Let f(x) = |x| and g(x) = |x| + a,where a 0 | vedclass

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(A) Given functions are $f(x) = |x|$ and $g(x) = |x| + a$ with $a > 0$.For $0 \leq x \leq b$,the region is defined by $g(x) \leq y \leq f(x)$,which im

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a parallelogram

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(A) Given functions are $f(x) = |x|$ and $g(x) = |x| + a$ with $a > 0$.For $0 \leq x \leq b$,the region is defined by $g(x) \leq y \leq f(x)$,which implies $|x| + a \leq y \leq |x|$.However,since $a > 0$,$|x| + a$ is always greater than $|x|$. Thus,the condition $g(x) \leq y \leq f(x)$ is impossible for $a > 0$ as it implies $|x| + a \leq y \leq |x|$,which is a contradiction.Re-evaluating the problem statement based on the provided image,the region is bounded by $x=0$,$x=b$,$y=|x|$,and $y=|x|+a$. For $x \geq 0$,$f(x) = x$ and $g(x) = x + a$.The region is bounded by $x=0$,$x=b$,$y=x$,and $y=x+a$.These lines are $x=0$,$x=b$,$y-x=0$,and $y-x=a$.The lines $y-x=0$ and $y-x=a$ are parallel,and the lines $x=0$ and $x=b$ are parallel.Therefore,the region is a parallelogram.

Let $f(x) = |x|$ and $g(x) = |x| + a$,where $a > 0$. For $0 \leq x \leq b$,the set $\{(x, y) \mid g(x) \leq y \leq f(x)\}$ represents all the points in the interior of:

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