Paper 4, Section I, HTopics in AnalysisPart II, 2019Show that π\piπ is irrational. [Hint: consider the functions fn:[0,π]→Rf_{n}:[0, \pi] \rightarrow \mathbb{R}fn:[0,π]→R given by fn(x)=xn(π−x)nsinx.]\left.f_{n}(x)=x^{n}(\pi-x)^{n} \sin x .\right]fn(x)=xn(π−x)nsinx.]