Important Questions For Grade 12 Maths Relations and Functions

What is the Relation?

Let us consider two sets A and B. Then a relation R from Set A into Set B is defined as the subset of A × B. The relation can be classified into three different types, such as:

• Reflexive Relation
• Symmetric Relation
• Transitive Relation

What is a Function?

A function is defined as the relation between two sets, in which every element in set 1 is associated with the elements in set 2. In other words, if  “f” is a function from A to B, then every element in set B is the image of an element in set A.  The different types of functions are:

• One to One function
• One to Many functions
• Onto function
• One to one Correspondence

Go through the solved problems given below. Here, all the important questions for class 12 chapter 1 maths problems are given. The problems given here are frequently asked in the previous board examinations. Practice these problems well and solve the practice problems provided here.

Q.1:  Show that the Signum Function f: R → R, given by

$\begin{array}{l}�(�)=\{\begin{array}{cc}1& ����>0\\ 0& ����=0������ℎ�����-����������\\ -1& ����<0\end{array}\end{array}$

Solution:

Check for one to one function:

For example:

f(0) = 0

f(-1) = -1

f(1) = 1

f(2) = 1

f(3) = 1

Since, the different elements say f(1), f(2) and f(3), shows the same image, then the function is not one to one function.

Check for Onto Function:

For the function,f: R →R

$\begin{array}{l}�(�)=\{\begin{array}{cc}1& ����>0\\ 0& ����=0\\ -1& ����<0\end{array}\end{array}$

In this case, the value of f(x) is defined only if x is 1, 0, -1

For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.

Thus, the function “f” is not onto function.

Hence, the given function “f” is neither one-one nor onto.

Q.2: If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).

Solution:

Given function:

f(x) = x2 − 3x + 2.

To find f(f(x))

f(f(x)) = f(x)2 − 3f(x) + 2.

= (x2 – 3x + 2)2 – 3(x2 – 3x + 2) + 2

By using the formula (a-b+c)2 = a2+ b2+ c2-2ab +2ac-2ab, we get

= (x2)2 + (3x)2 + 22– 2x2 (3x) + 2x2(2) – 2x2(3x) – 3(x2 – 3x + 2) + 2

Now, substitute the values

= x4 + 9x2 + 4 – 6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2

= x4 – 6x3 + 9x2 + 4x2 – 3x2 – 12x + 9x – 6 + 2 + 4

Simplify the expression, we get,

f(f(x)) = x4 – 6x3 + 10x2 – 3x

Q.3: Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.

Solution:

Check the binary operation * is commutative :

We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

L.H.S =(a, b) * (c, d)

=(a + c, b + d)

R. H. S = (c, d) * (a, b)

=(a + c, b + d)

Hence, L.H.S = R. H. S

Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

* is commutative (a, b) * (c, d) = (a + c, b + d)

Check the binary operation * is associative :

We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

L.H.S =  (a, b) * ( (c, d) * (x, y) ) =  (a+c+x, b+d+y)

R.H.S =  ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)

Thus, L.H.S = R.H.S

Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

Thus, the binary operation * is associative

Checking for Identity Element:

e is identity of * if (a, b) * e = e * (a, b) = (a, b)

where e = (x, y)

Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)

= (x + a , b + y) = (a, b)

Now, (a + x, b + y) = (a, b)

Now comparing these, we get:

a+x = a

x = a -a = 0

Next compare: b +y = b

y = b-b = 0

Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not natural number. Thus, the identity element does not exist.

Therefore, operation * does not have any identity element.

Q.4: Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.

Solution:

Checking for Inverse:

f(x) = 4x + 3

Let f(x) = y

y = 4x + 3

y – 3 = 4x

4x = y – 3

x = (𝑦 − 3)/4

Let g(y) = (𝑦 − 3)/4

where g: Y → N

Now find gof:

gof= g(f(x))

= g(4x + 3) = [(4𝑥 + 3) − 3]/4

= [4𝑥 + 3 − 3]/4

=4x/4

= x = IN

Now find fog:

fog= f(g(y))

= f [(𝑦 − 3)/4]

=4[(𝑦 − 3)/4] +3

= y – 3 + 3

= y + 0

= y = Iy

Thus, gof = IN and fog = Iy,

Hence, f is invertible

Also, the Inverse of f = g(y) = [𝒚 – 3]/ 4

Q. 5: Let A = R {3} and B = R – {1}. Consider the function f: A →B defined by f (x) = (x- 2)/(x -3). Is f one-one and onto? Justify your answer.

Solution:

Given function:

f (x) = (x- 2)/(x -3)

Checking for one-one function:

f (x1) = (x1– 2)/ (x1– 3)

f (x2) = (x2-2)/ (x2-3)

Putting f (x1) = f (x2)

(x1-2)/(x1-3)= (x2-2 )/(x2 -3)

(x1-2) (x2– 3) = (x1– 3) (x2-2)

x1 (x2– 3)- 2 (x2 -3) = x1 (x2– 2) – 3 (x2– 2)

x1 x2 -3x1 -2x2 + 6 = x1 x2 – 2x1 -3x2 + 6

-3x1– 2x2 =- 2x1 -3x2

3x2 -2x2 = – 2x1 + 3x1

x1= x2

Hence, if f (x1) = f (x2), then x1 = x2

Thus, the function f is one-one  function.

Checking for onto function:

f (x) = (x-2)/(x-3)

Let f(x) = y such that y B i.e. y ∈ R – {1}

So, y = (x -2)/(x- 3)
y(x -3) = x- 2

xy -3y = x -2

xy – x = 3y-2

x (y -1) = 3y- 2

x = (3y -2) /(y-1)

For y = 1, x is not defined But it is given that. y ∈ R – {1}

Hence, x = (3y- 2)/(y- 1) ∈ R -{3} Hence, f is onto.

                    Practice Problems 

1. Show that the function f: R → R is given by f(x) = x3 is injective.

2. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

3. Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.