Draw the graph of y = x - 1 2 (where · denotes the fractional part function)

We know that y = x - 1 2 is a parabola of vertex is 1 , 0 , domain is all real values and range is [ 0 , ∞ ) . We know that y = f x is a periodic function with period one. So, the graph of y = x - 1 2 is periodic with period one. Now, for the graph of y = x - 1 2 take the mirror image of y = x - 1 2 with respect to x -axis only for the portion that lies above the x -axis.

Draw the graph of y = 2 - 1 x - 1 .

We know that y = 1 x - 1 is the rectangular hyperbola whose asymptotes are x = 1 & y = 0 . For the graph of y = 1 x - 1 take the mirror image of y = 1 x - 1 with respect to x -axis only for the portion that lies below the x -axis. For the graph of y = - 1 x - 1 take the mirror image of y = 1 x - 1 with respect to x -axis. For the graph of y = 2 - 1 x - 1 , shift the graph of y = - 1 x - 1 two unit along positive y -axis. Now, for the graph of y = 2 - 1 x - 1 take the mirror image of y = 2 - 1 x - 1 with respect to x -axis only for the portion that lies below the x -axis.

Draw the graph of x - y ≥ 1

We know that y = x = x , x ≥ 0 - x , x < 0 . For the graph of y = x - 1 shift the graph of y = x one unit along negative y -axis. Now, y ≤ x - 1 is the region lies below the lines y = x - 1 & y = - x - 1 . For the graph of y ≤ x - 1 , take the mirror image of y = x - 1 with respect to x - axis only for the portion that lies above the x -axis. Now, the region of x - y ≥ 1 is show by shaded portion given below figure.

Draw the graph of x + y + x - y ≤ 4 .

We know that x = x , x ≥ 0 - x , x < 0 and x - y = x - y , x ≥ y - x - y , x ≤ y . Thus, x + y + x - y ≤ 4 ⇒ x + y + x - y ≤ 4 , x ≥ y x + y - x + y ≤ 4 , x < y ⇒ x ≤ 2 , x ≥ y y ≤ 2 , x < y ⇒ x ≤ 2 & y ≤ 2 So, the region of x + y + x - y ≤ 4 is the shaded portion when x ∈ - 2 , 2 & y ∈ - 2 , 2 is given below.

Draw the graph of y = cos x , x ∈ - 2 π , 2 π (where · denotes the greatest integer function)

We know that the domain, range and period of the function y = cos x is - ∞ , ∞ , - 1 , 1 & 2 π respectively. We also know that y = cos x is an even function i.e. the graph of y = cos x is symmetric about y -axis. We know that if y = k , then k ≤ y < k + 1 . We also know that y = f x represents only integral values of y . Now, draw the graph of y = cos x , x ∈ - 2 π , 2 π .

Draw the graph of y = 4 - x (where · denotes the greatest integer function)

We know that y = x = x , x ≥ 0 - x , x < 0 . Now, x = 0 , x ∈ - 1 , 1 1 , x ∈ ( - 2 , - 1 ] ∪ [ 1 , 2 ) 2 , x ∈ ( - 3 , - 2 ] ∪ [ 2 , 3 ) 3 , x ∈ ( - 4 , - 3 ] ∪ [ 3 , 4 ) 4 , x ∈ ( - 5 , - 4 ] ∪ [ 4 , 5 ) ⋮ ⋮ For the graph of y = 4 - x apply two steps Step 1 : Take the mirror image of y = x with respect to x -axis, so that it will become y = - x Step 2 : shift the graph of y = - x four units along positive y -axis. For the graph of y = 4 - x take mirror image of y = 4 - x with respect to x -axis only for the portion that lies above the x -axis. Now, apply the graphical transformation of y = f x from y = f x for the graph of y = 4 - x .

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XI>Chapter 2 - Relations and Functions>EXERCISE

Embibe Experts Solutions for Exercise 8: EXERCISE

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 8: EXERCISE

Attempt the practice questions from Exercise 8: EXERCISE with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.