Mathematics

In starting, numbers were used only for counting which created the system of Natural Number. Natural Numbers came into existence when man first learnt counting which can also be viewed as adding successively the number 1 to unity. If we add two natural numbers we get a natural number. N = {1,2,3,4,5,6...................................   } But Natural number had no solution of the equation x + 5 = 5 So, whole number system was evolved W = {0,1,2,3,4...................................  }  If we add two natural numbers we get a natural number.But it has no solution of the equation x + 8 = 3 So, now it became necessary to enlarge the system to get the solution. Thus to every natural number corresponds a unique negative integer designated -n and called the additive inverse of n, and the number '0' written such that n + (-n) = 0, and n + 0 = n for every natural number n.  Also n is the inverse of (-n). T he negative integers, the number 0 and the natural numbers(i

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 + 1)/2] + [(n + 2)/4] + [(n + 4)/8 + [(n + 8)/16] + ...... = n Sol: Since [x] = [x/2] +[(x + 1)/2]  therefore, [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.e. value of sin x is less than -1/2 between    ) Generalize the result for entire number line, To find the