Mathematics

1. Solve the equations,   and   where [.] denotes greatest integer function. Sol:  we know that [ f(x) ] = I (Integer) and [x + I] = [x] + I [sin x + [sin x]] = Integer and [sin x] = Integer Therefore, y = 1/3 [ sin x + [sin x + [sin x]]]    = 1/3 [sin x] + 1/3 [sin x + [sin x]]   = 1/3 [sin x] + 1/3[sin x] + 1/3[sin x]  y = [sin x]  .................(i) And, [y + [y]] = 2cos x [y] + [y] = 2cosx [y] = cos x or, [[sin x]] = cos x ( since from (i) y = [sin x]) or, [sin x] = cos x You can compare the graph of [sin x] and cos x i.e. the below graph, it is clear that in a period of 2pi, [sin x] is never equal to cos x Therefore, [sin x] = cos x has no solution. Therefore, given eqautions have no solution. 2. If [x] = [x/2] +[(x + 1)/2] , where [.] denotes the greatest integer function and n be a positive integer then show, [(n +...

Solved examples on Domain - I ( click here ) 5.  therefore,  or,    ( since   ) or, Therefore, by wavy cury method ,    or  When,  Since |x| is always positive. Therefore, it will never less than or equal to -1. Therefore,   i.e. no solution is possible When,  To know how to solve modulus inequality ( click here ) Clearly from graph  , when  or  Therefore,  5.  Since,  And we cannot find the root of negative number. Therefore, 1 + 2sin x > 0 sin x > -1/2 First we have to find the interval or values of 'x' for which sin x > -1/2 in a period of sin x i.e. in the interval of [0, 2pi] , then we generalise the result. see the graph of sin x, In a period of 2pi or in interval [0, 2pi] , sin x > -1/2 when    ( from the above graph you can see that between   ,the graph is below the line y = -1/2, i...