Relations And Functions - Class 12 Commerce Maths - Extra Questions

Given, $$f(x)=\dfrac { 8 }{ x } $$ and $$\text{ fog}(x)=4x$$

$$ f(g(x))=\dfrac { 8 }{ g(x) } \\ \Rightarrow 4x=\dfrac { 8 }{ g(x) } \\ \Rightarrow g(x)=\dfrac { 2 }{ x } $$

Thus $$g(f(x))=\dfrac { 2 }{ \dfrac { 8 }{ x }  } =\dfrac { x }{ 4 } $$

$$\Rightarrow  3=\dfrac { x }{ 4 } \\ \Rightarrow x=12$$

Given, $$f(x)=\dfrac { x-2 }{ 5 } $$

Thus $$ g(f(x))=x-3$$

$$\Rightarrow  g\left( \dfrac { x-2 }{ 5 }  \right) =x-3$$    ....(i)

$$\Rightarrow \dfrac { x-2 }{ 5 } =z\\ \Rightarrow x=5z+2$$

Substituting in (i), we get

$$\Rightarrow g\left( \dfrac { 5z+2+2 }{ 5 }  \right) =5z-1\\ \Rightarrow g(z)=5(z-1)\\ \Rightarrow g(x)=5x-1\\ $$

Let $$\int {\left( x \right) = y} $$

$$y = 4{x^2} + 12x + 15$$

$$4{x^2} + 12x + \left( {15 - y} \right) = 0$$

$$x = {{ - 12 \pm \sqrt {144 - 16\left( {15 - y} \right)} } \over 8}$$

$$ = {{ - 12 \pm \sqrt {16\left( {9 - 15 + y} \right)} } \over 8}$$

$$ = {{ - 12 \pm 4\sqrt {y - 6} } \over 8}$$$$\int {:N \to s} $$

$$so\,\,y \in s$$$$y \in $$ range of

$$ = {{ - 3 \pm \sqrt {y - 6} } \over 8}$$

As $$x \in n$$

$$x = {{ - 3 + \sqrt {y - 6} } \over 2}$$

Now $$\int {\left( x \right) = \int {\left( {{x_2}} \right)} } $$

$$ \Rightarrow 4{x_1}^2 + 12{x_1} + 15 = 4{x_2}^2 + 12{x_2} + 15$$

$$ \Rightarrow 4\left( {{x_1}^2 - {x_2}^2} \right) + 12\left( {{x_1} - {x_2}} \right) = 0$$

$$ \Rightarrow {x_1} = {x_2}$$  or  $$\,{x_1} + {x_2} + 3 = 0$$

So $${x_1} = {x_2}$$  as   $${x_1} + {x_2} + 3 \ne 0$$

$$\therefore f\;{\text{is}}\;{\text{one}}\;{\text{row,}}\;{\text{\& }}\;{\text{hence}}\;{\text{invertible}}\;{\text{and}}\;f\left( x \right) = \dfrac{{ - 3 + \sqrt {y - 6} }}{2}$$

$$f\left( x \right) = {{3x + 2} \over {5x + 3}},x \ne {{ - 3} \over 5}$$

One One:

$${{3{x_1} + 2} \over {5{x_1} + 3}} = {{3{x_2} + 2} \over {5{x_2} + 3}}$$

$$ \Rightarrow 15{x_1}{x_2} + 9{x_1} + 10{x_2} + 6 = 15{x_1}{x_2} + 9{x_2} + 10{x_1} + 6$$

$$ \Rightarrow {x_2} = {x_1}$$

f is one-one

Onto:

Let $$y = {{3x + 2} \over {5x + 3}}$$

$$ \Rightarrow 5xy + 3y = 3x + 2$$

$$ \Rightarrow \left( {5y - 3} \right)x = 2 - 3y$$

$$ \Rightarrow x = {{2 - 3y} \over {5y - 3}}$$

Now, $$f\left( {{{2 - 3y} \over {5y - 3}}} \right) = {{3\left( {{{2 - 3y} \over {5y - 3}}} \right) + 2} \over {5\left( {{{2 - 3y} \over {5y - 3}}} \right) + 3}}$$

$$ = {{6 - 9y + 10y - 6} \over {10 - 15y + 15y - 9}}$$

$$ = y$$

f is onto

and $${f^{ - 1}}\left( x \right) = {{2 - 3x} \over {5x - 3}}$$

Given that $$* $$ is a binary operation on Q defined by a*b = ab + 1 for all a,b∈Q.

We know that commutative property is p*q = q*p, where * is a binary operation.

Let’s check the commutativity of the given binary operation:

⇒ a*b = ab + 1

⇒ b*a = ba + 1 = ab + 1

⇒ b*a = a*b

∴ The commutative property holds for given binary operation on ‘Q’.

We know that associative property is (p*q)*r = p*(q*r)

Let’s check the associativity of given binary operation:

⇒ (a*b)*c = (ab + 1)*c

⇒ (a*b)*c = ((ab + 1)×c) + 1

⇒ (a*b)*c = abc + c + 1 ...... (1)

⇒ a*(b*c) = a*(bc + 1)

⇒ a*(b*c) = (a×(bc + 1)) + 1

⇒ a*(b*c) = abc + a + 1 ...... (2)

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation on ‘Q’.

$$\therefore \text{the given binary operation is commutative but not associative.}$$

Given $$f\left( x \right) =\sin ^{ 2 }{ x+\sin ^{ 2 }{ \left( x+\dfrac { \pi  }{ 3 }  \right)  }  } +\cos { x\quad \cos { \left( x+\dfrac { \pi  }{ 3 }  \right)  }  } $$

and $$g\left( \dfrac { 5 }{ 4 }  \right) =1$$

$$f\left( x \right) =\dfrac { 1 }{ 2 } \left[ 2\sin ^{ 2 }{ x } +2\sin ^{ 2 }{ \left( x+\dfrac { \pi  }{ 3 }  \right)  } +2\cos { x\quad \cos { \left( x+\dfrac { \pi  }{ 3 }  \right)  }  }  \right] $$

$$=\dfrac { 1 }{ 2 } \left[ 1-\cos { 2x\quad +1- } \cos { \left( 2x+\dfrac { 2\pi  }{ 3 }  \right)  } +\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  } +\cos { \dfrac { \pi  }{ 3 }  }  \right] $$

$$=\dfrac { 1 }{ 2 } \left[ \dfrac { 5 }{ 2 } -\cos { 2x\quad - } \cos { \left( 2x+\dfrac { 2\pi  }{ 3 }  \right)  } +\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  }  \right] $$

$$=\dfrac { 1 }{ 2 } \left[ \dfrac { 5 }{ 2 } -2\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  } \cos { \left( \dfrac { \pi  }{ 3 }  \right)  } +\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  }  \right] $$

$$=\dfrac { 1 }{ 2 } \left[ \dfrac { 5 }{ 2 } -\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  } +\cos { \left( 2x+\dfrac { \pi  }{ 3 }  \right)  }  \right] =\dfrac { 5 }{ 4 }$$

$$f\left( x \right)=\dfrac { 5 }{ 4 } $$ for all $$x\in R$$

Therefore for any $$x\in R$$  we have

$$gof\left( x \right) =g\left( f\left( x \right)  \right) =g\left( \dfrac { 5 }{ 4 }  \right) =1 $$

Thus $$gof \left( x \right) =1\quad for\quad all\quad x\in R$$

Hence $$gof(x)=1$$

The function $$f: {1,3,5}\rightarrow {1,2,5}$$ and $$g:{1,2,5} \rightarrow {1,3}$$ are defined as

$$f={(1,2), (3,5), (4,1)}$$ and $$g= {(1,3), (2,3), (5,1)}$$

$$gof(3)=g(f(3))=g(5)=1 $$   as  $$[f(3)=5$$ and $$g(5)=1]$$

It is seen that $$f(1)=f(2)=1$$, but $$1 \neq 2$$.

It is given that $$A= $${$$1,2,3$$}, $$B=$${$$4,5,6,7$$}.

(i)

To prove:

$$(f+g)oh=foh+goh$$

Consider:

$$=(f+g)(h(x))$$

$$=f(h(x))+g(h(x))$$

$$=(foh)(x)+(goh)(x)$$

$$={(foh)+ (goh)}(x)$$

$$\therefore ((f+g)oh)(x)= {(foh)+ (goh)}(x)$$    for all $$x \in R$$

$$(i)$$ $$f(x)=|x|$$ and $$g(x)=|5x-2|$$

$$\therefore (gof)(x)=g(f(x))=g(|x|)=|5|x|-2|$$

$$\therefore (gof)(x)=g(f(x))=g(8x^3)=(8x^3)^{\displaystyle \frac {1}{3}}=2x$$

Given, $$f(x)=2x+p$$

$$\Rightarrow  y=2x+p$$

$$ \Rightarrow y-p=2x$$

$$\Rightarrow x=\dfrac { y-p }{ 2 } $$

$$ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x-p }{ 2 } $$    .....(i)

Given $$f^{ -1 }\left( x \right) =m(4x+3)=4mx+3m$$    ....(ii)

Comparing (i) and (ii), we get

$$\Rightarrow \dfrac { x }{ 2 } -\dfrac { p }{ 2 } =4mx+3m\\ \Rightarrow 4m=\dfrac { 1 }{ 2 } \\ \Rightarrow m=\dfrac { 1 }{ 8 } $$

Therefore, $$-\dfrac { p }{ 2 } =3m$$

$$\Rightarrow  -\dfrac { p }{ 2 } =\dfrac { 3 }{ 8 } \\ \Rightarrow p=-\dfrac { 3 }{ 4 } $$

• Given $$f(x)={x}^{2}$$ and $$g(x)=4x-2$$
• $$(fof)(x)=f(f(x))=f({x}^{2})={x}^{4}$$

$$Given\  that\  ‘*’$$ $$\text{is an operation that is valid in the Natural Numbers ‘N’ and it is defined as given:}$$

$$a*b=a^b \ where \ a,b\  \in \ N$$

since $$a\in N \ and \ b\in N$$

$$\text{According to the problem it is given that on applying the operation}$$ $$‘*’ $$ $$\text{for two given natural numbers it gives}$$ $$\text{a natural number as a result of the operation,}$$

$$\text{We also know that}$$ $$p^q>0\ \ if\  p>0 \ and\ q>0 $$ 

$$\text{So, we can state that,}$$

$$\text{From (i) and (ii) we can see that both L.H.S and R.H.S gave only Natural numbers as a result.}$$

$$\text{Thus we can clearly state that}$$ $$‘*’$$ $$\text{ is a Binary Operation on ‘N’}$$

$$\text{Given that ‘O' is an operation that is valid in the Integers ‘Z’ and it is defined as given:}$$

$$Given \ 'x_6  'that$$ $$\text{is an operation that is valid for the numbers in the Set S = {1,2,3,4,5} and it is defined as given:}$$

$$a\ x_6b$$ = $$\text{Remainder when ab is divided by 6, where}$$ $$a,b\in S$$

$$since \ a\in S\ and\ b\in S,$$

$$\text{According to the problem it is given that on applying the operation}$$ $$‘*’$$ $$\text{for two given numbers in the set ‘S’ it}$$ $$\text{gives one of the numbers in the set ‘S’ as a result of the operation,}$$

$$\Rightarrow a 'x_6' \in S...............(1)$$

$$\text{Let us take the values of a = 3, b = 4,}$$

$$ab = 3\times4$$

$$ab = 12$$

$$\text{We know that 12 is a multiple of 6. So, on dividing 12 with 6 we get 0 as remainder which is not in the given set ‘S’}$$.

$$\therefore the \ operation\ 'x_6'$$  $$\text{does not define a binary operation on set S.}$$

$$\text{Given that * is a binary operation on the set I of integers}$$.

$$\text{the operation is defined by a*b = 2a + b – 3.}$$

$$\text{We need to find the value of 3*4.}$$

$$\text{Since 3 and 4 belongs to the set of integers we can use the binary operation.}$$

$$3*4 = (2×3) + 4–3$$

$$3*4 = 6 + 1$$

$$3*4 = 7$$

$$\therefore \text{the value of 3*4 is 7.}$$

$$The \ binary \ operations$$ $$'*'$$ $$\text{on a non-empty set A are functions from A × A to A.} $$

$$\therefore\  \ if\  a,b\in A\Rightarrow a*b \in A\ \forall \ a,b\in A.$$

$$\text{It is given that On}\ Z^+\ define\ '*'\ by\ a*b=a$$

$$\text{We can see that for each a, b}\ \in\ Z^+,there\ is\ an \ element\ in\ Z^+.$$ 

$$\Rightarrow\  '*'\text{carries each pair (a, b) to an element a}*b=a\ in \ Z^+$$ 

$$\therefore\ '*'$$ $$\text{is a binary operation on}\ Z^+$$

$$The \ binary \ operations$$ $$'*'$$ $$\text{on a non-empty set A are functions from A × A to A.} $$

$$\therefore\  \ if\  a,b\in A\Rightarrow a*b \in A\ \forall \ a,b\in A.$$

$$\text{It is given On}$$ $$Z^+ define \ by\ '*' \Rightarrow a*b =\begin{vmatrix}a-b\end{vmatrix}$$        

$$\text{We can see that for each a, b}\in Z^+  there\ is\ an\\element   \begin{vmatrix}a-b\end{vmatrix}\ in\ Z^+.$$ 

$$\Rightarrow \ '*'\text{carries each pair (a, b) to an element}\ a*b=\begin{vmatrix}a-b\end{vmatrix} \ in\ Z^+.$$

$$\therefore\ '*'$$ $$\text{is a binary operation on }\ Z^+.$$

$$Given \ that\ '\bigodot'$$  $$\text{is an operation that is valid in the Natural Numbers ‘N’ and it is defined as given:}$$

$$The \ binary \ operations$$ $$'*'$$ $$\text{on a non-empty set A are functions from A × A to A.} $$

$$\therefore\  \ if\  a,b\in A\Rightarrow a*b \in A\ \forall \ a,b\in A.$$

$$\text{It is given that on}$$  $$Z^+ $$ define $$'*' $$ by $$a*b =ab$$        

$$We\  can\  see \ that \ for \ each$$ $$a,b\in Z^+$$ , $$there\  is\  a \ unique \ element\  a\ b \ in\ Z^+$$

$$\Rightarrow '* '\text{carries each pair (a, b) to a unique element} \ a*b=ab\  in \ Z^+$$ 

$$\therefore \ ' *' \ is \ a \ binary\  operation.$$

$$Given \ that$$ $$‘*’$$ $$\text{is an operation that is valid in the Natural Numbers ‘N’ and it is defined as given:}$$

$$\Rightarrow a*b = a + b – 2,$$ $$where\  a,b\in N$$

$$since \ a\in N \ and \ b\in N,$$

$$\text{According to the problem it is given that on applying the operation}$$ $$‘*’ $$ $$\text{for two given natural numbers it gives a natural number as a result of the operation,}$$

$$\Rightarrow a\in N..........(1)$$

$$\text{Let us take the values of a = 1 and b = 1, substituting in the R.H.S side we get}$$,

$$a+ b-2=1+1-2$$

$$\text{From (2), we can see that a + b – 2 doesn’t give only Natural numbers as a result. So, this cannot be stated as a}$$ $$binary \ function.$$

$$\therefore \text{The operation ‘*’ does not define a binary operation on N}$$

$$\text{According to the problem it is given that on applying the operation}$$ $$‘*’$$ $$\text{for two given rational numbers it gives}$$ $$\text{a rational number as a result of the operation,}$$

$$ \text{Let the value of b = – 1 and a = 2}$$

$$\Rightarrow \dfrac{a-1}{b+1}= \dfrac{2-1}{-1+1}$$

$$The \ binary \ operations$$ $$'*'$$ $$\text{on a non-empty set A are functions from A × A to A.} $$

$$\therefore\  \ if\  a,b\in A\Rightarrow a*b \in A\ \forall \ a,b\in A.$$

$$\text{Given set S = {a,b}, we need to find the total number of binary operations.}$$

$$\text{We know that the total number of binary operations on a set ‘S’ with ‘n’ elements is given by}\ n^{n^n}$$ 

$$Here \ n = 2,$$

$$\Rightarrow n^{n^n}= 2^{2^2}$$

$$Given: a*b = LCM(a, b)$$

$$5*7 = LCM(5, 7)$$

$$= 5 × 7$$

$$= 35$$

$$Hence, 5*7 = 35$$

$$\text{Given set S = {a,b,c}, we need to find the total number of binary operations possible for the set ‘S’.}$$

$$\text{We know that the total number of binary operations on a set ‘S’ with ‘n’ elements is given by}\ n^{n^n}$$ 

$$Here \ n = 3,$$

$$\Rightarrow n^{n^n}= 3^{3^3}$$

$$Given: a*b = LCM(a, b)$$

Given that * is a binary operation on Z defined by a*b = a – b for all a,b∈Z.

We know that commutative property is p*q = q*p, where * is a binary operation.

Let’s check the commutativity of given binary operation:

⇒ a*b = a – b

⇒ b*a = b -a

⇒ b*a ≠ a*b

∴ The commutative property doesn’t hold for given binary operation on ‘Z’.

We know that associative property is (p*q)*r = p*(q*r)

Let’s check the associativity of given binary operation:

⇒ (a*b)*c = (a – b)*c

⇒ (a*b)*c = a – b – c ...... (1)

⇒ a*(b*c) = a*(b – c)

⇒ a*(b*c) = a – (b – c)

⇒ a*(b*c) = a – b + c ...... (2)

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation on ‘Z’.

$$\therefore \text{the given binary operation is neither commutative nor associative}$$

Given that * is a binary operation on R defined by a*b = a + b – 7 for all a,b∈R.

We know that commutative property is p*q = q*p, where * is a binary operation.

Let’s check the commutativity of given binary operation:

⇒ a*b = a + b – 7

⇒ b*a = b + a – 7 = a + b – 7

⇒ b*a = a*b

∴ Commutative property holds for given binary operation ‘*’ on ‘R’.

We know that associative property is (p*q)*r = p*(q*r)

Let’s check the associativity of given binary operation:

⇒ (a*b)*c = (a + b – 7)*c

⇒ (a*b)*c = a + b – 7 + c – 7

⇒ (a*b)*c = a + b + c – 14 ...... (1)

⇒ a*(b*c) = a*(b + c – 7)

⇒ a*(b*c) = a + b + c – 7 – 7

⇒ a*(b*c) = a + b + c – 14 ...... (2)

From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘R’.

$$\therefore \text{the given binary operation is commutative and associative both.}$$

Given that $$*$$ is a binary operation on Q defined by a$$*$$b = a $$+$$ ab for all a,b∈Q.

We know that commutative property is p$$*$$q = q$$*$$p, where $$*$$ is a binary operation.

Let’s check the commutativity of given binary operation:

⇒ a$$*$$b = a $$+$$ ab

.

⇒ b$$*$$a = b $$+$$ ba = b $$+$$ ab

⇒ b$$*$$a≠a$$*$$b

∴ Commutative property doesn’t holds for given binary operation ‘$$*$$’ on ‘Q’.

We know that associative property is (p$$*$$q)$$*$$r = p$$*$$(q$$*$$r)

Let’s check the associativity of given binary operation:

⇒ (a$$*$$b)$$*$$c = (a $$+$$ ab)$$*$$c

⇒ (a$$*$$b)$$*$$c = a $$+$$ ab $$+$$ ((a $$+$$ ab)×c)

⇒ (a$$*$$b)$$*$$c = a $$+$$ ab $$+$$ ac $$+$$ abc ...... (1)

⇒ a$$*$$(b$$*$$c) = a$$*$$(b $$+$$ bc)

⇒ a$$*$$(b$$*$$c) = a $$+$$ (a×(b $$+$$ bc))

⇒ a$$*$$(b$$*$$c) = a $$+$$ ab $$+$$ abc ...... (2)

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation ‘$$*$$’ on ‘Q’.

$$\therefore \text{the given binary operation is neither commutative nor associative.}$$

Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except 1 i.e., R – {1} defined as a*b = a + b – ab

Let us assume a + b – ab = 1

⇒ a – ab + b – 1 = 0

⇒ a(1 – b) – 1(1 – b) = 0

⇒ (1 – a)(1 – b) = 0

⇒ a = 1 or b = 1

But according to the problem, it is given that a≠1 and b≠1 so,

a + b + ab≠1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’.

(i) On $${{\bf{Z}}^ + }$$, the binary operation $$ * $$ defined by $$a * b = a - b$$ is not a binary operation because if the points are taken as $$\left( {1,2} \right)$$, then by applying binary operation, it becomes $$1 - 2 =  - 1$$ and $$ - 1$$ does not belong to $${{\bf{Z}}^ + }$$.

(ii) On $${{\bf{Z}}^ + }$$, the binary operation $$ * $$ defined by $$a * b = ab$$ is a binary operation because each element in $${{\bf{Z}}^ + }$$ has a unique element in $${{\bf{Z}}^ + }$$.

(iii) On $${\bf{R}}$$, the binary operation $$ * $$ defined by $$a * b = a{b^2}$$ is a binary operation because each element in $${\bf{R}}$$ has a unique element in$${\bf{R}}$$.

(iv) On $${{\bf{Z}}^ + }$$, the binary operation $$ * $$ defined by $$a * b = \left| {a - b} \right|$$ is a binary operation because each element in $${{\bf{Z}}^ + }$$ has a unique element in $${{\bf{Z}}^ + }$$.

The function f is one-one.

Reason: Let $$x_1, x_2\epsilon A$$ be such that $$f(x_1)=f(x_2)$$

$$\dfrac{4x_1+3}{6x_1-4}=\dfrac{4x_2+3}{6x_2-4}$$

$$24x_1x_2-16x_1+18x_2-12=24x_1x_2+18x_1-16x_2-12$$

$$\Rightarrow 34x_2=34x_1 \Rightarrow x_2=x_1\Rightarrow$$ f is one-one.

The function f is onto:

We shall find the range of the function f.

For the range of f, let y = f(x)

$$y=\dfrac{4x+3}{6x-4}\Rightarrow 6xy-4y=4x+3$$

$$\Rightarrow (6y-4)x=4y+3\Rightarrow x=\dfrac{4x+3}{6x-4}$$ but $$x\epsilon A$$

$$6y-4\neq 0\Rightarrow y\neq \frac{2}{3}$$

Range of $$f$$ is $$R-\dfrac{2}{3}=A$$

Range of $$f = $$Co-domain of $$f$$

$$f$$ is onto function.

Thus, the function f is one-one and onto, f is invertible.

To find $$f^{-1}$$:

Let $$f(x)=y$$

$$y=\dfrac{4x+3}{6x-4}$$

$$6xy-4y=4x+3$$

$$(6y-4)x=4y+3$$

$$x=\dfrac{4y+3}{6y-4}$$

Now, $$f(x) = y$$ and f is inversible.

$$f^{-1}(y)=x\Rightarrow f^{-1}(y)=\dfrac{4y+3}{6y-4}$$

Thus, the function $$f^{-1}:A\rightarrow A$$ is given by $$f^{-1}(y)=\dfrac{4y+3}{6y-4}$$

i.e., $$f^{-1}=\dfrac{4y+3}{6y-4}$$

So, $$f^{-1}=f$$