$f(x)=x^2+\frac{5}{12}$ and $g(x)=\left\{\begin{array}{cc}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{array}\right.$
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. . . . . .
$x^2+\frac{8 x}{3}+\frac{5}{12}-2=0$
$\begin{array}{l}12 x^2+32 x -19=0 \\ 12 x ^2+38 x -6 x -19=0 \\ 2 x (6 x +19)-1(6 x +19)=0 \\ (6 x +19)(2 x -1)=0 \\ x =\frac{1}{2} \\ \alpha=2 A _1+ A _2 \\ \alpha=2\left(\int_0^{1 / 2} x ^2+\frac{5}{12} dx +\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right) \\ \Rightarrow \alpha=2\left[\left(\frac{ x ^3}{3}+\frac{5 x }{12}\right)_0^{1 / 2}+\frac{1}{12}\right] \\ \Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right] \\ \Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12} \\ \Rightarrow 9 \alpha=6\end{array}$
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