Question

Let $f\left(x\right)=x^2+px+3$ and $g\left(x\right)=x+q,$ where $p,q\in R.$ If $F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+x^{n}g\left(x\right)}{1+x^n}$ is derivable at $x=1$, then the value of $p^2+q^2$ is

Answer 1

$F\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(x\right)+x^{n}g\left(x\right)}{1+x^n}$

$\Rightarrow F\left(x\right)=\{\begin{array}{cc}f\left(x\right);& 0\le x<1\\ \frac{f\left(1\right)+g\left(1\right)}{2};& x=1\\ g\left(x\right);& x>1\end{array}$

$\Rightarrow F\left(x\right)=\{\begin{array}{cc}{x}^2+px+3;& 0\le x<1\\ \frac{p+q+5}{2};& x=1\\ x+q;& x>1\end{array}$
Differentiating w.r.t $x$,
$\Rightarrow F^{\prime }\left(x\right)=\{\begin{array}{cc}2x+p;& 0<x<1\\ 1;& x>1\end{array}$
As the function is derivable, L.H.D = R.H.D,
$2+p=1\Rightarrow p=-1$

$\because F\left(x\right)$ is derivable, therefore continuous too.
Checking continuity at $x=1$,
$1+p+3=1+q\Rightarrow p+3=q\Rightarrow -1+3=q\Rightarrow q=2$
$\therefore p^2+q^2=5$