Functions - Class 11 Commerce Applied Mathematics - Extra Questions

We have $$\sqrt{x+2}$$ is defined for $$x+2\ge 0$$ or, $$x\ge -2$$........(1).

$$\begin{array} { *{ 20 }{ l } }{ f\left( x \right) =\dfrac { { 2x+3 } }{ { 3x-2 } }  } \\ { f\circ f\left( x \right) =f\left( { f\left( x \right)  } \right) =f\left( { \dfrac { { 2x+3 } }{ { 3x-2 } }  } \right)  } \\ { =\dfrac { { 2\left( { \dfrac { { 2x+3 } }{ { 3x-2 } }  } \right) +3 } }{ { 3\left( { \dfrac { { 2x+3 } }{ { 3x-2 } }  } \right) -2 } }  } \\ { =\dfrac { { 2x+6+9x-6 } }{ { 6x+9-6x+4 } }  } \\ { =\dfrac { { 13x } }{ { 13 } }  } \\ { =x } \\ { f\circ f\left( x \right) =x } \end{array}$$

i)$$\sqrt(x)-1$$

$$y= \pm \sqrt{25 -x^2}$$
where both x and y are integers.
Cleraly for $$x= 0,\pm 3, \pm 4, \pm 5$$, y will be integers as
$$y = \pm 5, \pm 4, \pm 3, 0$$
$$\therefore R = {(0, \pm 5), (\pm 3, \pm 4), (\pm 4, \pm 3), (\pm 5, 0) }$$
$$R^{-1} ={ ( \pm 5, 0), ( \pm 4, \pm 3), (\pm 3, \pm 4), (0, \pm 5)}$$
Each R and $$R^{-1}$$ consists of 2+ 4 + 4 +2 = 12 ordered pairs.
Domain of R ={$$ 0, \pm 3, \pm 4, \pm 5$$} = domain of $$R^{-1}$$.

$$f(x)=\sqrt{x-5}$$

Find the domain of $$\sin^{-1}\left[log_2\left(\cfrac{x}{2}\right)\right]$$

The function will be defined if $$x^2-4x>0$$

Given the function is $$f(x)=\dfrac{x^2}{\sqrt{(a-x)(x-b)}}$$ with $$b>a$$.

$$\sqrt { \left[ x ^ { 2 } \right] - 1 } + \sqrt { 3 - \left[ x ^ { 2 } \right] }$$

The function is defined only when $$25-x^2\geq0\\x^2\leq25\\|x|\leq 5\\x \in [-5,5]$$

The function is defined only when $$16-x^2\geq0\\x^2\leq16\\|x|\leq 4\\x \in [-4,4]$$

The function is defined only when $$9-x^2\geq0\\x^2\leq9\\|x|\leq 3\\x \in [-3,3]$$

$$f(x)=\dfrac1{x^2-x}\\f(3)=\dfrac1{3^2-3}=\dfrac16$$

The function $$f(x)=\dfrac{x}{\sqrt{1+x}}$$  is defined only when $$1+x>0\\x>-1\\x \in (-1,0)$$

$$\mathrm{(a)}y=1-\log_{10}x\implies x>0$$

The function $$y=\dfrac{\sqrt{x-25}}{x+5}$$ is defined only when $$x+5\neq0\\x\neq-5$$ and $$ x-25\geq 0\\x\geq25$$

The function is not defined when 

Given $$f(x)=\sqrt{4-9 x^2}$$

The function $$f(x)=\dfrac{1}{1+x^2}$$ is only defined when $$1+x^2\neq0\\x^2\neq-1\\x\in R$$

Given $$f(x)=\sqrt{9-x^2}$$

$$f(x)=x^2+3x+2\\f(5)=5^2+3(5)+2\\f(5)=25+15+2=42$$

The function $$f(x)=\dfrac{x-3}{2x+1}$$ is defined only when $$2x+1\neq0\\x\neq\dfrac{-1}{2}$$

the function is $$\sqrt{x^2-4}$$

The function $$f(x)=\dfrac{x^2}{1+x^2}$$ is only defined when $$1+x^2\neq0\\x^2\neq-1\\x\in R$$

Let $$g:A\to B$$ denote a mapping such that $$g=\left\{(2,2),(3,5),(4,5)\right\}$$ which is not an injective mapping.

Let $$h:B\to A$$ denote a mapping such that $$h=\left\{(2,2),(5,3),(6,4),(7,4)\right\}$$ which is mapping from $$A$$ to $$B$$.

Let $$f:N\to N$$ given by $$f(1)=f(2)=1$$ and $$f(x)=x-1$$ for every $$x > 2$$ is onto but not one-one $$f$$ is not one-one as $$f(1)=f(2)=1$$. But $$f$$ is onto.

The mapping $$f:R\to R$$ define as $$f(x)==x^2$$, is neither one-one not onto.

Here, 

$$\displaystyle y= \sqrt{\frac{\left ( x+1 \right )\left ( x-3 \right )}{x-2}}.$$

Taking set $$\{ 1,2,3\} $$

$$f\left( x \right) =\sqrt { 2x } $$
As $$\sqrt{2x} >0$$

$$f(x)=\sqrt{\dfrac{-x+1}{x+3}}$$

$${ x }^{ 3 }+{ x }^{ 2 }-2x\quad \neq 0\\ x\neq -2,0,1$$

$$f(x)=\sqrt{2x+1}$$

Domain of $$x-1$$ is all real numbers i.e. $$R$$

$$f(x)=e^{(x-4)}$$

$$f(x)=\dfrac{3x+4}{5x-7}, g(x)=\dfrac{7x+4}{5x-3}$$

A is set of n different elements.

$$A$$={$${x_1,x_2,x_3,.............x_n}$$}

For one element there are n functions possible.

So for n elements ,

No. of functions possible$$=n⋅n⋅n⋅n………..n $$  [n times]

∴ number of functions from A to A $$=n^n$$

Onto functions: when every element of the image set has an inverse image in the first set.

If ,for $$x_1$$ there are n functions possible.

For $$x_2$$ there are (n-1) functions possible

Similarly for $$x_3,x_4,.......x_n$$

Number of Onto functions are $$ =n\times(n-1) \times (n-2).............1=n!$$

∴ number of Onto functions from $$A$$ to $$A =n!$$

$$(1)$$  $$f(x)=1-2^{-x}$$

For $$n$$ being odd,

Let $${x}_{1}=1$$,  $${x}_{2}=-1$$ (any type of this example)
$$f({ x }_{ 1 })=1+{ 1 }^{ 2 }=2$$
$$f({ x }_{ 2 })=1+{ (-1) }^{ 2 }=2$$
$$f({ x }_{ 1 })=f({ x }_{ 2 })\quad $$
but $${ x }_{ 1 }\neq { x }_{ 2 }$$
$$\therefore$$ $$f$$ is not $$1-1$$
To show $$f$$ is not onto
Let $$-11\in R$$ (co domaine) no $$x$$ in $$R$$ (domaine)
Such that $$f(x)=-11$$
hence Range of $$f\ne$$ co domain of $$f$$
$$\therefore$$ $$f$$ is not on to.

$$f(x)=\sqrt { [x]-1 } +\sqrt { 4-[x] } \\ [x]-1\ge 0\\ [x]\ge 1\\ x\epsilon [1,\infty )\quad -(i)\\ 4-[x]\ge 0\\ [x]\le 4\\ x<5\quad -(ii)\\ from(i)\& (ii)\\ x\epsilon [1,5)$$

Since $$g:B \rightarrow C$$ is on-to.

$$f\left( x \right)={ x }^{ 2 }-6x+5$$

$$f\left(x\right)=\sqrt{2-x-x^{2}}$$ 

$$f{\left( x \right)} = \sqrt{\sin{x}} + \sqrt{16 - {x}^{2}}$$

We must have
$$\log_{2}\log_{\pi/2}(tan^{-1}x)^{-1} > 0$$
or $$\log_{\pi/2}(tan^{-1}x)^{-1} > 1$$
or $$0 < (tan^{-1} x)^{-1} < \frac{2}{\pi}$$
or $$\frac{\pi}{2} < tan^{-1}x < \infty$$, which is not possible. Hence the domain is $$\phi$$

A function is bijective if it is both one-one and onto.

Consider the given function.

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

$$ g\left( x \right)=9x-6 $$

 

If $$f\left( x \right)=g\left( x \right)$$, then

$$ 3{{x}^{2}}-2x=9x-6 $$

$$ 3{{x}^{2}}-11x+6=0 $$

$$ 3{{x}^{2}}-2x-9x+6=0 $$

$$ x\left( 3x-2 \right)-3\left( 3x-2 \right)=0 $$

$$ \left( x-3 \right)\left( 3x-2 \right)=0 $$

$$ x=\dfrac{2}{3},3 $$

 

Therefore,

Required domain $$=\left\{ \dfrac{2}{3},3 \right\}$$

We have,

$$\begin{array}{l}\left[ x \right] - 1 \ge 0\\4 - \left[ x \right] \ge 0\\ = 1 \le \left[ x \right] \le 4\\ = 1 \le x < 5\\x \in \left( {1,5} \right)\end{array}$$

To find domain.

Onto means in $$f:R$$ that range is $$R$$ 

Consider the problem

Let $$x$$ be any real and $$y = 0$$.

$$\begin{array}{l} f\left( x \right) =\sqrt { { x^{ 2 } }-5x+6 } +\sqrt { 2x+8-{ x^{ 2 } } }  \\ Domain\, of\, f\left( x \right) ={ x^{ 2 } }-5x+\ge 0\& 2x+8-{ x^{ 2 } }\ge 0 \\ =\left( { x-2 } \right) \left( { x-3 } \right) \ge 0\& { x^{ 2 } }-2x-8\le 0 \\ x\in (-\infty ,2]\cup [3,\infty )\& \left( { x-4 } \right) \left( { x+2 } \right) \le 0 \\ x\in (-\infty ,2]\cup [3,\infty )\& x\in \left[ { -2,4 } \right]  \\ \Rightarrow \left\{ { -2 } \right\} \cup \left[ { 3,4 } \right]  \end{array}$$

$$f(x)=\dfrac{x^2-8}{x-2}$$

$$f(g(x))$$ and $$g(f(x))$$ are identity functions $$\implies f^{-1}(x)=g(x)$$

$$f(x)=\sqrt{[x]-x}$$......... [ Given ]

$$f(x)=\sqrt{x-[x]}$$.........  [ Given ]

$$f:A\rightarrow B$$ and $$g:B\rightarrow C$$ are onto. .........  [ Given ]

Case $$1:$$

We have, $$f(x)=\sqrt{25-x^2}$$

Let $$f(n)= \sin (\sin^{-1} n)$$
let $$n_1 n_2 \in [-1,1]$$
$$f(n_1)=f(n_2) \Rightarrow \sin (\sin^{-1} n_1)=\sin [\sin^ 1 (n_2)]$$
$$\Rightarrow n_1=n_2$$
hence $$f$$ is one-one
let $$y=\sin (\sin^{-1}n)=n$$
$$\forall y \in [-1,1] \exists [-1,1] f(n)=n$$
hence $$f$$ is onto
since $$f$$ is one-one and onto it is bijective.

Find the domain by finding where the equation is defined. The range is the set of values that correspond with the domain.

{( 1 ,a)(2 , b)( 1 ,b)(2 ,a)}
It is not a function because element 1 corresponds to two elements a and b .

$$f(x)=|x-1|$$
$$f(x)$$ is real for all $$x\in R$$
$$\therefore $$ Domain of $$f=R$$
Also $$f(x)=|x-1| \ge 0$$
$$\therefore$$ Range of $$f=[0, \infty)$$.

The function $$f:M \to R; f(x)=x$$ for all $$\parallel  x\in R$$ is called an identity function on $$R$$.

$$x$$	$$-2$$	$$-1$$	$$0$$	$$1$$	$$2$$
$$f(x)$$	$$-2$$	$$-1$$	$$0$$	$$1$$	$$2$$

Domain of $$f=R$$
Range of $$f=R$$

{(a , 0)(b ,0), (c ,1)( d , 1)}
It is a function because under this each elements corresponds to one and only one element.

{(1 ,2),(2 ,3),(3 , 4), ( 2 , 1)}
It is not a function because element 2 corresponds to two elements 3 and 1.

Let $$c$$ be a fixed real number. Then, the function $$f: R \to R, f(x)=c$$ for all $$x\in R$$ is called the constant function.

$$x$$	$$-2$$	$$-1$$	$$0$$	$$1$$	$$2$$
$$f(x)=c$$	$$c$$	$$c$$	$$c$$	$$c$$	$$c$$

$$f(x)$$ is defined for all real number, 
$$\therefore $$ Domain $$=M$$
Range $$=\left\{ c \right\}$$

{( a ,a)(b , b)( c ,c)}
It is a function because first element of ordered  pair set is not same.

{( a ,b)}
It is a function because a corresponds to b

$$ \left \{ ( x ,y): x , y \in R , y^{2} = x \right \} $$
Here $$ y^{2} =  x \Rightarrow y = \pm \sqrt{x} and  if x =  4  then  y = \pm 2$$ Hence, element of  y  is related with 2 and - 2 so , it is not a function.

$$ \left \{ (x , y) : x , y \in R , x = y^{3} \right \} $$ 
It is a function because $$ y = x ^{1 /3} , \forall  x \in R $$ unique image  is the set B.

{( 4 , 1)(4,2),(4 , 3), ( 4 , 4)}
It is not a function because first element or ordered pair set is same.

{( 1, 4)(2 , 4),(3 , 4), ( 4 , 4 )}
It is a function because first elements of ordered pair set is unequal .

$$ \left \{ ( x , y) : x , y \in R , x^{2} = y \right \} $$ 
It is a function because for $$ y = x^{2} $$ , each real value of x there is a unique image in R for each elements of R

Since, $$f:R\rightarrow R$$ is an onto mapping.
$$\therefore $$ Range of f=R
$$\Rightarrow $$   $$\displaystyle \frac{\alpha x^{2}+6x-8}{\alpha +6x-8x^{2}}$$ assumes all real values of x.
Let $$\displaystyle y=\frac{\alpha x^{2}+6x-8}{\alpha +6x-8x^{2}}$$
Then, y assumes all real values for real values of x.
$$\Rightarrow $$   $$\alpha y+6xy-8x^{2}y=\alpha x^{2}+6x-8$$, $$\forall y\epsilon R$$
$$\Rightarrow $$   $$x^{2}\left ( \alpha +8y \right )+6x\left ( 1-y \right )-\left ( 8+\alpha y \right )=0$$, $$\forall y\epsilon R$$
we know, above equation assumes all real values
$$\Rightarrow $$   $$D\geq 0$$
So, $$36\left ( 1-y \right )^{2}+4\left ( \alpha +8y \right )\left ( 8+\alpha y \right )\geq 0$$
$$\Rightarrow $$   $$4\left [ 9\left ( 1-2y+y^{2} \right )+\left ( 8\alpha +\alpha ^{2}y+64y+8\alpha y^{2} \right ) \right ]\geq 0$$
$$\Rightarrow $$   $$\left [ 9-18y+9y^{2}+8\alpha +\alpha ^{2}y+64y+8\alpha y^{2} \right ]\geq 0$$
$$\Rightarrow $$   $$\left [ y^{2}\left ( 8\alpha +9 \right )+y\left ( \alpha ^{2}+46 \right )+\left ( 8\alpha +9 \right ) \right ]\geq 0$$
We know, if $$ax^{2}+bx+c> 0\forall x$$ and a> 0$$\Rightarrow $$D< 0
So, $$\left ( \alpha ^{2}+46 \right )^{2}-4\left ( 8\alpha +9 \right )\left ( 8\alpha +9 \right )\leq 0$$ and $$\left ( 8\alpha +9 \right )> 0$$
$$\Rightarrow $$   $$\left ( \alpha ^{2}+46 \right )^{2}-\left [ 2\left ( 8\alpha +9 \right ) \right ]^{2}\leq 0$$ and $$\alpha > -9/8$$
$$\Rightarrow $$   $$\left ( \alpha ^{2}+46-16\alpha -18 \right )\left ( \alpha ^{2}+46+16\alpha +18 \right )\leq 0$$ and $$\alpha > -9/8$$
$$\left ( \alpha ^{2}-16\alpha +28 \right )\left ( \alpha ^{2}+16\alpha +64 \right )\leq 0$$ and $$\alpha > -9/8$$
$$\Rightarrow $$   $$\left ( \alpha -14 \right )\left ( \alpha -2 \right )\left ( \alpha +8 \right )^{2}\leq 0$$ and $$\alpha > -9/8$$
$$\alpha \epsilon \left [ 2, 14 \right ]\cup \left \{ -8 \right \}$$ and $$\alpha > -9/8$$
Thus, $$\alpha \epsilon \left [ 2, 14 \right ]$$

Given, $$f\left( x \right) =\cfrac { x-1 }{ x-3 } $$
Thus, $$f\left( x \right) $$ is real if $$x-3\neq 0$$

$$f(x)=\ln({ x }^{ 2 }−9)$$
Thus, $$f(x)$$ is real valued if $$\left( { x }^{ 2 }-9 \right) >0$$
$$\implies { x }^{ 2 }>9$$  and  $$\left| x \right| >3$$
$$\implies x<-3$$ or $$x>3$$

	$$x$$	$$y$$	$$y=mx+c$$
$$\left(1,1\right)$$	$$1$$	$$1$$	$$1=m\times 1+c\Rightarrow 1=m+c$$
$$\left(2,3\right)$$	$$2$$	$$3$$	$$3=m\times 2+c\Rightarrow 3=2m+c$$

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