Question

Let ƒ be defined by $f(x) = |x| + |cos({\frac{\pi }{2} - x }), x \, \, \epsilon \, \,(-\frac{\pi }{2},\frac{\pi }{2}).$ Then

• A. f is continuous on $(-\frac{\pi }{2},0) \bigcup(0,\frac{\pi }{2})$
• B. f is differentiable at x=0.
• C. f is differentiable everywhere except x=0.
• D. $ lim_{x⇢0}f(x)=0.$

Answer 1

The Correct Option is A, C, D

To solve the given problem, we need to analyze the function \(f(x) = |x| + |\cos\left(\frac{\pi}{2} - x\right)|\) over the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

1. Understanding the Function:
   • The function is composed of two parts: \(|x|\) and \(|\cos\left(\frac{\pi}{2} - x\right)|\).
   • The term \(|x|\) is the absolute value of \(x\), and is continuous for all \(x\).
   • The term \(\cos\left(\frac{\pi}{2} - x\right)\) simplifies to \(\sin(x)\), thus making the function \(f(x) = |x| + |\sin(x)|\).
2. Continuity:
   • The function \(|x|\) is continuous everywhere, and \(|\sin(x)|\) is continuous everywhere as \(\sin(x)\) is continuous everywhere.
   • Therefore, \(f(x)\) is continuous on the entire interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).
3. Differentiability:
   • The function \(|x|\) is not differentiable at \(x=0\) because it has a cusp there.
   • The function \(f(x)\) is smooth and differentiable elsewhere on \((-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})\).
   • Thus, \(f(x)\) is differentiable everywhere except \(x=0\).
4. Limit at \(x=0\):
   • We need to find \(\lim_{x \to 0} f(x)\).
   • As \(x \to 0\), \(|x| \to 0\) and \lim_{x \to 0} f(x) = 0 + 0 = 0.

Conclusion: The correct answers are:

• f is continuous on \((-\frac{\pi}{2},0) \cup (0,\frac{\pi}{2})\),
• f is differentiable everywhere except \(x=0\),
• \(\lim_{x \to 0} f(x) = 0\).