Question:
Let ƒ be defined by $f(x) = |x| + |cos({\frac{\pi }{2} - x }), x \, \, \epsilon \, \,(-\frac{\pi }{2},\frac{\pi }{2}).$ Then
Let ƒ be defined by $f(x) = |x| + |cos({\frac{\pi }{2} - x }), x \, \, \epsilon \, \,(-\frac{\pi }{2},\frac{\pi }{2}).$ Then
Updated On: Oct 1, 2024
- f is continuous on $(-\frac{\pi }{2},0) \bigcup(0,\frac{\pi }{2})$
- f is differentiable at x=0.
- f is differentiable everywhere except x=0.
- $ lim_{x⇢0}f(x)=0.$
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The Correct Option is A, C, D
Solution and Explanation
The correct option is (A): f is continuous on $(-\frac{\pi }{2},0) \bigcup(0,\frac{\pi }{2})$ (C): f is differentiable everywhere except x=0. and (D): $ lim_{x⇢0}f(x)=0.$
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