Question

Let f(x) =((x-1)^2.e^x)/(1+x^2)^2 , then which of the following statement(s) is (are) correct?

• A. f(x) is strictly increasing in (-∞, -1)
• B. f(x) is strictly decreasing in (1,∞)
• C. f(x) has two points of local extremum
• D. f(x) has a point of local minimum at some x ∈ (-1,0)

Answer 1

Answer: (A), (C)

A, C

Answer 2

To analyze the function $f(x) = \frac{(x-1)^2 e^x}{(1+x^2)^2}$, we need to find its first derivative $f'(x)$ and determine its sign.

1. Find the derivative $f'(x)$:

We use the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.
Let $u = (x-1)^2 e^x$ and $v = (1+x^2)^2$.

First, find $u'$ using the product rule:
$u' = \frac{d}{dx}((x-1)^2 e^x) = 2(x-1)e^x + (x-1)^2 e^x = e^x(x-1)[2 + (x-1)] = e^x(x-1)(x+1)$.

Next, find $v'$ using the chain rule:
$v' = \frac{d}{dx}((1+x^2)^2) = 2(1+x^2) \cdot (2x) = 4x(1+x^2)$.

Now, substitute $u, u', v, v'$ into the quotient rule formula:
$f'(x) = \frac{e^x(x-1)(x+1)(1+x^2)^2 - (x-1)^2 e^x(4x(1+x^2))}{((1+x^2)^2)^2}$
Factor out common terms from the numerator, which are $e^x(x-1)(1+x^2)$:
$f'(x) = \frac{e^x(x-1)(1+x^2) [(x+1)(1+x^2) - 4x(x-1)]}{(1+x^2)^4}$
Simplify the denominator:
$f'(x) = \frac{e^x(x-1) [(x+1)(1+x^2) - 4x(x-1)]}{(1+x^2)^3}$

Simplify the expression inside the square brackets:
$(x+1)(1+x^2) - 4x(x-1) = (x + x^3 + 1 + x^2) - (4x^2 - 4x)$
$= x^3 + x^2 + x + 1 - 4x^2 + 4x$
$= x^3 - 3x^2 + 5x + 1$.

So, the derivative is: $f'(x) = \frac{e^x(x-1)(x^3 - 3x^2 + 5x + 1)}{(1+x^2)^3}$

2. Analyze the critical points and the sign of $f'(x)$:

The sign of $f'(x)$ depends on the sign of $(x-1)(x^3 - 3x^2 + 5x + 1)$, because $e^x > 0$ and $(1+x^2)^3 > 0$ for all real $x$.

Let $h(x) = x^3 - 3x^2 + 5x + 1$.
To find the roots of $h(x)$, analyze its derivative:
$h'(x) = 3x^2 - 6x + 5$.
The discriminant of this quadratic is $\Delta = (-6)^2 - 4(3)(5) = 36 - 60 = -24$.
Since $\Delta < 0$ and the leading coefficient ($3$) is positive, $h'(x)$ is always positive for all $x \in \mathbb{R}$. This means $h(x)$ is a strictly increasing function.
As a cubic polynomial, a strictly increasing function has exactly one real root.
Evaluate $h(x)$ at some integer values:
$h(-1) = (-1)^3 - 3(-1)^2 + 5(-1) + 1 = -1 - 3 - 5 + 1 = -8$.
$h(0) = 0^3 - 3(0)^2 + 5(0) + 1 = 1$.
Since $h(-1) < 0$ and $h(0) > 0$, the unique real root of $h(x)=0$, let's call it $\beta$, lies in the interval $(-1, 0)$.

The critical points for $f'(x)=0$ are $x=1$ (from $x-1=0$) and $x=\beta$ (from $h(x)=0$).

Now, let's analyze the sign of $f'(x)$ in different intervals determined by $\beta$ and $1$:

• For $x < \beta$: $(x-1)$ is negative, and $h(x)$ is negative (since $h(x)$ is increasing and $h(\beta)=0$). Therefore, $(x-1)h(x)$ is positive, so $f'(x) > 0$. $f(x)$ is strictly increasing.
• For $\beta < x < 1$: $(x-1)$ is negative, and $h(x)$ is positive. Therefore, $(x-1)h(x)$ is negative, so $f'(x) < 0$. $f(x)$ is strictly decreasing.
• For $x > 1$: $(x-1)$ is positive. $h(1) = 1^3 - 3(1)^2 + 5(1) + 1 = 1 - 3 + 5 + 1 = 4$. Since $h(1)=4>0$ and $h(x)$ is increasing, $h(x)$ is positive for $x>1$. Therefore, $(x-1)h(x)$ is positive, so $f'(x) > 0$. $f(x)$ is strictly increasing.

3. Evaluate the given statements:

• A. $f(x)$ is strictly increasing in $(-\infty, -1)$.
  From our analysis, $f(x)$ is strictly increasing in $(-\infty, \beta)$. Since $\beta \in (-1, 0)$, the interval $(-\infty, -1)$ is a subset of $(-\infty, \beta)$. Thus, $f(x)$ is strictly increasing in $(-\infty, -1)$.
  Statement A is Correct.

• B. $f(x)$ is strictly decreasing in $(1, \infty)$.
  From our analysis, $f(x)$ is strictly increasing in $(1, \infty)$.
  Statement B is Incorrect.

• C. $f(x)$ has two points of local extremum.
  At $x=\beta$, $f'(x)$ changes from positive to negative, indicating a local maximum at $x=\beta$.
  At $x=1$, $f'(x)$ changes from negative to positive, indicating a local minimum at $x=1$.
  Thus, $f(x)$ has two points of local extremum.
  Statement C is Correct.

• D. $f(x)$ has a point of local minimum at some $x \in (-1,0)$.
  We found a local minimum at $x=1$. The point of local extremum in $(-1,0)$ is a local maximum (at $x=\beta$).
  Statement D is Incorrect.

The correct statements are A and C.

The final answer is $\boxed{A, C}$.