The chapter Relations and Functions contain important questions in CBSE 12th Standard Mathematics. It also covered with most important questions in Relations and Functions.
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Questions:

Generic 2

12 x 2 = 24

Which of the following binary operations are commutative?
(i) * On Z defined by a*b = a²+ b²
(ii) * On Q defined by a*b = a²+2b

If the binary operation * on the set of integers Z is defined by a*b = a + 3b² then find the value of 8*3

Let f and g be real function be $f(x)=\sqrt { x+4 } ,x\ge 4$ find the function fg, $\frac { f }{ g } $

Define Transitive Relation. Give one example.

Define Reflexive. Give one example.

State the reason why the Relation R = [(a, b) : $a\le b^{ 2 }$ on the set R of the real numbers is not reflexive

Show that the absolute value function : R$\rightarrow$R given by f(x) = |x| is neither one-one nor onto.

Prove that the greatest integer function f : R $\rightarrow$R, given by f(x) = [x] is neither one-one nor onto. Where [x] denotes the greatest integer less than or equal to x

Let f:$R\rightarrow R$ is defined by f(x) = x². Is f one-one?

10.

If f:$R\to R$ and g:$R\to R$are given by f(x) = sin x and g(x) = 5x²find gof(x).

11.

Let * be a binary operation on N given by a*b=HCF(a,b), $a,b\in N$. Write the value of 22*4.

12.

If the binary operation * on the set of integers Z is defined by a*b=a+3b²then find the value of 2 * 4.

Generic 3

5 x 3 = 15

13.

Determine whether or not each of the definition of '*' given below gives a binary operation. In the event that '*' is not a binary operation, give justification for this:
(i) On $Z^{ + }$, define '*' by a *b=a-b
(ii) On $Z^{ + }$, define '*' by a *b=ab
(iii) On R, define '*' by a * b=$ab^{ 2 }$
(iv) On $Z^{ + }$, define '*' by a *b=$\left| a-b \right| $
(v) On $Z^{ + }$, define '*' by a *b=a

14.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) $f:R\rightarrow R$ defined by f(x) = 3 - 4x
(ii) $f:R\rightarrow R$ defined by $f(x)=1+x^{ 2 }$ .

15.

Show that $*:R\times R\rightarrow R$ given by $a*b\rightarrow a+2b$ is not associative.

16.

Show that subtraction and division are not binary operations on N.

17.

Show that if $f:A\rightarrow B$ and $g:B\rightarrow C$ are onto, then $gof:A\rightarrow C$ is also onto.

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Answers Key:

Generic 2

12 x 2 = 24

(i) At a,b $\in Q$
a*b = a²+ 2b
and b*a = b²+ 2a
$\Rightarrow b*a\neq b*a$
* is not commutative on Q.

Given a * b = a+3b²
(8 * 3) = 8 + 3(3)²
= 8 + 27 = 35
⇒ (8 ∗ 3) = 35

(i) f(g) = f(x)g(x)
$fg=(\sqrt { x+4 } )(\sqrt { x-4 } )=\sqrt { x^{ 2 }-4 } $
(ii) $\frac { f }{ g } =\frac { f(x) }{ g(x) } =\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } $
$=\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \times \frac { \sqrt { x-4 } }{ \sqrt { x-4 } } $
$=\frac { \sqrt { x^{ 2 }-16 } }{ x-4 } $

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) $\in R$ then (a, c) $\in R$ , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c $\in A$ such that (a, b)(b, c) $\in R$ implies (a, c) . $\in R$
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 $\Rightarrow $(3, 1) $\in R$
$\therefore$ A is transitive.

Reflexive Relation : A relation R on a set A is called reflexive relation if aRa for every $a\in A$ ; if (a, a) $\in $ R, for every $a\in A$
Example let
A = [1, 2, 3]
A x A =(1, 1) (1, 2)(1, 3) (2, 1) (2, 2) (2, 3) (3, 1)(3, 2) (3, 3) $\in R$
Since (a, a) $\in R$ for every $a\in A$

$\frac{1}{2}>\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}, \frac{1}{2}\right) \in R$
Hence R is reflexive.

$ f(x)=|x|=\left\{\begin{array}{l} x \text { if } x>0 \\ -x \text { if } x<0 \end{array}\right.\\ f(-1)=|-1|=1, f(1)=|1|=1 \therefore f(-1)=f(1) ,$

$f$ is not one-one
Now, consider $- 1 \in R$ .
but it is known that $f (x) = | x|$ is always non-negative.
Thus, there does not exist any element $x$ in domain $R$ such that $f (x) = | x| = - 1$
$∴ f$ is not onto.
Hence, the modulus function is neither one-one nor onto.

Given) f : R →→ R defined by f(x) = [x]
For one-one: We know by definition that for
a ≤ x i.e. for x₁,x²∈ (a, a+1) ,
x₁ ≠ x₂⇒ f(x₁) = f(x₂) = a
Hence, not one-one.
For onto: For y(non integer) in co-domain there does not exist x ∈ R in domain such that f(x) = y. Hence, not onto.

No, as $f(-2)=(-2)^2=4$ and $f(2)=(2)^2=4$
i.e $x_1 \neq x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)$.
Not one-one

10.

$ \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\ and\ \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} defined by\\ f(x)=\sin x \text { and } g(x)=5 x^2 \\ g \circ f(x)=g[f(x)]=g(\sin x)=5(\sin x)^2=5 \sin ^2 x $

Given f : R→R and g : R→R defined by f (x) = sin x and g(x) = 5x2 gof(x) = g [f(x)] = g (sin x) = 5 (sin x)2 = 5 sin2 xRead more on Sarthaks.com - https://www.sarthaks.com/53682/if-f-r-r-and-g-r-r-are-given-by-f-x-sin-x-and-g-x-5x-2-find-gof-x

11.

22 * 4 = HCF (22,4) = 2

12.

$ a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 $

Generic 3

5 x 3 = 15

13.

(i) When $a,b\in Z^{ + }$ , then a-b may not belong to $Z^{ + }$.
$\left[ For\quad Ex.\quad 5-7=-2\notin Z^{ + } \right] $
Hence "*' is not a binary operation on $Z^{ + }$.
(ii) When $a,b\in Z^{ + }$ , then $a,b\in Z^{ + }$
Hence, "*' is a binary operation on $Z^{ + }$.
(iii) When $a,b\in R$ , then $ab^{ 2 }$ $\in R$ .
Hence, '*' is a binary operation on R.
(iv) When $a,b\in Z^{ + }$, $\left| a-b \right| \in Z^{ + }$.
$\left[ \because \quad \left| a-b \right| \ge 0 \right] $
Hence, '*' is a binary operation on $Z^{ + }$.
(v) When $a,b\in Z^{ + }$ , then $a,\in Z^{ + }$.
Hence, '*' is a binary operation on $Z^{ + }$ .

14.

(i) Let $x_{ 1 },x_{ 2 }y\in R$.
Now  $f(x_{ 1 })=f(x_{ 2 })$
$\Rightarrow $  $3-4_{ x1 }=3-4_{ x2 }$
$\Rightarrow $ $x_{ 1 }=x_{ 2 }\Rightarrow f$ is one-one.
Let $y\in R$. Let $y=f(x_{ 0 })$.
Then $3-4x_{ 0 }=y\Rightarrow x_{ 0 }=\frac { 3-y }{ 4 } $.
Now $y\in R\Rightarrow \frac { 3-y }{ 4 } \in R\Rightarrow x_{ 0 }\in R$
$f(x_{ 0 })=3-4x_{ 0 }=3-4\frac { 3-y }{ 4 } =3-3+y=y$.
$\because $ For each $y\in R$ , there exists $x_{ 0 }\in R$ such that
$f(x_{ 0 })=y$
$\because $  $f$ is onto
Hence, 'f' is ne-one and onto or bijective.
(ii) Here f(1) = 1 + 1 = 2,
f(-1) = 1 + 1 = 2.
Now $1\neq -1$ but f(1) = f(-1)
$\because $ $f$ is onto
Also range of $f$ is $[1,\infty )\neq R$
$\because $  $f$ is onto.
Hence, 'f' os not bijective

15.

Take 3,5 and 8 as real numbers.
Now (3 * 5) * 8 = (3 + 10) * 8 = 13 * 8 = 13 + 16 =2 9
and 3 * (5 * 8) = 3* (5 + 16) = 3 * 21 = 3 + 42 = 45.
Thus (3 * 5) * 8 $\neq $ 3 * (5 * 8) $\left[ \because 29\neq 45 \right] $
Hence, "*' is not associative.

16.

(i) $N\rightarrow N$ is given by:
$(x,y)\rightarrow x-y$ , which is not binary operation.
$\left[ \because Image\ of\ (4,6)\ under'-is\ 4-6=2\notin N \right] $
(ii) $\div :N\rightarrow N;$ is given by:
$(x,y)\rightarrow x\div y$ , which is not a binary operation.
$\left[ \because Image\ of\ (4,6)\ under'-is\ 4\div 6=\frac { 2 }{ 3 } \notin N \right] $

17.

Let $z\in C$ be the given arbitrary element.
Then there exists pre-image y of z under g such that
$g(y)=z.$ $\left[ \because \quad g\quad is\quad onto \right] $
Again for $y\in B$ , there exists pre-image x of y under f such that $f(x)=y$.
$\left[ \because \quad f\quad is\quad onto \right] $
$\because $ $gof(x)=g(f(x)=g(y)=z$
Hence, gof is onto.