Question

Statement $1:$ Let $f\left(x\right)=\underset{m\to \infty }{lim}\left(\underset{n\to \infty }{lim}{\cos }^{2m}(n!\pi x)\right),$ and $g\left(x\right)=\{\begin{array}{c}0,\text{if}x\text{is rational}\\ 1,\text{if}x\text{is irrrational}\end{array}$. Then $h\left(x\right)=f\left(x\right)+g\left(x\right)$ is continuous for all $x.$
Statement $2:$ $f\left(x\right)$ and $g\left(x\right)$ are discontinuous functions for $x\in R.$

• A. Both the statements are true and Statement $2$ is the correct explanation of Statement $1.$
• B. Both the statements are true and Statement $2$ is not the correct explanation of Statement $1.$
• C. Statement $1$ is true and Statement $2$ is false
• D. Statement $1$ is false and Statement $2$ is true

Answer 1

The correct option is B Both the statements are true and Statement $2$ is not the correct explanation of Statement $1.$

If $x$ is rational $(\frac{p}{q},q\ne 0),$ then $n!x\in Z$ when $n\to \infty$, so
$f\left(x\right)=\underset{m\to \infty }{lim}\left(\underset{n\to \infty }{lim}{\cos }^{2m}(n!\pi x)\right)=1$

If $x$ is irrational, then
${\cos }^2(n!\pi x)\in (0,1)\Rightarrow f\left(x\right)=0f\left(x\right)=\{\begin{array}{c}1,\text{if}x\text{is rational}\\ 0,\text{if}x\text{is irrrational}\end{array}\therefore h\left(x\right)=1\forall x\in R$
So, $h\left(x\right)$ is a continuous function for all $x\in R$

Clearly it can be observed that $f(x$ and $g\left(x\right)$ are discontinuous functions, but statement $2$ is not correct explanation for statement $1$