A problem from Paper 3 in 2011

[Dinstruction]

This is a question that has stumped even my teacher.

This was derived from a previous part in the question.

Hence, show

I'm not even sure where to begin! Letting x = 1 and moving the -1 to the other side doesn't seem to lead anywhere.

[Lero]

I remember attempting this question and getting stumped at a certain step.

As you said adding one to both sides and combining the fractions leads to:

e^(x) < 2x + (2 - x) / (2 - x) = (2 + x)/(2 - x).

In order to isolate e, raise both sides of the inequality to the power of 1/x.

This should yield: e < [(2 + x)/(2 - x)]^(1/x).

I couldn't get any further than this at the time but markscheme suggests substituting n = 1/x.

Should be able to get the answer from there.

[Dinstruction]

I got it. If we use the substitution n=1/x we get:

e< ((1/n+2)/(2-1/n))^1/n

And by multiplying both 2's by (n/n) we get

e<((1/n+2n/n)/(2n/n-1/n))^1/n

And then we simplify.

e<((2n+1)/(2n-1))^1/n

The 1/n can be reduces and can be cancelled.

What gets me is how anyone is expected to come up with that substitution, let alone during a one hour exam. I swear, the math exam isn't a test over what you learned, but a test over how well you can magically think up substitutions.