Find the number of solutions of 2 cos x = | sin x | , when x ∈ [ 0 , 2 π ] .

We have to find the number of solutions of 2 cos x = | sin x | , when x ∈ [ 0 , 2 π ] . So, we can draw a graph for both the functions for x ∈ [ 0 , 2 π ] and the number of points they cut is the number of solutions . First, we have to generalise y = sin x using the property of a modulus function. y = sin x : x ∈ 0 , π - sin x : x ∈ π , 2 π We know that | sin x | is periodic with period π . sin x is an even function and symmetric about the y -axis. 2 cos x is periodic with period 2 π . At x = 0 , y = 2 1 y = 2 At x = π 2 , y = 2 0 y = 1 At x = π , y = 2 - 1 y = 1 2 Now, if we take common graph, we get: Number of solutions is the intersection point of both the graphs. The total number of intersection points is four. Hence, the number of solutions of 2 cos x = | sin x | is four.

Find the number of solutions for the equation x 2 1 - | x - 2 | = 1 graphically.

We need to find the number of solutions for the equation x 2 1 - | x - 2 | = 1 graphically. So, x 2 1 - | x - 2 | = 1 x 2 1 - | x - 2 | - 1 = 0 x 2 - ( 1 - | x - 2 | ) 1 - | x - 2 | = 0 This is possible when, x 2 - ( 1 - | x - 2 | ) = 0 x 2 = 1 - | x - 2 | We need to draw the graphs for x 2 and 1 - | x - 2 | and the number of points at which they intersect is number of solutions. For the graph of x 2 : y = x 2 is a parabola with vertex at the origin and opening upwards. At x = 2 , y = 4 . At x = 0 , y = 0 x 2 is an even function which is symmetric about the y -axis. Now, y = 1 - | x - 2 | When x ≥ 2 , y = 3 - x At x = 2 , y = 1 At x = 3 , y = 0 When x < 2 , y = x - 1 . At x = 0 , y = - 1 At x = 1 , y = 0 Now, when we draw the graphs of x 2 and 1 - | x - 2 | on a combined graph where the minimum value of x 2 is 0 at x = 0 and the maximum value of 1 - x - 2 is 1 at x = 2 , we get, Since, there is no intersection point, there is no solution.