Indefinite IntegrationHard
Question
Let f′(x) = 3x2 sin 
- x cos 
, if x ≠ 0; f(0) = 0 and f(1/π) = 0 then :
- x cos , if x ≠ 0; f(0) = 0 and f(1/π) = 0 then : Options
A.f(x) is continuous at x = 0
B.f(x) is non derivable at x = 0
C.f′(x) is continuous at x = 0
D.f′(x) is non derivable at x = 0
Solution
f′(x) = 3x2 sin 
- x cos
Now
f′(x) = 0 - 0 = 0
∴ at x = 0, f(x) is derivable so continuous also, and f′(x) is continuous at x = 0
now f″(x) = 6x sin
+ 3x2 cos 
so f″(x) = 6x sin
at x = 0, cos
is not define so at x = 0, f″(x) is not continuous or f′(x) in not differentiable x = 0.
- x cosNow
f′(x) = 0 - 0 = 0∴ at x = 0, f(x) is derivable so continuous also, and f′(x) is continuous at x = 0
now f″(x) = 6x sin
+ 3x2 cos so f″(x) = 6x sin
at x = 0, cos
is not define so at x = 0, f″(x) is not continuous or f′(x) in not differentiable x = 0. Create a free account to view solution
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