Question

Let \( f(x) = |x| \) and \( g(x) = |x| + a \), \( (a>0) \). For \( 0 \le x \le b \), \( \{(x, y) / g(x) \le y \le f(x)\} \) represents all the points in the interior of

• A. a parallelogram
• B. a triangle
• C. a square
• D. a circle

Hint:

To determine the shape of a region defined by inequalities, it's helpful to find the vertices of the region by considering the intersection of the boundary lines. For \( 0 \le x \le b \), \( |x| = x \). The inequalities \( x \le y \le x + a \) along with \( 0 \le x \le b \) define a region bounded by the lines \( y = x \), \( y = x + a \), \( x = 0 \), and \( x = b \). Identifying the vertices helps in recognizing the geometric shape. In this case, the vertices form a parallelogram.

Answer 1

The Correct Option is A

We are given two functions \( f(x) = |x| \) and \( g(x) = |x| + a \), where \( a>0 \).
We are considering the region defined by \( g(x) \le y \le f(x) \) for \( 0 \le x \le b \).
Since \( 0 \le x \le b \), we have \( |x| = x \).
Thus, the functions become \( f(x) = x \) and \( g(x) = x + a \).
The region is defined by \( x + a \le y \le x \) for \( 0 \le x \le b \).
Let's analyze the boundaries of this region: The lower boundary is \( y = x + a \).
The upper boundary is \( y = x \).
The left boundary is \( x = 0 \).
The right boundary is \( x = b \).
Consider the vertices of this region.
At \( x = 0 \), the lower bound for \( y \) is \( y = 0 + a = a \) and the upper bound is \( y = 0 \).
This gives the interval \( [a, 0] \) for \( y \), which is impossible since \( a>0 \).
This indicates that the condition \( g(x) \le y \le f(x) \) cannot be satisfied for any \( y \) when \( a>0 \), because \( x + a>x \).
However, let's re-examine the question.
It seems there might be a misunderstanding in the inequality.
If the region was defined by \( f(x) \le y \le g(x) \), then we would have \( |x| \le y \le |x| + a \).
For \( 0 \le x \le b \), this becomes \( x \le y \le x + a \).
Let's consider the vertices of the region bounded by \( x = 0 \), \( x = b \), \( y = x \), and \( y = x + a \).
At \( x = 0 \): \( 0 \le y \le a \).
The vertices are \( (0, 0) \) and \( (0, a) \).
At \( x = b \): \( b \le y \le b + a \).
The vertices are \( (b, b) \) and \( (b, b + a) \).
The four vertices of the region are \( (0, 0) \), \( (b, b) \), \( (b, b + a) \), and \( (0, a) \).
Let's check if this forms a parallelogram.
The vector from \( (0, 0) \) to \( (b, b) \) is \( \langle b, b \rangle \).
The vector from \( (0, a) \) to \( (b, b + a) \) is \( \langle b - 0, b + a - a \rangle = \langle b, b \rangle \).
These two opposite sides are parallel and equal in length.
The vector from \( (0, 0) \) to \( (0, a) \) is \( \langle 0, a \rangle \).
The vector from \( (b, b) \) to \( (b, b + a) \) is \( \langle b - b, b + a - b \rangle = \langle 0, a \rangle \).
These two opposite sides are also parallel and equal in length.
Therefore, the region represents a parallelogram.