Investigating Hard Inequalities: Understanding Part (f)

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If (f) is the part you are having trouble with, then presumably you have already proved that 2x_n^2- (2n-1)x- (n+1)= 0 (part (e)). Now you want to find the smallest n such that x_n< n+ 0.05. You could, for example, solve that using the quadrative formula and compare the solutions to n+ 0.05. Have you calculated some values of x_n? What are x_0 x_1, etc.?
 
Thanks for the help. I am still confused as to how the markscheme answers have come about which I attached above.

Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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