Let α∈R be such that the function $$f(x)=$$$$cos$$$$-11-x2sin-11-x{x}-{x}3$$,x≠$$0$$α,$$x=0$$ is continuous at $$x=0$$, where $${x}=x-$$[x], [x] is the greatest integer less than or equal to x. Then :

Question

Let α∈R be such that the function $$f(x)=$$$$cos$$$$-11-x2sin-11-x{x}-{x}3$$,x≠$$0$$α,$$x=0$$ is continuous at $$x=0$$, where $${x}=x-$$[x], [x]  is the greatest integer less than or equal to x. Then :

Infinity Learn · Accepted Answer

$$RHL=limx$$→$$0+$$$$cos$$$$-11-x2sin-1(1-x)x1-x2$$         $$=$$π$$2limx$$→$$0+$$$$cos$$$$-11-x2x$$         $$=$$π$$2limx$$→$$0+-11-1-x22(-2x)$$ (L$$'Hospital $$Rule)          $$=$$πlimx→$$0+x2x2$$−$$x4=$$πlimx→$$0+12$$−$$x2=$$π$$2LHL=limx$$→$$0+$$$$cos$$$$-11-(1+x)2sin-1(-x)(1+x)-(1+x)3$$         $$=$$π$$2limx$$→$$0+$$$$sin$$$$-1x(1+x)(1+x)2-1$$         $$=$$π$$2limx$$→$$0+$$$$sin$$$$-1xx2+2x$$          $$=$$π$$212=$$π$$4$$  As LHL≠RHL so $$f(x)$$ is not continuous at $$x=0$$. Therefore no such α exists

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Let α∈R be such that the function f(x)=cos-11-x2sin-11-x{x}-{x}3,x≠0α,x=0 is continuous at x=0, where {x}=x-[x], [x] is the greatest integer less than or equal to x. Then :

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