Question

Let $f\left(x\right)=x^2+ax+3,g\left(x\right)=x+b$ and $F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+x^{2n}g\left(x\right)}{1+x^{2n}}$. If $F\left(x\right)$ is continuous $\forall x\in R,$ then

• A. $a=5,b=4$
• B. $a=1,b=3$
• C. $a=4,b=1$
• D. $a=1,b=4$

Answer 1

The correct option is D $a=1,b=4$

$F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+x^{2n}g\left(x\right)}{1+x^{2n}}\Rightarrow F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+\left(x^2{)}^{n}g\right(x)}{1+(x^2{)}^n}\Rightarrow F\left(x\right)=\{\begin{array}{c}f\left(x\right),0\le x^2<1\\ \frac{f\left(x\right)+g\left(x\right)}{2},x^2=1\\ g\left(x\right),x^2>1\end{array}\Rightarrow F\left(x\right)=\{\begin{array}{c}g\left(x\right),x<-1\\ \frac{f(-1)+g(-1)}{2},x=-1\\ f\left(x\right),-1<x<1\\ \frac{f\left(1\right)+g\left(1\right)}{2},x=1\\ g\left(x\right),x>1\end{array}$

If $F\left(x\right)$ is continuous $\forall x\in R$, then
$F\left(x\right)$ must be continuous at $x=\pm 1$

For continuity at $x=-1$
$f(-1)=g(-1)=\frac{f(-1)+g(-1)}{2}\Rightarrow 4-a=b-1\Rightarrow a+b=5\cdots \left(1\right)$

For continuity at $x=1$
$f\left(1\right)=g\left(1\right)=\frac{f\left(1\right)+g\left(1\right)}{2}\Rightarrow 1+a+3=1+b\Rightarrow a-b=-3\cdots \left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$, we get
$a=1,b=4$