Question

The function $F\left(x\right),$ defined as $F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+x^{2n}g\left(x\right)}{1+x^{2n}}$ shall be continuous everywhere, if

• A. $f\left(1\right)=g\left(1\right)$
• B. $f(-1)=g(-1)$
• C. $f\left(1\right)=-g\left(1\right)$
• D. $f(-1)=-g(-1)$

Answer 1

Answer: (A), (B)

The correct options are
A $f(-1)=g(-1)$
B $f\left(1\right)=g\left(1\right)$

We have,

$F\left(x\right)=f\left(x\right),\left|x\right|<1$ $[\underset{n\to \infty }{lim}{x}^{2n}=0$ for $\left|x\right|<1]$

$\frac{f\left(x\right)+g\left(x\right)}{2},\left|x\right|=1$

$=\underset{n\to \infty }{lim}\frac{x^{-2n}f\left(x\right)+g\left(x\right)}{x^{-2n}+1}=g\left(x\right),\left|x\right|>1$

$[=\underset{n\to \infty }{lim}{x}^{-2n}=0$ for $\left|x\right|>1]$

Thus, the function $F\left(x\right)$ shall be continuous everywhere if $f\left(x\right)$ and $g\left(x\right)$ are continuous everywhere and if $F\left(x\right)$ is continuous at $x=\pm 1,$ we have

$f\left(1\right)=\frac{f\left(1\right)+g\left(1\right)}{2}=g\left(1\right)\Rightarrow f\left(1\right)=g\left(1\right)$

and, $f(-1)=\frac{f(-1)+g(-1)}{2}=g(-1)\Rightarrow f(-1)=g(-1)$