Study

Minimum value by Vivek Pandey

  Find the minimum value of    over  Solution: Let , . Here the function  is continuous and differentiable on . Therefore   For critical points , which implies   The first equation gives The second equation gives or squaring we get , for some . Now we get the values…

Function

: Show that the function given by is a one-to-one function Solution: We know that a function is a one-to-one function iff for all Case 1: Let be even integers Case 2: Let be odd integers Case 3: Let be odd and be even then and i.e. is a false…

Trigonometric Inequalities

: Show that for each real. Sol. Consider the equation .     Taking +ve   sign, we get, which is not possible Taking – ve sign  we get, , which is again not possible Hence the equation     has no solution. Now consider the function The function is continuous…