Does $(a,b,1)$ lie on a line or a plane?

Question

The following question is taken from Cengage book by G.Tiwani. This book is used for the preparation of IIT-JEE exam.

Question:

A function $f:\mathbb R \to \mathbb R$ is defined as $$f(x)=\lim_{n\to\infty}\frac{ax^2+bx+c+e^{nx}}{1+ce^{nx}},$$ where $f$ is continuous on $\mathbb R$ then

(A) points $(a,b,c)$ lie on a line in the 3-dimensional coordinate system.

(B) points $(a,b)$ represent the 2-dimensional Cartesian plane

(C) locus of points $(a,c)$ and $(c,b)$ intersect at one point.

(D) points $(a,b,c)$ lie on a plane in the 3-dimensional coordinate system.

My Attempt:

$$f(x)=\begin{cases} ax^2+bx+c;& x\lt0 \\ \frac{c+1}{1+c};&x=0 \\ \frac1c;&x\gt0\end{cases}$$

Therefore, $c=1, a, b \in \mathbb R$

So, I think the options B,D are correct.

But the answer given is A,B,C.

How to approach this?

Answer 1

As noticed in the comments, starting from your attempt we have 2 cases

• $c\neq -1$

$$f(x)=\begin{cases} ax^2+bx+c;& x\lt0 \\ \frac{c+1}{1+c};&x=0 \\ \frac1c;&x\gt0\end{cases} \implies c=1,\ f(x)=\begin{cases} ax^2+bx+1;& x\lt0 \\ 1;&x\ge0\end{cases}$$

• $c= -1$

$$ f(x)=\begin{cases} ax^2+bx-1;& x\lt0 \\ -1;&x\ge0\end{cases}$$

So it seems that $(B)$ is always true and $(C)$, $(D)$ are also true but only separately for each case with $c=1$ and $c=-1$.