Question

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

Answer 1

The answer is 6.

Answer 2

\(\begin{aligned} &Given,\\& x^2+\frac{5}{12}=\frac{2-8x}{3}\\& x^2+\frac{8x}{3}+\frac{5}{12}-2=0\\&\text{Solve for quadratic equation} \\&12 x^2+32 x-19=0 \\ & 12 x^2+38 x-6 x-19=0 \\ & 2 x(6 x+19)-1(6 x+19)=0 \\ & (6 x+19)(2 x-1)=0 \\ & x=\frac{1}{2} \\ &\text{The area a is calculated using definite integration}\\ & \alpha=2 A_1+A_2 \\ & \alpha=2\left(\int_0^{1 / 2} x^2+\frac{5}{12} d x+\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right) \\ & \Rightarrow \alpha=2\left[\left(\frac{x^3}{3}+\frac{5 x}{12}\right)_0^{1 / 2}+\frac{1}{12}\right] \\ & \Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right] \\ & \Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12} \\ & \Rightarrow 9 \alpha=6 \end{aligned}\)

S, the answer is 6.