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piz help A Level Pure maths

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Which question are you troubled by, what have you tried to solve it?
Reply 2
Original post by gdunne42
Which question are you troubled by, what have you tried to solve it?



questions 1 , 2 , 9 and10
Original post by stalleyzim
questions 1 , 2 , 9 and10


Q1 and Q3 - since these cubics have a repeated root, you can denote this root by x=αx=\alpha, and then note that f(α)=0f(\alpha) = 0. Then consider the derivative of these cubics. At x=αx=\alpha you must have f(α)=0f'(\alpha) = 0. This gives another equation which you can use with the first and solve for the value of α\alpha

Q2 - Simply sub x=0 and show that the equality does not hold. Dividing it through by x2x^2 and rearranging it a bit yields 5(x2+x2)16(x+x1)42=05(x^2+x^{-2})-16(x+x^{-1})-42=0. So given the substitution, express this eq. in terms of yy. The only bit of work you really need to do here is work out what x2+x2x^2+x^{-2} is in terms of yy

Q9 - begin by squaring both sides to yield 4x+2(3x+1)(x1)=7x+14x+2\sqrt{(3x+1)(x-1)}=7x+1 then rearrange for the root and square once more to get rid off it. Solve the quadratic at hand. Disregard solutions (if any) for which any of the expressions 3x+13x+1, x1x-1, or 7x+17x+1 are negative.

Q10 - You can sketch both sides of the equation on the same graph and work from there. Otherwise, note that xx2|x| \equiv \sqrt{x^2} so really you're just dealing with the eq. x2=3(1x)2\sqrt{x^2}=3-\sqrt{(1-x)^2} which can be dealt in a similar fashion as Q9. OR yet another alternative - search for solutions in the regions x<0x<0, 0<x<10<x<1 and x>1x>1 individually by noting, for example, that for x<0x<0 we have x=x|x|=-x and 1x=1x|1-x|=1-x, hence the eq. for this region is just x=3(1x)-x=3-(1-x)
(edited 6 years ago)
Reply 4
thank you
Reply 5
Screenshot (9).pngPlease help me especially the Induction trainning

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