Relations and Functions Test

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Relations and Functions Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f(x) = \frac{ax}{x + 1}, x \neq -1$$. Then for what value of $$\alpha$$ is $$f(f(x)) = x$$? (IIT-JEE, 2001)

A
$$\sqrt 2$$

B
$$- \sqrt 2$$

C
1

D
-1

Solution
Question 2
1 / -0

If $$f(x + 1) + f(x - 1) = 2f(x) $$ and $$f(0) = 0$$, then $$f(n+1), n \space \epsilon \space N$$, is

A
$$(n+1)f(1)$$

B
$$\left \{ f(1)\right \}^{n}$$

C
0

D
None of these

Solution

Putting $$x = 1, f(2) + f(0) = 2f(1) \Rightarrow f(2) = 2f(1)$$
Putting $$x = 2, f(3) + f(1) = 2f(2)$$
$$\Rightarrow f(3) = 2 \times 2f(1) - f(1) = 3f(1)$$, and so on.
$$\therefore f(n) = nf(1)$$, for $$n = 1, 2,........., n$$
$$f(n + 1) + f(n - 1) = 2f(n)$$
$$\Rightarrow f(n + 1) + (n - 1)f(1) = 2nf(1)$$
$$\Rightarrow f(n + 1) = (n + 1) f(1)$$
Question 3
1 / -0

If $$ f : R^{+} \rightarrow R, f(x) + 3xf \left (\frac {1} {x} \right ) = 2 \left ( x + 1 \right )$$, then $$f(99)$$ is equal to

A
40

B
30

C
50

D
60

Solution

$$f(x) + 3xf \left (\frac {1} {x} \right ) = 2 \left ( x + 1 \right )$$ (1)
Replacing $$x$$ by $$\frac{1} {x}$$, we get
$$ f \left (\frac {1} {3} \right )+ 3 \frac{1} {x} f \left (x \right ) = 2 \left ( \frac{1} {x} + 1 \right )$$
$$\Rightarrow x f\left ( \frac{1} {x} \right ) + 3f(x) = 2\left ( x + 1 \right )$$ (2)
From (1) and (2), we have $$f(x) = \frac{ x + 1} {2}$$
$$\Rightarrow f(99) = 50$$
Question 4
1 / -0

Let $$f:R\to R$$ be the function defined by $$f(x)=\left\{ \begin{matrix} 2x:x>3 \\ { x }^{ 2 }:1<x\le 3 \\ 3x:x\le 1 \end{matrix} \right.$$ Then $$f(-1)+f(2)+f(4)$$ is

A
$$9$$

B
$$14$$

C
$$5$$

D
$$none\ of\ these$$

Solution

Given that $$f(x)=\left\{ \begin{matrix} 2x:x>3 \\ { x }^{ 2 }:1<x\le 3 \\ 3x:x\le 1 \end{matrix} \right.$$
$$f(-1)+f(2)+f(4)=3(-1)+(2)^2 +2\times 4$$
$$=-3+4+8=9$$
Question 5
1 / -0

If A = { a , b , c , d} and B = { p , q ,r ,s} then relation from A and B is

A
{(a ,p) , ( b, r), (c , r)}

B
{(a , p), (p, q) , (c, r) , ( s , d)}

C
{{ b , a) , (q , b ) , (c , r) }

D
{ ( c,s) , ( d , s ,) ( r, a) , ( q , b )}

Solution

Because $$ { ( a , p), ( b ,r) ( c ,r)} \subseteq A \times B $$
Question 6
1 / -0

Rule from of relation {(1 ,2) , ( 2 , 5) , (3 , 10) , ( 4 , 17), ......} in N

A
$$ {(x ,y) : x , y \in N = y = 2x + 1 } $$

B
$$ {( x , y ) : x , y \in N , y = x^{2} + 1 } $$

C
$$ {( x , y ) : x , y \in N , y = 3x - 1} $$

D
$$ {( x , y ) : x , y \in N , y = x + 3} $$

Solution
Question 7
1 / -0

A relation R in N is defined such that $$ xRy \Leftrightarrow x + 4y = 16 ,$$ then the range of R is

A
{ 1 , 2 , 4}

B
{ 1 , 3 , 4}

C
{ 1 , 2 , 3}

D
{ 2 , 3 , 4}

Solution
Question 8
1 / -0

If A = { a ,b , c} , then numbers of possible non zero relations in A is

A
$$ 511$$

B
$$ 512 $$

C
$$ 7 $$

D
$$ 8 $$

Solution
Question 9
1 / -0

Directions For Questions

A function $$y=f(x)$$ is said to be symmetric about $$x=a$$ if $$f(a-x)=f(a+x)$$.
A function $$y=f(x)$$ is symmetric about $$y=x$$ if $$f(f(x))=x$$.

...view full instructions

A real valued function $$f(x)$$ satisfies the functional equation $$f(x-y)=f(x)f(y)-f(a-x)f(a+y)$$, where $$a$$ is a given constant and $$f(0)=1$$, then

A
$$f(2a-x)=f(x)$$, $$f(x)$$ is symmetric about $$x=a$$

B
$$f(2a-x)=0=f(x)$$, $$f(x)$$ is symmetric about $$x=a$$

C
$$f(2a-x)+f(x)=0$$, $$f(x)$$ is not symmetric about $$x=a$$

D
$$f(x)=0$$, $$f (x)$$ is symmetric about $$x=a$$

Solution

Using the given relation
$$f(2a-x)=f(a-(x-a))$$

Using the given expression for $$f(x-y)$$, take $$x = a$$ and $$y = x-a$$
$$f(2a-x)=f(a)f(x-a)-f(a-a)f(a+x-a)$$
$$f(2a-x)=f(a)f(x-a)-f(0)f(x)$$

When $$x=0, y=0 \Rightarrow f(0)=(f(0))^2-(f(a))^2$$
It is given that $$f(0) = 1$$
$$\Rightarrow f^2(a)=0\Rightarrow f(a)=0$$

Substitute this in the expression,
$$\Rightarrow f(2a-x)=0(f(x-a))-1(f(x))$$
$$\Rightarrow f(2a-x) = -f(x)$$
$$\Rightarrow f(2a-x) +f(x)=0$$
Take $$x = a-x$$
$$\therefore f(2a-(a-x)) = -f(a-x)$$
$$\Rightarrow f(a+x) = -f(a-x)$$
Hence, $$f(x)$$ is not symmetric about $$x=a$$.
Question 10
1 / -0

If $$f(x) + 2f(1 - x) = x^2 + 2 $$; $$\forall x \in R$$, then $$f(x)$$ is given by

A
$$\displaystyle \frac{(x-1)^2}{3}$$

B
$$-\displaystyle \frac{(x-2)^2}{3}$$

C
$$x^2-1$$

D
$$x^2-2$$

Solution

$$\Rightarrow$$$$f(x)+2f(1-x)=x^2+2\cdots\cdots(i)$$

Put $$x=1-x$$

$$\Rightarrow$$$$f(1-x)+2f(x)=(1-x)^2+2$$

Multiply by $$2$$

$$\Rightarrow$$$$2f(1-x)+4f(x)= 2(1-x)^2+4\cdots\cdots(ii)$$

Subtract $$(ii)$$ from $$(i)$$

$$-3f(x)=x^2-2-2(1-x)^2$$

$$-3f(x)=x^2-2-2x^2+4x$$

$$-3f(x)=-(x^2-4x+4)$$

$$f(x)=-\dfrac{(x-2)^2}{3}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 49

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Relations and Functions Test - 49

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SHARING IS CARING

Question 1

$$\sqrt 2$$

$$- \sqrt 2$$

1

-1

Solution

Question 2

$$(n+1)f(1)$$

$$\left \{ f(1)\right \}^{n}$$

0

None of these

Solution

Question 3

40

30

50

60

Solution

Question 4

$$9$$

$$14$$

$$5$$

$$none\ of\ these$$

Solution

Question 5

{(a ,p) , ( b, r), (c , r)}

{(a , p), (p, q) , (c, r) , ( s , d)}

{{ b , a) , (q , b ) , (c , r) }

{ ( c,s) , ( d , s ,) ( r, a) , ( q , b )}

Solution

Question 6

$$ {(x ,y) : x , y \in N = y = 2x + 1 } $$

$$ {( x , y ) : x , y \in N , y = x^{2} + 1 } $$

$$ {( x , y ) : x , y \in N , y = 3x - 1} $$

$$ {( x , y ) : x , y \in N , y = x + 3} $$

Solution

Question 7

{ 1 , 2 , 4}

{ 1 , 3 , 4}

{ 1 , 2 , 3}

{ 2 , 3 , 4}

Solution

Question 8

$$ 511$$

$$ 512 $$

$$ 7 $$

$$ 8 $$

Solution

Question 9

$$f(2a-x)=f(x)$$, $$f(x)$$ is symmetric about $$x=a$$

$$f(2a-x)=0=f(x)$$, $$f(x)$$ is symmetric about $$x=a$$

$$f(2a-x)+f(x)=0$$, $$f(x)$$ is not symmetric about $$x=a$$

$$f(x)=0$$, $$f (x)$$ is symmetric about $$x=a$$

Solution

Question 10

$$\displaystyle \frac{(x-1)^2}{3}$$

$$-\displaystyle \frac{(x-2)^2}{3}$$

$$x^2-1$$

$$x^2-2$$

Solution

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