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First of all, we need to show that the statement is true for the base case n=1. For this case the expression becomes: 1+p≥1+p, which is clearly true as both sides are equal and hence solve the inequality....

1.Differentiating: f'(x)= 2cosh(x)+3sinh(x) At a stationary point, we know f'(x)=0. Therefore 2cosh(x)+3sinh(x)=0. (easy to forget that unlike nromal trig there is no change in sign) Rearranging gives tan...

Say arctan of x is equal to a value y. Now take the tangent of both sides; x now equals tan of y! Easy from here, differentiate both sides wrt x. Now 1 equals sec^2y dy/dx, and you can rearrange to find d...

The line of intersection will be parallel to both planes. We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. We can then read off the normal vecto...

Consider the case n=1. Then 1(1+1)(2*1+1)/6 = 1 = 1^2 and so the claim is true for n=1. Suppose the claim is true for some positive integer n, so that 1+4+9+...+n^2 = n(n+1)(2n+1)/6. Then by the inductive...