Question

Let $f:A \rightarrow Bf(x)=\frac{x+a}{bx^2+cx+2}$, where A represent domain set and B represent range set of function $f(x), a, b, c \in R, f(-1)=0$ and $y=1$ is an asymptote of $y=f(x)$ and $y=g(x)$ is the inverse of $f(x)$. Area bounded between the curves $y=f(x)$ and $y=g(x)$ is

Answer 1

3\sqrt{5}-8\ln\left(\frac{1+\sqrt{5}}{2}\right)

Answer 2

The problem asks for the area bounded between the curves $y=f(x)$ and $y=g(x)$, where $g(x)$ is the inverse of $f(x)$. The area bounded by a function and its inverse is twice the area bounded by the function and the line $y=x$, provided the function is monotonic.

First, let's determine the function $f(x)$. Given $f(x)=\frac{x+a}{bx^2+cx+2}$.

1. Condition 1: $f(-1)=0$ Substituting $x=-1$ into $f(x)$: $f(-1) = \frac{-1+a}{b(-1)^2+c(-1)+2} = 0$ $\frac{a-1}{b-c+2} = 0$ For this fraction to be zero, the numerator must be zero (and the denominator non-zero). So, $a-1=0 \implies a=1$. The function becomes $f(x) = \frac{x+1}{bx^2+cx+2}$.

2. Condition 2: $y=1$ is an asymptote of $y=f(x)$ For a rational function $f(x) = \frac{P(x)}{Q(x)}$, if the degree of the numerator $P(x)$ is less than the degree of the denominator $Q(x)$, the horizontal asymptote is $y=0$. If the degree of $P(x)$ is equal to the degree of $Q(x)$, the horizontal asymptote is $y = \frac{\text{leading coefficient of } P(x)}{\text{leading coefficient of } Q(x)}$. If $b \neq 0$, the degree of the denominator ($2$) is greater than the degree of the numerator ($1$). In this case, the horizontal asymptote would be $y=0$. However, the problem states that $y=1$ is an asymptote. This implies that the degree of the numerator and denominator must be the same. This can only happen if the coefficient of $x^2$ in the denominator is zero, i.e., $b=0$. With $b=0$, the function becomes $f(x) = \frac{x+1}{cx+2}$. Now, the degree of the numerator (1) is equal to the degree of the denominator (1). The horizontal asymptote is $y = \frac{\text{coefficient of } x \text{ in numerator}}{\text{coefficient of } x \text{ in denominator}} = \frac{1}{c}$. Given that the asymptote is $y=1$, we have $\frac{1}{c}=1 \implies c=1$.

So, we have found $a=1$, $b=0$, and $c=1$. The function is $f(x) = \frac{x+1}{x+2}$.

Next, we need to find the intersection points of $y=f(x)$ and $y=x$. The area bounded by $y=f(x)$ and $y=g(x)$ is $2 \int_{x_1}^{x_2} |f(x)-x| dx$, where $x_1$ and $x_2$ are the intersection points. Set $f(x)=x$: $\frac{x+1}{x+2} = x$ $x+1 = x(x+2)$ $x+1 = x^2+2x$ $x^2+x-1 = 0$ Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1+4}}{2}$ $x = \frac{-1 \pm \sqrt{5}}{2}$ Let $x_1 = \frac{-1-\sqrt{5}}{2}$ and $x_2 = \frac{-1+\sqrt{5}}{2}$.

Now, let's check the monotonicity of $f(x)$: $f(x) = \frac{x+1}{x+2} = \frac{(x+2)-1}{x+2} = 1 - \frac{1}{x+2}$. $f'(x) = 0 - (-1)(x+2)^{-2} = \frac{1}{(x+2)^2}$. Since $f'(x) > 0$ for all $x \neq -2$, the function $f(x)$ is strictly increasing on its domain. This confirms that the area calculation method is valid.

To determine the sign of $f(x)-x$ in the interval $(x_1, x_2)$, let's pick a test point, say $x=0$. $f(0) = \frac{0+1}{0+2} = \frac{1}{2}$. At $x=0$, $y=x$ is $0$. Since $f(0) = 1/2 > 0$, and $x=0$ is between $x_1 \approx -1.618$ and $x_2 \approx 0.618$, we have $f(x) > x$ in the interval $(x_1, x_2)$.

The area $A$ is given by: $A = 2 \int_{x_1}^{x_2} (f(x)-x) dx = 2 \int_{x_1}^{x_2} \left(\frac{x+1}{x+2} - x\right) dx$ $A = 2 \int_{x_1}^{x_2} \left(1 - \frac{1}{x+2} - x\right) dx$ $A = 2 \left[ x - \ln|x+2| - \frac{x^2}{2} \right]_{x_1}^{x_2}$ $A = 2 \left[ \left( x_2 - \ln|x_2+2| - \frac{x_2^2}{2} \right) - \left( x_1 - \ln|x_1+2| - \frac{x_1^2}{2} \right) \right]$ $A = 2 \left[ (x_2-x_1) - (\ln|x_2+2| - \ln|x_1+2|) - \frac{1}{2}(x_2^2-x_1^2) \right]$

From $x^2+x-1=0$, we know: Sum of roots: $x_1+x_2 = -1$ Product of roots: $x_1x_2 = -1$ Difference of roots: $x_2-x_1 = \frac{(-1+\sqrt{5}) - (-1-\sqrt{5})}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$. Also, $x^2 = 1-x$. So $x_2^2 = 1-x_2$ and $x_1^2 = 1-x_1$. Thus, $x_2^2-x_1^2 = (1-x_2)-(1-x_1) = x_1-x_2 = -(x_2-x_1) = -\sqrt{5}$.

Substitute these values into the area formula: $A = 2 \left[ \sqrt{5} - \ln\left|\frac{x_2+2}{x_1+2}\right| - \frac{1}{2}(-\sqrt{5}) \right]$ $A = 2 \left[ \sqrt{5} - \ln\left|\frac{x_2+2}{x_1+2}\right| + \frac{\sqrt{5}}{2} \right]$ $A = 2 \left[ \frac{3\sqrt{5}}{2} - \ln\left|\frac{x_2+2}{x_1+2}\right| \right]$ $A = 3\sqrt{5} - 2 \ln\left|\frac{x_2+2}{x_1+2}\right|$

Now, let's calculate the ratio $\frac{x_2+2}{x_1+2}$: $x_2+2 = \frac{-1+\sqrt{5}}{2} + 2 = \frac{-1+\sqrt{5}+4}{2} = \frac{3+\sqrt{5}}{2}$. $x_1+2 = \frac{-1-\sqrt{5}}{2} + 2 = \frac{-1-\sqrt{5}+4}{2} = \frac{3-\sqrt{5}}{2}$. $\frac{x_2+2}{x_1+2} = \frac{(3+\sqrt{5})/2}{(3-\sqrt{5})/2} = \frac{3+\sqrt{5}}{3-\sqrt{5}}$ Rationalize the denominator: $\frac{3+\sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}} = \frac{(3+\sqrt{5})^2}{3^2-(\sqrt{5})^2} = \frac{9+5+6\sqrt{5}}{9-5} = \frac{14+6\sqrt{5}}{4} = \frac{7+3\sqrt{5}}{2}$.

Now, we need to simplify $\ln\left(\frac{7+3\sqrt{5}}{2}\right)$. Recall the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$. Let's compute powers of $\phi$: $\phi^2 = \left(\frac{1+\sqrt{5}}{2}\right)^2 = \frac{1+5+2\sqrt{5}}{4} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$. $\phi^3 = \phi \cdot \phi^2 = \left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{3+\sqrt{5}}{2}\right) = \frac{3+\sqrt{5}+3\sqrt{5}+5}{4} = \frac{8+4\sqrt{5}}{4} = 2+\sqrt{5}$. $\phi^4 = \phi \cdot \phi^3 = \left(\frac{1+\sqrt{5}}{2}\right) (2+\sqrt{5}) = \frac{2+\sqrt{5}+2\sqrt{5}+5}{2} = \frac{7+3\sqrt{5}}{2}$. So, $\frac{7+3\sqrt{5}}{2} = \phi^4$. Therefore, $\ln\left(\frac{7+3\sqrt{5}}{2}\right) = \ln(\phi^4) = 4\ln(\phi)$.

Substitute this back into the area formula: $A = 3\sqrt{5} - 2 \left(4\ln(\phi)\right)$ $A = 3\sqrt{5} - 8\ln(\phi)$ $A = 3\sqrt{5} - 8\ln\left(\frac{1+\sqrt{5}}{2}\right)$.

The final answer is $\boxed{3\sqrt{5}-8\ln\left(\frac{1+\sqrt{5}}{2}\right)}$.

Explanation of the solution:

1. Determine the function f(x):
   • Use $f(-1)=0$ to find $a=1$.
   • Use $y=1$ as an asymptote to deduce $b=0$ and $c=1$.
   • This gives $f(x) = \frac{x+1}{x+2}$.
2. Verify Monotonicity:
   • Calculate $f'(x) = \frac{1}{(x+2)^2}$, which is positive. So $f(x)$ is strictly increasing.
3. Find Intersection Points:
   • Solve $f(x)=x \implies \frac{x+1}{x+2}=x \implies x^2+x-1=0$.
   • Roots are $x_1 = \frac{-1-\sqrt{5}}{2}$ and $x_2 = \frac{-1+\sqrt{5}}{2}$.
4. Set up Area Integral:
   • Area $= 2 \int_{x_1}^{x_2} (f(x)-x) dx$. (Since $f(0)=1/2 > 0$, $f(x)>x$ in the interval).
   • $A = 2 \int_{x_1}^{x_2} \left(1 - \frac{1}{x+2} - x\right) dx$.
5. Evaluate the Integral:
   • $A = 2 \left[ x - \ln|x+2| - \frac{x^2}{2} \right]_{x_1}^{x_2}$.
   • Substitute limits and use properties of roots: $x_2-x_1=\sqrt{5}$, $x_1+x_2=-1$, $x_2^2-x_1^2 = -(x_2-x_1) = -\sqrt{5}$.
   • Simplify to $A = 3\sqrt{5} - 2 \ln\left|\frac{x_2+2}{x_1+2}\right|$.
6. Simplify Logarithmic Term:
   • Calculate $\frac{x_2+2}{x_1+2} = \frac{7+3\sqrt{5}}{2}$.
   • Recognize this as $\phi^4$, where $\phi = \frac{1+\sqrt{5}}{2}$ (golden ratio).
   • So, $\ln\left(\frac{7+3\sqrt{5}}{2}\right) = 4\ln(\phi)$.
7. Final Area:
   • Substitute back: $A = 3\sqrt{5} - 2(4\ln(\phi)) = 3\sqrt{5} - 8\ln\left(\frac{1+\sqrt{5}}{2}\right)$.

Answer:

The area bounded between the curves $y=f(x)$ and $y=g(x)$ is $3\sqrt{5}-8\ln\left(\frac{1+\sqrt{5}}{2}\right)$.