Let $ f:(-1,1) \rightarrow \mathbb{R}$ be the function defined by $f(x)=x^2e^{1/(1-x^2)}$. Then

Question

A. $f$ is decreasing in $(-1,0)$
B. $f$ is increasing in $(0,1)$
C. $f(x)=1$ has two solutions in $(-1,1)$
D. $f(x)=1$ has no solutions in $(-1,1)$

It's a multiple select question

i find that there is no solution in $(-1,1)$ but the answer says option A,B,C are correct.

Answer 1

Observe that $$f'(x)=\frac{2x^3}{(1-x^2)}e^\frac{1}{1-x^2}+2xe^\frac{1}{1-x^2}=2xe^\frac{1}{1-x^2}\left(\frac{x^2}{1-x^2}+1\right)=\frac{2xe^\frac{1}{1-x^2}}{1-x^2}$$

For $0<x<1$, $f'(x)>0$ and hence in this interval, we have $f(x)$ to be increasing.

Similarly, for $0>x>-1$, $f'(x)<0$ and hence in this interval, we have $f(x)$ to be decreasing.

In the interval $(-1,1)$, $f(x)$ is continuous and to be strictly precise, $f(-1) \to \infty$ and $f(1) \to\infty$ and $f(0)=0$. Taking into matter the facts that $f(x)$ first decreases from $x=-1$ to $x=0$ and then starts increasing to $x=1$. So, in $(-1,0)$, $f(x)=1$ has $1$ solution and in $(0,-1)$, $f(x)=1$ has $1$ solution. This shows that , in that interval, $f(x)=1$ has $2$ solutions.

So, options (A),(B) and (C) are correct.

Answer 2

$$f(x)=x^2e^{1/(1-x^2)}\implies f'(x)=2xe^{1/(1-x^2)}\left(1+\frac{x^2}{(1-x^2)^2}\right).$$ Since $2e^{1/(1-x^2)}\left(1+\frac{x^2}{(1-x^2)^2}\right)>0$ we have that sign$(f'(x))=$sign$(x)$. So, $f'<0$ on $(-1,0)$ which shows it's decreasing and $f'>0$ on $(0,1)$ which shows it's increasing. So, $A$ and $B$ are correct.

Now, $f(0)=0,$ $f$ in increasing on $(0,1)$ and $$\lim_{x\to 1^-} f(x)=+\infty$$ implies that there is one point (and only one) $x_0\in (0,1)$ such that $f(x_0)=1.$ Since $f(-x)=f(x)$ we get that $f(-x_0)=1.$ So, $C$ holds.