Question

Let f(x) =((x-1)^2.e^x)/(1+x^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

The problem asks us to analyze the properties of the function $f(x) = \frac{(x-1)^2 e^x}{1+x^2}$ and determine which of the given statements are correct. To do this, we need to find the first derivative of $f(x)$, $f'(x)$, and analyze 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 $v = 1+x^2$.

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

Next, find $v'$: $v' = \frac{d}{dx}(1+x^2) = 2x$.

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

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

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

2. Analyze the critical points and the sign of $f'(x)$: The sign of $f'(x)$ determines whether $f(x)$ is increasing or decreasing. Since $e^x > 0$ and $(1+x^2)^2 > 0$ for all real $x$, the sign of $f'(x)$ depends solely on the sign of $(x-1)(x^3 - x^2 + 3x + 1)$.

Let $g(x) = x^3 - x^2 + 3x + 1$. To find the roots of $g(x)$, let's analyze its derivative: $g'(x) = 3x^2 - 2x + 3$. The discriminant of this quadratic is $\Delta = (-2)^2 - 4(3)(3) = 4 - 36 = -32$. Since $\Delta < 0$ and the leading coefficient ($3$) is positive, $g'(x)$ is always positive for all $x \in \mathbb{R}$. This means $g(x)$ is a strictly increasing function. As a cubic polynomial, since $g(x)$ is strictly increasing, it has exactly one real root. Let's evaluate $g(x)$ at some integer values: $g(-1) = (-1)^3 - (-1)^2 + 3(-1) + 1 = -1 - 1 - 3 + 1 = -4$. $g(0) = 0^3 - 0^2 + 3(0) + 1 = 1$. Since $g(-1) < 0$ and $g(0) > 0$, the unique real root of $g(x)=0$, let's call it $\alpha$, lies in the interval $(-1, 0)$.

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

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

Interval	$x-1$	$g(x)$ (since $g(\alpha)=0$ and $g(x)$ is increasing)	$(x-1)g(x)$	$f'(x)$	Behavior of $f(x)$
$x < \alpha$	Negative	Negative	Positive	Positive	Strictly Increasing
$\alpha < x < 1$	Negative	Positive	Negative	Negative	Strictly Decreasing
$x > 1$	Positive	Positive (since $g(1)=4>0$ and $g(x)$ is increasing)	Positive	Positive	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 $(-\infty, \alpha)$. Since $\alpha \in (-1, 0)$, the interval $(-\infty, -1)$ is a subset of $(-\infty, \alpha)$. Therefore, $f(x)$ 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 $x=\alpha$, $f'(x)$ changes from positive to negative, indicating a local maximum at $x=\alpha$. 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 maximum at $x=\alpha \in (-1,0)$. The local minimum is at $x=1$. Statement D is Incorrect.

Therefore, statements A and C are correct.