Question

If $\int \frac{x(x+1)(2x^2-x+1)}{(x^3+x^2+x-1)^3}dx = \frac{1}{Af^2(x)}+C$ where f(1) = 2 then

• A. A = -2
• B. Range of function f is $(-\infty, \infty)$
• C. A = 2
• D. f is many one function

Answer 1

Answer: (A), (B), (D)

A = -2, Range of function f is $(-\infty, \infty)$, f is many one function

Answer 2

Let the given integral be $I$.

$I = \int \frac{x(x+1)(2x^2-x+1)}{(x^3+x^2+x-1)^3}dx$

We are given that $I = \frac{1}{Af^2(x)}+C$. Differentiating the right-hand side with respect to $x$:

$\frac{d}{dx}\left(\frac{1}{Af^2(x)}\right) = \frac{1}{A} \frac{d}{dx}(f(x))^{-2} = \frac{1}{A} (-2) (f(x))^{-3} f'(x) = \frac{-2f'(x)}{A(f(x))^3}$.

Comparing this with the integrand, we have:

$\frac{x(x+1)(2x^2-x+1)}{(x^3+x^2+x-1)^3} = \frac{-2f'(x)}{A(f(x))^3}$.

Let $D(x) = x^3+x^2+x-1$. We observe that $D(x)$ can be written as $x(x^2+x+1-\frac{1}{x})$.

Let's propose $f(x) = x^2+x+1-\frac{1}{x}$. Then $D(x) = x \cdot f(x)$.

Substituting this into the comparison equation:

$\frac{x(x+1)(2x^2-x+1)}{(x f(x))^3} = \frac{-2f'(x)}{A(f(x))^3}$

$\frac{x(x+1)(2x^2-x+1)}{x^3 (f(x))^3} = \frac{-2f'(x)}{A(f(x))^3}$

$\frac{(x+1)(2x^2-x+1)}{x^2} = \frac{-2f'(x)}{A}$

$\frac{x(x+1)(2x^2-x+1)}{x^3} = \frac{-2f'(x)}{A}$.

$\frac{(x+1)(2x^2-x+1)}{x^2} = \frac{-2f'(x)}{A}$.

$(1+\frac{1}{x})(2-\frac{1}{x}+\frac{1}{x^2}) = \frac{-2f'(x)}{A}$.

The left side is $2-\frac{1}{x}+\frac{1}{x^2}+\frac{2}{x}-\frac{1}{x^2}+\frac{1}{x^3} = 2+\frac{1}{x}+\frac{1}{x^3}$.

Now, let's find $f'(x)$ for $f(x) = x^2+x+1-\frac{1}{x}$:

$f'(x) = 2x+1+\frac{1}{x^2}$.

Let's use the relation $N(x) = \frac{-2}{A} f'(x) \left(\frac{D(x)}{f(x)}\right)^3$.

If $f(x) = x^2+x+1-\frac{1}{x}$, then $D(x) = x(x^2+x+1-\frac{1}{x}) = x f(x)$.

So $\frac{D(x)}{f(x)} = x$.

Therefore, $N(x) = \frac{-2}{A} f'(x) x^3$.

The numerator $N(x) = x(x+1)(2x^2-x+1) = (x^2+x)(2x^2-x+1) = 2x^4-x^3+x^2+2x^3-x^2+x = 2x^4+x^3+x$.

And $f'(x) = \frac{d}{dx}(x^2+x+1-\frac{1}{x}) = 2x+1+\frac{1}{x^2}$.

So, $N(x) = \frac{-2}{A} (2x+1+\frac{1}{x^2}) x^3 = \frac{-2}{A} (2x^4+x^3+x)$.

Comparing $2x^4+x^3+x$ with $\frac{-2}{A} (2x^4+x^3+x)$, we get:

$1 = \frac{-2}{A} \implies A = -2$.

So, option (A) is correct.

Now let's verify the condition $f(1)=2$ and check other options using $f(x) = x^2+x+1-\frac{1}{x}$.

1. $f(1) = 1^2+1+1-\frac{1}{1} = 1+1+1-1=2$. The condition $f(1)=2$ is satisfied.

2. Option (B): Range of function $f$ is $(-\infty, \infty)$.

We analyze $f(x) = x^2+x+1-\frac{1}{x}$.

The derivative is $f'(x) = 2x+1+\frac{1}{x^2}$.

For $x>0$: $2x>0$, $1>0$, $\frac{1}{x^2}>0$. So $f'(x)>0$ for all $x>0$.

Thus $f(x)$ is strictly increasing for $x>0$.

As $x \to 0^+$, $f(x) \to 0+0+1-\infty = -\infty$.

As $x \to \infty$, $f(x) \to \infty+\infty+1-0 = \infty$.

So for $x \in (0, \infty)$, the range of $f(x)$ is $(-\infty, \infty)$.

For $x<0$: $f'(x) = 2x+1+\frac{1}{x^2}$.

Let's find critical points by setting $f'(x)=0$: $2x+1+\frac{1}{x^2}=0 \implies 2x^3+x^2+1=0$.

By inspection, $x=-1$ is a root: $2(-1)^3+(-1)^2+1 = -2+1+1=0$.

So $f'(-1)=0$.

$f(-1) = (-1)^2+(-1)+1-\frac{1}{(-1)} = 1-1+1+1=2$.

To determine if $x=-1$ is a local extremum, we check the sign of $f'(x)$ around $x=-1$.

$f'(x)$ is a polynomial in terms of $x$ and $1/x^2$.

As $x \to -\infty$, $f'(x) \to -\infty$.

As $x \to 0^-$, $f'(x) \to \infty$.

Since $f'(-1)=0$ and $f'(x)$ goes from $-\infty$ to $\infty$, $x=-1$ is a local minimum.

$f(-\infty) = \lim_{x \to -\infty} (x^2+x+1-\frac{1}{x}) = \infty$.

$f(0^-) = \lim_{x \to 0^-} (x^2+x+1-\frac{1}{x}) = -\infty$.

So for $x \in (-\infty, 0)$, $f(x)$ decreases from $\infty$ to a minimum of $f(-1)=2$, and then increases from $2$ to $-\infty$.

The range of $f(x)$ for $x \in (-\infty, 0)$ is $(-\infty, \infty)$.

Combining both cases, the range of $f(x)$ is $(-\infty, \infty)$. So option (B) is correct.

3. Option (D): $f$ is many one function.

Since $f(1)=2$ and $f(-1)=2$, the function takes the same value for different inputs. Therefore, $f(x)$ is a many-one function. So option (D) is correct.

Options (A), (B), and (D) are all correct.