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for n=1 2^1=2  2>1 hence true for n=1 assume true for n then 2^n >n we need to show 2^n+1 > n+1 since 2^n >n 2^n+1 >2n =n+n >n+1 for n>1 hence by induction since true for n= 1 and if ...

To integrate this function integration by parts is used where u = lnx and v = 1. The process of integration by parts is then carried out with the final answer being xlnx -x +c. 

The formula for int...

An easy way to integrade and differentiate simple equations is by bringing up all the denominators, in this case rewrite 3/x^2 into 3x^-2. Thus it will be easier to manipulate the powers. For Integration ...

i) Using the power rule, f'(x) = 3x² - 6x ii) To find stationary points, set f'(x) to 0: 3x² - 6x = 0. 3x(x - 2) = 0. x = 0  or  x = 2 So the co-ordinates are (0,f(0)) = (0, 1), and ...