STEP 1 maths

Hi, I don't understand how they derived the lines in brackets, right at the end of the solution. Pleas help...

When x is roughly zero, shouldn't g''(x) roughly equal k^2 and not k?

How did they derive g''(x) = k(tanx)^2 when x is roughly pi/2? this makes no sense as tan x and sec x will tend to infinity, and e^(-ktanx) will tend to zero?

(edited 6 months ago)

step 1.jpg234.3KB

Reply 1

mqb2766

Theyre possibly just concentrating on the middle term as the other two terms are > 0, but if so, it does stretch the usual meaning of approximately equals.

(edited 6 months ago)

Reply 2

DFranklin

Original post by mqb2766

Theyre possibly just concentrating on the middle term as the other two terms are > 0, but if so, it does stretch the usual meaning of approximately equals.

I think they mean what you've said; it's more an incorrect explanation that a misuse of

\approx

Reply 3

mqb2766

Original post by aramis8

Just for completeness, you have a quadratic in tan(x), so
k tan^2(x) - 2tan(x) + k
where k>0 and you want to find where its positive. As k>0, its a "u" and the tangent to the curve when it crosses the y-axis is
-2tan(x) + k
so the intercept is positive (k) and has a negative gradient (-2). The two roots, if they exists will therefore be when tan(x) is positive (positive product and positive sum), and the ms derives them, but a bit of thought would give that when k=1, then the quadratic is
(tan(x)-1)^2
so it touches the x-axis at tan(x)=1. For k<1, the roots are as calculated and as noted in the examiners report, a reasonable number of the kids who got this far lost a few marks as they didnt note the two intervals where its positive and should have sketched the quadratic.

I guess thats what they were trying to convey with the two "approximately equals" expressions, though a decent sketch would suffice.