.
$$\therefore$$ $$A$$ needs to contain $$-1, 0$$ and $$1$$.
$$\therefore$$ Since $$n(A)=3 \Rightarrow A=\left \{ -1,0,1 \right \}$$
Question 6
1 / -0

If $$\displaystyle f(a)=\log\left ( \frac{2+a}{2-a} \right ), 0< a<2$$ then $$ \displaystyle \frac{1}{2}f\left ( \frac{8a}{4+a^{2}} \right )=$$

A
$$f(a)$$

B
$$2f(a)$$

C
$$\displaystyle \frac{1}{2}f(a)$$

D
$$-f(a)$$

Solution

Given $$\displaystyle f\left ( a \right )=\log \left ( \frac{2+a}{2-a} \right )$$

Now, $$\displaystyle f\left ( \frac{8a}{4+a^{2}} \right )=\log\left (\frac{2+\frac{8a}{4+a^{2}}}{2-\frac{8a}{4+a^{2}}} \right )$$
$$\displaystyle =\log \left ( \frac{8+2a^{2}+8a}{8+2a^{2}-8a} \right )$$
$$\displaystyle =\log \frac{\left ( 2+a \right )^{2}}{\left ( 2-a \right )^{2}}$$
$$\displaystyle =2\log \left ( \frac{2+a}{2-a} \right )$$
$$=2f(a)$$
$$\therefore \displaystyle \frac{1}{2}f\left ( \frac{8a}{4+a^{2}} \right )=f\left ( a \right )$$
Question 7
1 / -0

If $$f(x)=ax+\beta $$ and $$ f=\left \{ (1,1),(2,3),(3,5),(4,7) \right \} $$ then the values of $$ \alpha$$ and $$\beta $$ are

A
$$2, -1$$

B
$$-2, 1$$

C
$$3, -1$$

D
$$-2, -1$$

Solution

$$f(1)=\alpha+\beta =1$$

$$f(2)=\alpha\times2+\beta=3$$

$$\therefore\alpha=2$$, $$\beta=-1$$
Checking for rest of values using above $$\alpha $$ and $$\beta $$ also satisfies the function's input and output.
Thus option A is correct.
Question 8
1 / -0

If $$f(x)=2^{x}$$ then $$\dfrac{f(x+3)}{f(x-1)}=$$

A
$$f(x)$$

B
$$\dfrac{1}{f(x)}$$

C
$$f(4)$$

D
$$f(2)$$

Solution

Let$$f\left ( x \right )=2^{x}$$
$$f\left ( x+3 \right )=2^{x+3}=2^{x}\times2^{3}$$
$$f\left ( x-1 \right )=2^{x-1}=\dfrac{2^{x}}{2} \quad \quad \left [ \because a^{m+n}=a^{m}\times a^{n} \right ]$$
$$\Rightarrow

\dfrac { f\left( x+3 \right) }{ f\left( x-1 \right) } =\dfrac { 2^{ x

}\times2^{ 3 } }{ \dfrac { 2^x }{ 2 } } =16=2^{ 4 }=f\left( 4 \right) $$
Question 9
1 / -0

$$ f \left ( x +\dfrac{1}{x} \right ) = x^{2} + \dfrac{1}{x^{2}} $$
Then $$f(x)$$ can be

A
$$ 2-x^{2} $$

B
$$ x^{2} - 2 $$

C
$$ x^{2} + 4 $$

D
$$ 4x^{2} -2 $$

Solution

We will try to represent $$\displaystyle f\left(x+\dfrac{1}{x}\right)$$ in the form of $$\displaystyle x+\dfrac{1}{x}$$ and substitute $$x$$ in the place $$x+\dfrac{1}{x}$$.

Given $$\displaystyle f\left(x+\dfrac{1}{x}\right)=x^{2}+\dfrac{1}{x^2}$$

$$\displaystyle=\left(x+\dfrac{1}{x}\right)^{2}-2$$

$$\Rightarrow \displaystyle f\left(x+\dfrac{1}{x}\right)=\left(x+\dfrac{1}{x}\right)^{2}-2$$
Substituting $$x$$ in place of $$\displaystyle x+\dfrac{1}{x}$$, we get
$$f\left(x\right)=x^2-2$$
Question 10
1 / -0

$$f:R\rightarrow R$$ is defined as $$f(x)=2x+\left | x \right |$$ then $$f(2x)-f(-x)-4x=$$

A
$$f(x)$$

B
$$-f(x)$$

C
$$f(-x)$$

D
$$2f(x)$$

Solution

$$f\left ( 2x \right )=2(2x)+| 2x |$$

$$=4x+2\left | x \right |$$

$$f\left ( -x \right )=2\left ( -x \right )+\left | \left ( -x \right ) \right |$$

$$=-2x+\left | x \right |$$

$$f\left ( 2x \right )-f\left ( -x \right )=6x+\left | x \right |$$

$$f\left ( 2x \right )-f\left ( -x \right )-4x=2x+\left | x \right |$$

$$=\left ( 2x+\left | x \right | \right )$$

$$=f\left ( x \right )$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 19

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Relations and Functions Test - 19

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SHARING IS CARING

Question 1

$$\left\{ 2,4,8 \right\} $$

$$\left\{ 2,4,6,8 \right\} $$

$$\left\{ 2,4,6 \right\} $$

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

Solution

Question 2

2

0

-1

1

Solution

Question 3

$$\frac{2}{3}$$

$$\frac{1}{9}$$

$$\frac{1}{3}$$

$$\frac{4}{9}$$

Solution

Question 4

$$f(0)=1$$

$$f\left (\dfrac{\pi} 4\right )=2$$

$$f\left (\dfrac{\pi} 2\right )=-1$$

$$f(\pi )=1$$

Solution

Question 5

$$A=\left\{ -1,0 \right\}$$

$$A=\left\{ 0,1 \right\}$$

$$A=\left\{ -1,0,1 \right\}$$

$$A=\left\{ -1,1 \right\}$$

Solution

Question 6

$$f(a)$$

$$2f(a)$$

$$\displaystyle \frac{1}{2}f(a)$$

$$-f(a)$$

Solution

Question 7

$$2, -1$$

$$-2, 1$$

$$3, -1$$

$$-2, -1$$

Solution

Question 8

$$f(x)$$

$$\dfrac{1}{f(x)}$$

$$f(4)$$

$$f(2)$$

Solution

Question 9

$$ 2-x^{2} $$

$$ x^{2} - 2 $$

$$ x^{2} + 4 $$

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

Solution

Question 10

$$f(x)$$

$$-f(x)$$

$$f(-x)$$

$$2f(x)$$

Solution

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