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The examiner is not an alien from space! It is important that you are clear how you got from one line of working to the next however you do not need to do every little part as a separate line of working. ...

First, take the first differential: y' = 4x^2 -12x. At x=3 y'= 0 so therefore the function is at a point of inflection. Taking the second derivative: y'' = 8x -12. At x=3 y''= 12. As 12 is greater than 0,...

To differentiate this, we use the power rule and the chain rule. First we differentiate the outside part, which equals 2((x^2)+1). However, because the inside of the square is a function, we have to diff...

For x!=0, multiply the equation by x to get x^4+5x^2+1=0. Then substitute t=x^2 where t>=0. So the equation has a form t^2+5t+1. Then find the discriminant and two roots. One of the roots t2<0 doesn...

Get a line a form y=-ax/b-c/b, then substitute into a cirle equation (x-p)^2 +(y-s)^2=r^2. Get a quadratic and find whether a discriminant is equal to zero. If it is then the line is tangent to a circle. ...