Question

If x is a real number in [0,11], then the value of $f\left(x\right)=\underset{n\to \infty }{lim}\underset{n\to \infty }{lim}\{1+co{s}^{2m}(n!\pi x\left)\right\}$ is given by

• A. 2 or 1 according as x is rational or irrational
• B. 1 or 2 according as x is rational or irrational
• C. 1 for all x
• D. 2 or 1 for all x

Answer 1

The correct option is A 2 or 1 according as x is rational or irrational

$\because 0<co{s}^2(n!\pi x)\le 1If0<co{s}^2(n!\pi x)<1Then,f\left(x\right)=\underset{n\to \infty }{lim}=\underset{n\to \infty }{lim}\{1+(co{s}^2(n!\pi x){)}^m\}=1+0=1Whenxisirrationalandifco{s}^2(n!\pi x)=1\Rightarrow n!\pi x=r\pi \Rightarrow x=\frac{r}{n!}=rationalThen,f(x)=\underset{m\to \infty }{lim}\underset{n\to \infty }{lim}\{1+(co{s}^{2}n!\pi x{)}^m\}=1+1=2Hence,f\left(x\right)=\{\begin{array}{cc}2,& xisrational\\ 1,& xisirrational\end{array}$