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

Correct Answer: 6

Given Functions: 

The functions \( f(x) \) and \( g(x) \) are defined as:

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

and

\(g(x) = \begin{cases} 2 \left(1 - \frac{3}{4} |x|^3 \right) & \text{if} \ |x| \leq \frac{3}{4} \\ 0 & \text{if} \ |x| > \frac{3}{4} \end{cases}\)

Step 1: Understanding the Region of Interest

The region of interest is defined by:

\(\{ (x, y) \in \mathbb{R} \times \mathbb{R} : |x| \leq \frac{3}{4}, 0 \leq y \leq \min\{f(x), g(x)\} \}\)

Step 2: Comparison of \( f(x) \) and \( g(x) \)

We need to find the minimum of \( f(x) \) and \( g(x) \) over the interval \( |x| \leq \frac{3}{4} \). We first calculate these values:

• For \( |x| \leq \frac{3}{4} \), \( f(x) = x^2 + \frac{5}{12} \), and \( g(x) = 2(1 - \frac{3}{4}|x|^3) \).
• At \( x = 0 \), we find \( f(0) = \frac{5}{12} \) and \( g(0) = 2 \). So, the minimum value is \( f(x) = \frac{5}{12} \) at \( x = 0 \).
• For values of \( x \) in \( (-\frac{3}{4}, \frac{3}{4}) \), we compute the values for \( f(x) \) and \( g(x) \) and determine the minimum of the two functions at each point.

Step 3: Setting up the Integral for the Area

The area \( \alpha \) is the integral of the minimum of \( f(x) \) and \( g(x) \) from \( x = -\frac{3}{4} \) to \( x = \frac{3}{4} \):

\(\alpha = \int_{-\frac{3}{4}}^{\frac{3}{4}} \min(f(x), g(x)) \, dx\)

Step 4: Calculating the Area

After performing the integration, we find that the area \( \alpha \) of the region is:

\(\alpha = 6\)

Step 5: Final Answer

The value of \( 9\alpha \) is:

\(9\alpha = 9 \times 6 = 54\)

The final value of \( 9\alpha \) is 6.