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Q1. If f(x) = | x | and g(x) = | 3x – 4 |, find fog(x) and gof(x).

Solution

fog(x) = f(g(x)) = f(| 3x – 4 |) = | | 3x – 4 | | = | 3x – 4 | and gof(x) = g(f(x)) = g(|x|) = | 3| x | – 4 |.
Q2. For the binary operation * defined by a * b = a + b + 1. Find the identity element for operation *.

Solution

Let e be the identity element for operation * and we know that a * e = a = e * a.   or a + e + 1 = a   and  e + a +1 = a or e = – 1 and e = – 1. Thus, – 1 is the identity element for operation *.
Q3. If P = {5, 6} and Q = {1, 6, 7}, then find P x Q and Q x P.

Solution

P x Q = {(5, 1), (5, 6), (5, 7), (6, 1), (6, 6), (6, 7)} and Q x P = {(1, 5), (1, 6), (6, 5), (6, 6), (7, 5), (7, 6)}
Q4. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min (a, b). Write the multiplication table for binary operation *. Also find (2 * 3) * (4 * 5).

Solution

 (a) We have a * b = min (a, b) on the set {1, 2, 3, 4, 5}. * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5   (b) (2 * 3) * (4 * 5) = min (2, 3) * min (4, 5)                               = 2 * 4 = min (2, 4) = 2.
Q5. A binary operation is defined by a * b = ab/2. Is the binary operation commutative?

Solution

Yes. Here, a * b = ab/2, and b * a = ba/2 = ab/2 = a * b. Thus, binary operation * is commutative.
Q6. Let S be the set of rational numbers, and aRb, if and only if a=b. Show that R is an equivalence relation.

Solution

R e f l e x i v i t y colon
L e t space x element of S
A s space x equals x comma space x R x
H e n c e space R space i s space r e f l e x i v e.
S y m m e t r y colon
L e t space x element of A comma space y element of A space a n d space x R y
rightwards double arrow x equals y
rightwards double arrow y equals x
rightwards double arrow y R x
T h u s comma space x R y rightwards double arrow y R x
H e n c e space R space i s space s y m m e t r i c
T r a n s i t i v i t y colon
L e t space x element of A comma space y element of A comma space z element of A space a n d space x R y comma space y R z
rightwards double arrow x equals y space a n d space y equals z
rightwards double arrow x equals y equals z
rightwards double arrow x equals z
rightwards double arrow x R z
T h u s comma space x R y comma space y R z rightwards double arrow x R z
H e n c e space R space i s space t r a n s i t i v e
Q7. Consider the binary operation * on the set {1,2,3,4,5} defined by a * b = min {a,b}. Write the operation table of the operation *.

Solution

The binary operation * on the set {1,2,3,4,5} is defined by a * b = min {a,b} The operation table for the given operation * on the given set is as follows:   * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5  
Q8. Define bijective function.

Solution

Bijective Function: A one-one and onto function is also called a bijective function or bijection.
Q9. A binary operation is defined by x * y = 2x – 3y. Find *(3, 2).

Solution

Here, x * y = 2x – 3y, then *(3, 2) = 3 * 2 = 2(3) – 3(2) = 6 – 6 = 0.
Q10. Let A = {1, 2, 3} and B = {- 5, - 6, - 1, 0, -3} and let f(x) = {(1, - 3), (2, - 6), (3, 0)} be a function from A to B. Show that f is a one-one function.

Solution

Here, f open parentheses 1 close parentheses not equal to f open parentheses 2 close parentheses not equal to f open parentheses 3 close parentheses. And, 1 not equal to 2 not equal to 3 So, function f is a one-one function


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