Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 14

A probability density function must have total integral one. Put $h=\pi /6$. The term $ax$ is an odd function, soThus $\boxed{b=1}$.

With $b=1$, nonnegativity of the probability density function at both $t$ and $-t$ requiresThis is equivalent to $|a|\le {\inf }_{0<t\le h}\cos t/t$. The derivativeshows that this infimum occurs at $t=h$. Therefore the necessary and sufficient restriction isSufficiency follows from the same decreasing-ratio argument at every interior point. The endpoint values of $a$ are allowed: the probability density function merely vanishes at one endpoint.

For $h=\pi /6$, the expected value isbecause $x\cos x$ is an odd function. Comparing with the specified expected value gives $\boxed{a=1}$, which lies in the permitted parameter range.

Here $a=0$, so the probability density function is the even function $\cos x$ on $[-h,h]$, where $h=\pi /6$. Hence $\mathbb{E}[X]=0$ by oddness, and the variance is just $\mathbb{E}[X^2]$. Two applications of integration by parts giveThus

Keep $a$ general initially, since the final part does not explicitly fix it. For $-h\le y\le h$, where $h=\pi /6$, the cumulative distribution function isThe normalization constant in the probability density function of $Y$ must beEquivalently, the integrated cumulative distribution function on a bounded interval identity gives $M=h-\mathbb{E}[X]$. If the assumption $a=0$ from part (c) is retained, the requested special case is $\boxed{M=\pi /6}$. For the value $a=1$ in part (b), it is $M=\pi /6-{\pi }^3/324$.