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Mathematics for Economy Question on Differential Equations

Let ƒ be defined by f(x)=∣x∣+∣cos(π2−x),x  ϵ  (−π2,π2).f(x) = |x| + |cos({\frac{\pi }{2} - x }), x \, \, \epsilon \, \,(-\frac{\pi }{2},\frac{\pi }{2}). Then

A

f is continuous on (−π2,0)⋃(0,π2)(-\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

limx⇢0f(x)=0. lim_{x⇢0}f(x)=0.

Answer

f is continuous on (−π2,0)⋃(0,π2)(-\frac{\pi }{2},0) \bigcup(0,\frac{\pi }{2})

Explanation

Solution

The correct option is (A): f is continuous on (−π2,0)⋃(0,π2)(-\frac{\pi }{2},0) \bigcup(0,\frac{\pi }{2}) (C): f is differentiable everywhere except x=0. and (D): limx⇢0f(x)=0. lim_{x⇢0}f(x)=0.