Which of the following curves represents the function y = ln(|e^(|x| - pi)|) for |x|

Question

Which of the following curves represents the function y = ln(|e^(|x| - pi)|) for |x| <= 2*pi? Here, x represents the abscissa and y represents the ordinate.

A. (curve showing V-shape touching zero at x = +/-pi with smooth symmetric profile)
B. (curve with two peaks and a sharp dip)
C. (curve with |x|-pi shape: V-shaped touching zero at +/-pi, linear rise on both sides)
D. (curve with inverted shape)

Accepted Answer

y = |x| - pi; the correct curve is option C (V-shaped, vertex at y = -pi, crossing zero at x = +/-pi). The function is y = ln(|e^(|x| - pi)|). Since e^t > 0 for all real t, the absolute value inside is redundant: |e^(|x|-pi)| = e^(|x|-pi). Therefore y = ln(e^(|x|-pi)) = |x| - pi. This is a simple V-shaped function that equals 0 when |x| = pi (at x = +/-pi) and rises linearly from there.