Consider the functions f, g : R → R defined by $f\left(x\right)=x^2+\frac{5}{12}$ and $g\left(x\right)=\{\begin{array}{ll}2(1-\frac{4\left|x\right|}{3}),& \left|x\right|\le \frac{3}{4}\\ 0,& \left|x\right|>\frac{3}{4}\end{array}$. If  α is the area of the region $\left\{\right(x,y)\in ℝ\times ℝ:|x|\le \frac{3}{4},0\le y\le min\{f\left(x\right),g\left(x\right)\left\}\right\}$ then the value of 9α is _______ .

The correct answer is 6.

We are given two real-valued functions defined as follows:

$f\left(x\right)=x^2+\frac{5}{12}$

and

$g\left(x\right)=\{\begin{array}{ll}2(1-\frac{4\left|x\right|}{3}),& \left|x\right|\le \frac{3}{4}& 0,& \left|x\right|>\frac{3}{4}\end{array}$

We need to find the area α of the region defined by:

$\left\{\right(x,y)\in \mathrm{\&mathbb;R;}\times \mathrm{\&mathbb;R;}:|x|\le \frac{3}{4},0\le y\le min\{f\left(x\right),g\left(x\right)\left\}\right\}$

And finally, calculate the value of 9α.

Step 1: Symmetry of the Functions

Notice that both $f\left(x\right)$ and $g\left(x\right)$ are even functions of $x$ because $f(-x)=f\left(x\right)$ and $g(-x)=g\left(x\right)$. Thus, the region is symmetric with respect to the y-axis.

Therefore, the total area α is twice the area of the region in the first quadrant where $0\le x\le \frac{3}{4}$:

$\alpha =2{\int }_0^{3/4}min\left\{f\right(x),g(x\left)\right\}dx$

Step 2: Finding the Point of Intersection

For $0\le x\le \frac{3}{4}$, we have $g\left(x\right)=2-\frac{8x}{3}$.

To find where $f\left(x\right)=g\left(x\right)$, set the expressions equal to each other:

$x^2+\frac{5}{12}=2-\frac{8x}{3}$

Multiply the entire equation by 12 to eliminate fractions:

$12x^2+5=24-32x$

Rearranging all terms to one side gives the quadratic equation:

$12x^2+32x-19=0$

We can solve this quadratic equation by splitting the middle term:

$12x^2+38x-6x-19=0$

$2x(6x+19)-1(6x+19)=0$

$(2x-1)(6x+19)=0$

Since $x\ge 0$, the only valid solution is:

$x=\frac{1}{2}$

Step 3: Determining min(f(x), g(x)) on Intervals

For $x\in [0,1/2]$: $f\left(0\right)=5/12$ and $g\left(0\right)=2$, so $f\left(x\right)\le g\left(x\right)$. Thus, $min\left\{f\right(x),g(x\left)\right\}=f\left(x\right)$.

For $x\in [1/2,3/4]$: $f(3/4)=9/16+5/12=47/48$ and $g(3/4)=0$, so $g\left(x\right)\le f\left(x\right)$. Thus, $min\left\{f\right(x),g(x\left)\right\}=g\left(x\right)$.

Step 4: Evaluating the Area Integrals

Split the integral for area α into two parts:

$\alpha =2\left[{\int }_0^{1/2}\left(x^2+\frac{5}{12}\right)dx+{\int }_{1/2}^{3/4}\left(2-\frac{8x}{3}\right)dx\right]$

Evaluate the first integral $I_1$:

$I_1={\left[\frac{x^3}{3}+\frac{5x}{12}\right]}_0^{1/2}=\frac{1}{24}+\frac{5}{24}=\frac{6}{24}=\frac{1}{4}$

Evaluate the second integral $I_2$:

$I_2={\left[2x-\frac{4x^2}{3}\right]}_{1/2}^{3/4}$

At upper limit $x=3/4$:

$2\left(\frac{3}{4}\right)-\frac{4}{3}{\left(\frac{3}{4}\right)}^2=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}$

At lower limit $x=1/2$:

$2\left(\frac{1}{2}\right)-\frac{4}{3}{\left(\frac{1}{2}\right)}^2=1-\frac{1}{3}=\frac{2}{3}$

Thus,

$I_2=\frac{3}{4}-\frac{2}{3}=\frac{1}{12}$

Step 5: Calculating α and 9α

Adding the two integrals together:

$\alpha =2(I_1+I_2)=2\left(\frac{1}{4}+\frac{1}{12}\right)=2\left(\frac{4}{12}\right)=2\left(\frac{1}{3}\right)=\frac{2}{3}$

Now, calculate $9\alpha$:

$9\alpha =9\times \frac{2}{3}=6$