Sketch the graph of the curve y = sin x + π 2 .

We have y = sin x + π 2 . We know the graph of y = sin x + π 2 is obtained by shifting the graph of y = sin x by π 2 units towards left along the x -axis. Hence, the graph of y = sin x + π 2 is given below

Sketch the graph of the curve y = sin x + π 2 .

We have y = sin x + π 2 . We know the graph of y = sin x + π 2 is obtained by shifting the graph of y = sin x by π 2 units towards left along the x -axis. Hence, the graph of y = sin x + π 2 is given below

Sketch the graph of the curve y = - x , ( where · represents the fractional part function )

We have y = - x , ( where · represents the fractional part function ) . We know that, the value of x lies between [ 0 , 1 ) . Hence, the value of - x lies between ( - 1 , 0 ] . Thus, the graph of y = - x is given below .

Sketch the graph of the curve y = x + x - x , ( where · denotes the greatest integer function ) .

Given y = x + x - x . We know that x - x = x , where · denotes the greatest integer function and · denotes the fractional part function. Thus, the function become y = x + x . We know that, the range of the fractional part function i.e. x is [ 0 , 1 ) , hence, the range of x is also [ 0 , 1 ) . And, the range of the greatest integer function i.e. x is all integers. Hence, x = ⋮ ⋮ - 1 , - 1 ≤ x < 0 0 , 0 ≤ x < 1 1 , 1 ≤ x < 2 2 , 2 ≤ x < 3 ⋮ ⋮ . Thus, the value of the function y = x + x is y = ⋮ ⋮ - 1 + x , - 1 ≤ x < 0 0 + x , 0 ≤ x < 1 1 + x , 1 ≤ x < 2 2 + x , 2 ≤ x < 3 ⋮ ⋮ . Hence, the graph of the function y = x + x is given below,

Sketch the graph of the curve y = 2 - x 2 .

We know that the square root of a positive number is always positive and the function under square root is defined for non-negative values. Hence, the value y = 2 - x 2 is always positive and the function is defined for 2 - x 2 ≥ 0 , ⇒ x 2 ≤ 2 ⇒ - 2 ≤ x ≤ 2 . Also, we know that the graph of - x 2 is a parabola open downward and vertex at origin. The graph of y = f x + k , k > 0 is obtained by shifting the graph of y = f x , k units upwards along the y -axis. Hence, the graph of 2 - x 2 is also a parabola open downward and vertex at 0 , 2 . Thus, the graph of y = 2 - x 2 is given below

If A = 1 , 3 , 5 , 9 and B = 2 , 11 , find A × B .

Given that A = 1 , 3 , 5 , 9 and B = 2 , 11 . We know that, the cross product A × B = a , b , a ∈ A , b ∈ B . ∴ A × B = 1 , 2 , 1 , 11 , 3 , 2 , 3 , 11 , 5 , 2 , 5 , 11 , 9 , 2 , 9 , 11 .

If A × B = a , 1 , b , 3 , a , 3 , b , 1 , a , 2 , b , 2 . Find A and B .

We know that, the cross product of two sets is defined as A × B = x , y , x ∈ A , y ∈ B . Also, if n A = p and n B = q then n A × B = p q , where n A denotes the number of elements in the set A . Given that A × B = a , 1 , b , 3 , a , 3 , b , 1 , a , 2 , b , 2 . Hence A = set of first elements in A × B i.e. A = a , b . And B = set of second elements in A × B i.e. B = 1 , 2 , 3 .

Let A and B be two sets such that A × B consists of 6 elements. If three elements of A × B are 1 , 4 , 2 , 6 , 3 , 6 . Find A .

We know that, the cross product A × B = x , y , x ∈ A , y ∈ B . Also, if n A = p and n B = q then n A × B = p q , where n A denotes the number of elements in the set A . Given three elements of A × B are 1 , 4 , 2 , 6 , 3 , 6 . Hence, the first element belongs to the set A and the second element belongs to the set B . So, we can say that A = 1 , 2 , 3 and B = 4 , 6 .