Relations and Functions Test

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Relations and Functions Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f(x) =\frac {ax+b} {cx+d}$$. Then fof(x) = x provided that.

A
d =- a

B
d = a

C
a = b = c = d = 1

D
a = b = 1

Solution

$$f(f(x))=\dfrac{af(x) + b}{cf(x)+d}$$

$$f(f(x))=\dfrac{a\dfrac{ax+b}{cx+d} + b}{c\dfrac{ax+b}{cx+d}+d}$$

$$f(f(x))=\dfrac{a^2x + ab+bcx+bd}{acx+bc+d^2+dcx}=x$$

$$\Rightarrow a^2x+ab+bcx+bd=acx^2+bcx+d^2x+dcx^2$$

Given $$a=-d$$, the above equation can also be verified
Question 2
1 / -0

If $$f(x) =\dfrac {1}{1-x}, x \neq 0, 1$$ then the graph of the function $$y = f[f\{f(x)\}]$$ for $$x > 1 $$ is

A
a straight line

B
a circle

C
an ellipse

D
a pair of straight lines

Solution

Given $$f(x) = \cfrac{1}{1-x}$$
$$\displaystyle \Rightarrow f\{f(x)\} = \frac{1}{1-f(x)} = \frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}=1-\frac{1}{x}$$
$$\displaystyle \therefore f[f\{f(x)\}] = \frac{1}{1- f\{f(x)\}}=\frac{1}{1-1+\frac{1}{x}}=x$$, which is a straight line.
Question 3
1 / -0

If $$f : R\rightarrow R, f(x) = 3x-4$$, then $$f^{-1}(x)$$ is equal to

A
$$\dfrac {1}{3} (x-4)$$

B
$$\dfrac {1}{3} (x+4)$$

C
$$3x + 4$$

D
undefined

Solution

Let $$y=f^{ -1 }\left( x \right) $$

$$ \Rightarrow x=f\left( y \right) =3y-4$$

$$ \Rightarrow y=\dfrac { x+4 }{ 3 } $$
Question 4
1 / -0

If $$f : [0, \Pi ] \rightarrow [-1, 1]$$, f(x) = cosx, then f is.

A
one-one

B
onto

C
one-one onto

D
none of these

Solution

f takes all values from $$[-1,1]$$ while travelling from $$[0,\pi]$$. None of the values are repeated either.

Hence it is one-one and onto.
Question 5
1 / -0

If $$f(x) = \left\{\begin{matrix} 1&x \in Q \\ 0 &x \notin Q\end{matrix}\right.$$ then $$fof(\sqrt 3 )$$ is equal to

A
$$0$$

B
$$1$$

C
$$\sqrt 3$$

D
none of these

Solution

We know $$\sqrt{3}$$ is an irrational number

Thus $$f(\sqrt{3}) =0$$ and $$0$$ is a rational number

Hence $$fof(\sqrt 3 )= f(0) = 1$$
Question 6
1 / -0

If $$f : R\rightarrow R $$ and $$f(x) = x^2 - 5x + 9$$, then $$f^{-1}(9)$$ equals

A
$$\{0\}$$

B
$$\{5\}$$

C
$$\{0, 5\}$$

D
none of these

Solution

$$f^{-1}(9)$$ will be the roots of

$$x^2-5x+9 = 9$$

$$\Rightarrow x(x-5)=0$$

$$\Rightarrow x =0, 5$$

Hence $$f^{-1}(9) = 0,5$$
Question 7
1 / -0

If functions $$f\left ( x \right )$$ and $$g\left ( x \right )$$ are defined on $$R\rightarrow R$$ such that
$$f(x)=x+3, x$$ $$\in $$ rational
$$ =4x, x$$ $$\in $$ irrational
$$g(x)=x+\sqrt{5}$$, x$$\in $$ irrational
$$ =-x, x$$ $$\in $$ rational
then $$\left ( f-g \right )\left ( x \right )$$ is

A
one-one & onto

B
neither one-one nor onto

C
one-one but not onto

D
onto but not one-one

Solution

$$f\left( x \right) =x+3, x\epsilon$$ rational
$$=4x, x\epsilon$$ irrational

$$g\left( x \right) =x+\sqrt { 5 } , x \epsilon$$ irrational
$$= -x,x\epsilon$$ rational

$$(f-g)(x)=2x+3, x\ \epsilon$$ rational
$$+3x-\sqrt { 5\quad } x\quad \epsilon$$ irrational

For one-one;
We know that one one function is a $${ f }^{ \underline { n } }$$ for which every element of the range of the function corresponds to exactly one element of the domain.

But in this case this is not true as for all rational nos. the$$ { f }^{ \underline { n } }$$ is one one but for irrational $${ f}^{ \underline { n } }$$, every elements of the range does not corresponds to exactly one element of the domain.

$$(f-g)(x)\neq (f-g)({ x }^{ \prime })$$ when $$x \epsilon$$ rational and
$$x \epsilon$$ irrational
for onto:
The $${ f }^{ \underline { n } }(f-g)(x)$$ does not cover the whole range of the function.
Question 8
1 / -0

Let $$f:\left [ 1, \infty \right )\rightarrow \left [ 3, \infty \right )$$ & $$f\left ( x \right )=\left ( \log_{2}x \right )^{2}+2\log_{2}x+3,$$ then $$f^{-1}\left ( x \right )$$ is equal to

A
$$2^{\sqrt{x-2}-1}$$

B
$$2^{-1-\sqrt{x-2}}$$

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

D
$$4^{\sqrt{x-3}}$$
Question 9
1 / -0

If $$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$$, then find $$fof(x)$$

A
$$\dfrac1x$$

B
$$x$$

C
$$x^2$$

D
$$x^3$$

Solution

Given, $$\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$$
Therefore, $$\displaystyle fof\left ( x \right )=f\left [ f\left ( x \right ) \right ]$$
$$\displaystyle =f\left ( \left ( 1-x^{3} \right )^{\frac{1}{3}} \right )=\left [ 1-\left \{ \left ( 1-x^{3} \right )^{\frac{1}{3}} \right \}^{3} \right ]^{\frac{1}{3}}$$
$$\displaystyle =\left [ 1-\left ( 1-x \right )^{3} \right ]^{\frac{1}{3}}=\left ( x^{3} \right )^{\frac{1}{3}}=x$$
Question 10
1 / -0

Let $$\displaystyle f\left ( x \right )=\frac{3}{2}+\sqrt{x-\frac{3}{4}}$$ be a function and $$g\left ( x \right )$$ be another function such that $$g\left ( f\left ( x \right ) \right )=x,$$ then the value of $$g\left ( 20 \right )$$ will be

A
$$333$$

B
$$335$$

C
$$338$$

D
$$343$$

Solution

$$g\left ( x \right )$$ is inverse of $$f\left ( x \right )$$

$$\Rightarrow $$ $$\displaystyle x=\frac{3}{2}+\sqrt{y-\frac{3}{4}}$$

$$\Rightarrow $$ $$\displaystyle y=\left ( x-\frac{3}{2} \right )^2+\frac{3}{4}=f^{-1}(x)=g(x)$$

$$\therefore $$ $$\displaystyle g\left ( 20 \right )=\left ( 20-\frac{3}{2} \right )^{2}+\frac{3}{4}=343$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 51

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Relations and Functions Test - 51

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SHARING IS CARING

Question 1

d =- a

d = a

a = b = c = d = 1

a = b = 1

Solution

Question 2

a straight line

a circle

an ellipse

a pair of straight lines

Solution

Question 3

$$\dfrac {1}{3} (x-4)$$

$$\dfrac {1}{3} (x+4)$$

$$3x + 4$$

undefined

Solution

Question 4

one-one

onto

one-one onto

none of these

Solution

Question 5

$$0$$

$$1$$

$$\sqrt 3$$

none of these

Solution

Question 6

$$\{0\}$$

$$\{5\}$$

$$\{0, 5\}$$

none of these

Solution

Question 7

one-one & onto

neither one-one nor onto

one-one but not onto

onto but not one-one

Solution

Question 8

$$2^{\sqrt{x-2}-1}$$

$$2^{-1-\sqrt{x-2}}$$

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

$$4^{\sqrt{x-3}}$$

Question 9

$$\dfrac1x$$

$$x$$

$$x^2$$

$$x^3$$

Solution

Question 10

$$333$$

$$335$$

$$338$$

$$343$$

Solution

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