Seeing as people seem to have asked for a thread for this for quite a while I thought I might as well make one.
I guess you're all familiar what these threads are for by now; post your solutions to the older STEP papers as there are no solutions available freely on the net - plus it's good revision Feel free to post alternative solutions, and please point out when you see a weakness or mistake in a solution.
If you look in the STEP megathread you will find a link to the papers - if you still have problems accessing them, PM me or someone else who has them.
h(x)=xlogx h′(x)=−x2logx+x21=x21(1−logx) For max/min set h'(x)=0 i.e. 0=x21(1−logx)⇒x=e To find the nature of the stationary point, consider e.g. h'(e-0.2)=positive and h'(e+0.2)=negative hence a (global) maximum. For sketching purposes might be worth noting that x→0+limh(x)=−∞ and
Unparseable latex formula:
\displastyle\lim_{x\to\infty}h(x)=0
. Graph see attached (Mathematica because of lack of scanner).
Solving nm=mn is equivalent of solving lognm=logmn⇔mlogn=nlogm⇔nlogn=mlogm (last step valid as neither m nor n can be 0). m=n is the trivial solution. Consider the line f=c (i.e. a constant). For 0<c<e1 there are two solutions. There's one solution when c=e1 or c≤0. And for c>e1 there are no solutions.
Now if m=n then mn=k (k=1).
nlogn=knlogkn logn=nnk(logn−logk)=k(logn−logk) logn=k−1klogk Thus n=kk−1k and m=kk−11.
(What am I missing here, this looks trivial for a STEP question...)
Let f(x)=4+cos(2x)+2sin(x) f′(x)=−2sin(2x)+2cos(x)=2cos(x)(−2sin(x)+1) Now WLOG g′(x)=0⇔f′(x)=0or(f(x))2=1
f′(x)=0⇔2cos(x)(1−2sin(x))=0 Has solutions cos(x)=0 and sin(x)=1/2 i.e. x=2π,23π,6π,65π
f(x)=1⇔1=4+1−2sin2(x)+2sin(x)⇔0=2(2−u2+u) where u=sin(x) By quadratic formula the solutions are u=21±1+4×2 i.e. u=-1, 2 sin(x)=2 has no real solutions and sin(x)=−1 has the root x=23π (which we have already found).
f(x)=−1⇔−1=4+1−2sin2(x)+2sin(x)⇔0=2(3−u2+u) where u=sin(x) Solutions to this are u=21±1+4×3 and as 21±1+4×3>1 sin(x)=u will have no real solutions.
f′(2π−16π)<0, f′(2π+16π)>0 so a (local)min f(2π)=4−1+2=5⇒g(2π)=5+51 f′(23π−16π)<0, f′(23π+16π)>0 so a (global) min f(23π)=4−1−2=1⇒g(23π)=2 f′(6π+18π)<0, f′(6π−18π)>0 so a (global) max f(6π)=4+22+1=5+22⇒g(6π)=5+22+5+221 f′(65π+18π)<0,f′(65π−18π)>0 so a (local) max f(65π)=4+21+1=211⇒g(65π)=211+112
Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
High, quick question: What programme are you using to write out the Mathematics? I've seen such quite alot here but have no idea to do so myself. :|
Thanks.
LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
Ah, yes. *feels stupid*
Consider the line f=c (i.e. a constant). For 0<c<e1 there are two solutions. There's one solution when c=e1 or c≤0. And for c>e1 there are no solutions.
Now if m=n then mn=k (k=1).
nlogn=knlogkn logn=nnk(logn−logk)=k(logn−logk) logn=k−1klogk Thus n=kk−1k and m=kk−11.
Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.
Noticing the denominator of y, we find that there is an asymptote at x = -1.
And because of the exponential properties of e^-x, y tends to zero as x→+∞
y tends to −∞ as x→−∞ since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.
i) ∫1ex2logxdx Recall that ∫logxdx=x(log(x)−1)+c, now we can use this to integrate by parts. I=∫x2logxdx=x21x(log(x)−1)−∫−2x3x(log(x)−1)dx −I=x1(log(x)−1)+2x1=xlogx+1 Putting in the limits gives −I=e2−1= So I=1−e2
ii) ∫sin(x)1+sin(x)cos(x)dx Let u2=1+sin(x)⇒2udxdu=cos(x)⇔dx=cos(x)2udu So the integral becomes
I've got III/3 out. Will post later. Suprisingly easy, its more like a FP3 question to be fair. Might need a check over because my complex transformations are more than a little rusty.