Relations and Functions Test

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Relations and Functions Test - 48

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Question 1
1 / -0

Let $$ f : R \rightarrow R : f(x) = x^2 + 1 $$ Find the range of eqation

A
[ 0,1]

B
$$ ( - \infty , \infty ) $$

C
$$ [ 1 , \infty) $$

D
$$ (- \infty , 0 ] $$

Solution

Given function is,

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

As we know that square of a real number is always greater than $$0$$
$$\Rightarrow x^2\ge0$$

$$\Rightarrow x^2+1\ge1\implies f(x)\ge1$$

range of $$f(x)=[1,\infty)$$
Question 2
1 / -0

Let $$ f : R \rightarrow R : f (x) = x^2 + 1 $$ . Find the domain of the equation

A
$$ ( - \infty , \infty ) $$

B
$$ ( - \infty , 0 ] $$

C
$$ [ 0 , \infty ) $$

D
$$ [ -1, 1] $$

Solution

Given function is,

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

So we can put any value of $$x$$ for which $$f(x)$$ has real value.

Domain of $$f(x)=(-\infty,\infty)$$
Question 3
1 / -0

f,g and h are three function defined from R to R as follows:
1)$$f(x)=x^2$$
2)$$g(x)=sinx$$
3)$$h(x)=x^2+1$$
what is the common part in the range of these equation

A
0

B
1

C
[0 , 1]

D
$$ ( 0 , \infty )

Solution

Given function are

1)$$f(x)=x^2$$
2)$$g(x)=sinx$$
3)$$h(x)=x^2+1$$

(i) As we know that square of a real number is always greater than $$0$$
$$\Rightarrow x^2\ge0,f(x)\ge0$$
Range of first function $$=[0,\infty)$$

(ii) As we know that tehe domain of sin function is as
$$-\dfrac{\pi}{2}<x<\dfrac{pi}{2}$$

$$sin(-\dfrac{\pi}{2})<sinx<sin(\dfrac{\pi}{2})$$
$$\Rightarrow -1<sinx<1$$
Range of second function $$=[-1,1]$$

(iii) As we know that square of a real number is always greater than $$0$$
$$\Rightarrow x^2\ge0$$

$$\rightarrow x^2+1\ge1,f(x)\ge1$$

Range of third function $$=[1,\infty)$$

the coomon part in given function is $$1$$.
Question 4
1 / -0

Find the domain and range of $$ f(x) = \dfrac {3x -2}{ x+ 2} $$.

A
Domian of $$f(x)=R-[-2]$$
Range can be any real value except $$3$$

B
Domian of $$f(x)=R-[2]$$
Range can be any real value except $$3$$

C
Domian of $$f(x)=R-[0]$$
Range can be any real value except $$5$$

D
Domian of $$f(x)=R-[-2]$$
Range can be any real value except $$0$$

Solution

Given function,
$$f(x)=\dfrac{3x-2}{x+2}$$

Domian of $$f(x)=R-[-2]$$

Domain can be any real value except $$-2$$

Let $$f(x)=y$$
$$\implies y=\dfrac{3x-2}{x+2}$$

$$\implies yx+2y=3x-2\implies 2y+2=(3-y)x$$

$$x=\dfrac{2y+2}{3-y}$$

Range can be any real value except $$3$$
Question 5
1 / -0

Given the function $$f(x) = \frac{a^{x} + a^{-x}} {2} $$ (where $$a > 2$$). Then $$ f(x + y) + f(x - y)$$ =

A
$$ 2f(x).f(y)$$

B
$$ f(x).f(y)$$

C
$$\frac {f(x)} {f(y)}$$

D
None of these

Solution

We have $$ f(x + y) + f(x - y)$$
$$ = \frac {1} {2} \left [ a^{x + y} + a^{-x - y} + a^{x - y} + a^{-x + y} \right ]$$
$$ = \frac {1} {2} \left [ a^{x} (a ^{y} + a^{-y}) + a^{-x}(a^{y} + a^{-y}) \right ]$$
$$ =\frac{1} {2} (a^{x} + a^{-x}) (a^{y} + a^{-y}) = 2f(x) f(y)$$
Question 6
1 / -0

The function f (x) =$$ \dfrac{4x-x^{2}}{4x-x^{3}} $$ is

A
Discontinous at only one point.

B
Discontinous exactly at two points

C
Discontinous exactly at three points

D
None of these

Solution

We have $$f(x) =$$$$ \dfrac{4-x^{2}}{x(4-x^{2})} =\dfrac{4-x^2}{x(2-x)(2+x)}$$
$$f(0)=\dfrac{4-0^2}{0(2-0)(2+0)}=\infty$$
$$f(2)=\dfrac{4-2^2}{2(2-2)(2+2)}=\infty$$
$$f(-2)=\dfrac{4-(-2)^2}{(-2)(2-(-2))(2+(-2))}=\infty$$

Clearly, there are three points of discontinuity , viz$$,0,2,-2$$
Question 7
1 / -0

If $$ F(n + 1) = \dfrac {2F(n) + 1} {2}, \ \ \ n = 1, 2,.......$$ and f(1) = 2 then f(101) equals

A
52

B
49

C
48

D
51

Solution

$$ F(n + 1) = \frac {2F(n) + 1} {2} \Rightarrow F(n + 1) - F(n) = \frac{1} {2}$$

Put n = 1, 2, 3, ........, 100 and add, we get

$$F(101) - F(1) = 100 \times \frac{1} {2}$$

$$\Rightarrow F(101) = 52$$
Question 8
1 / -0

If f is a function such that $$f(0) = 2, f(1) = 3, f(x + 2) = 2f(x) - f(x +1)$$ for every real x, then $$f(5)$$ is

A
7

B
13

C
1

D
5

Solution

Put $$ x = 0 \Rightarrow f(2) = 2 f(0) - f(1) = 2 \times 2 - 3 = 1$$
$$ x = 1 \Rightarrow f(3) = 6 - 1 = 5$$
$$ x = 2 \Rightarrow f(4) = 2 f(2) - f(3) = 2 \times 1 - 5 = -3$$
$$ x = 3 \Rightarrow f(5) = 2 f(3) - f(4) = 2(5) - (- 3) = 13$$
Question 9
1 / -0

Directions For Questions

Let $$ f: N \rightarrow R$$ be a function satisfying the following conditions,
$$f(1) = \dfrac{1}{2}$$ and $$f(1) + 2. f(2) + 3. f(3) + .........+ n.f(n) = n(n + 1). f(n)$$ for $$n \geq 2$$.

...view full instructions

The value of $$f(1003) = \dfrac{1}{k}$$, where $$k$$ equals

A
1003

B
2003

C
2005

D
2006

Solution

We have $$f(1)+2f(2)+...+nf(n) = n(n+1)f(n)\ \ \ \ ... (i) $$

For $$k=n+1$$ we get:

$$f(1)+2f(2)+...+n f(n)+(n+1) f(n+1) = (n+1)(n+2)f(n+1)\ \ \ \ ...(ii) $$

Subtracting $$(i)$$ from $$(ii)$$ we get:

$$(n+1) f(n+1)=(n+1)(n+2) f(n+1)-n(n+1)f(n)$$

$$\Rightarrow (n+1)(n+2) f(n+1) – (n+1) f(n+1) = n(n+1)f(n)$$

$$\Rightarrow (n+1)f(n+1)\{n+2 – 1\} = n(n+1)f(n)$$

$$\Rightarrow (n+1)^2 f(n+1) = (n+1) f(n)$$

$$\Rightarrow (n+1)f(n+1) = n f(n)$$

Hence we get that for all $$k,\ kf(k) = C$$ where $$C $$ is a constant.
When $$k=1$$, we see that
$$1f(1)=\dfrac{1}{2}=C$$

So $$f(k) = \dfrac{1}{2k}$$ for all $$k$$

Hence,
$$f(1003)=\dfrac{1}{2\times 1003}=\dfrac{1}{2006}$$
So, $$k=2006$$
Question 10
1 / -0

If $$f(x + f(y)) = f(x) + y \space \forall \space x, y \space \epsilon \space R$$ and $$f(0) = 1$$, then the value of $$f(7)$$ is

A
1

B
7

C
6

D
8

Solution

$$f(x + f(y)) = f(x) + y, f(0) = 1$$
Putting $$y = 0$$, we get $$f(x + f(0)) = f(x) + 0$$
$$\Rightarrow f(x+1) = f(x) \space \forall \space x \space \epsilon \space R$$
Thus, $$f(x)$$ is the period with 1 as one of its period.
$$\Rightarrow f(7) = f(6) = f(5) = ...... = f(1) = (0) = 1$$.

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Relations and Functions Test - 48

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Relations and Functions Test - 48

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Question 1

[ 0,1]

$$ ( - \infty , \infty ) $$

$$ [ 1 , \infty) $$

$$ (- \infty , 0 ] $$

Solution

Question 2

$$ ( - \infty , \infty ) $$

$$ ( - \infty , 0 ] $$

$$ [ 0 , \infty ) $$

$$ [ -1, 1] $$

Solution

Question 3

0

1

[0 , 1]

$$ ( 0 , \infty )

Solution

Question 4

Domian of $$f(x)=R-[-2]$$Range can be any real value except $$3$$

Domian of $$f(x)=R-[2]$$Range can be any real value except $$3$$

Domian of $$f(x)=R-[0]$$Range can be any real value except $$5$$

Domian of $$f(x)=R-[-2]$$Range can be any real value except $$0$$

Solution

Question 5

$$ 2f(x).f(y)$$

$$ f(x).f(y)$$

$$\frac {f(x)} {f(y)}$$

None of these

Solution

Question 6

Discontinous at only one point.

Discontinous exactly at two points

Discontinous exactly at three points

None of these

Solution

Question 7

52

49

48

51

Solution

Question 8

7

13

1

5

Solution

Question 9

1003

2003

2005

2006

Solution

Question 10

1

7

6

8

Solution

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Domian of $$f(x)=R-[-2]$$
Range can be any real value except $$3$$

Domian of $$f(x)=R-[2]$$
Range can be any real value except $$3$$

Domian of $$f(x)=R-[0]$$
Range can be any real value except $$5$$

Domian of $$f(x)=R-[-2]$$
Range can be any real value except $$0$$