Zadanie 1e, Paper 3, maj 2024, TZ2, Matematyka AA IB HL

$g^{\prime}=(\cos(x/2)-\sin(x/2))/2$, and $2g^{\prime}-g=-2\sin(x/2)\ne0$. Thus the ratio is $1/4-(1/4)\cot(x/2)$. Its integral is $x/4-(1/2)\ln\sin(x/2)$, since sin(x/2)>0 on the domain. Exponentiating gives $f=Ae^{x/4}[\sin(x/2)]^{-1/2}$. At x=π the sine is 1 and the normalization gives A=1. Therefore $h(x)=1/\sqrt{\sin(x/2)}$.

A1: correct g′. M1: form the prescribed ratio. A1: simplify to 1/4−cot(x/2)/4. M1: attempt an indefinite integral of their nonzero constant multiple of cot(x/2), or the equivalent cosine/sine ratio; the half-angle factor need not yet be correct for this method mark. A1: x/4−(1/2)ln sin(x/2). A1: correctly exponentiate and simplify the logarithmic factor. A1: final normalized h=1/√sin(x/2). Modulus signs may be retained, since sin(x/2)>0 here; arbitrary constants may appear in working but must be fixed in the final answer.