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If f(x)=[x^{2}]+\sqrt{\left \{ x \right \}^{2}}, where \left [ \cdot \right ] and \left \{ \cdot \right \}denote the greatest integer and fractional part functions respectively, then-

 

Option: 1

f(x) is continuous at all integral points except 0
 


Option: 2

f(x) is continuous and differentiable at x = 0

 


Option: 3

f(x) is discontinuous for all x\: \in I-\left \{ 1 \right \}

 


Option: 4

f(x) is not differentiable for all x\: \in I


Answers (1)

best_answer

 

Continuity at a point -

A function f(x)  is said to be continuous at  x = a in its domain if 

1.  f(a) is defined  : at  x = a.

2. \lim_{x\rightarrow a}\:f(x)\:exists\:means\:limit\:x\rightarrow a

of  f(x) at  x = a exists from left and right.

3. \lim_{x\rightarrow a}\:f(x)=f(a)\:then\:the\:limit\:equals \:the\:value\:at\:x=a

-

 

 

Geometrical interpretation of continuity at a point -

When a graph breaks at a particular point when it approaches from left and right.

\because \lim_{x\rightarrow a^{-}}\:f(x)=\lim_{x\rightarrow a^{+}}\:f(x)

So limit exist but not continuous: but when it is equal to f(a) at x = a then f(x) is continuous.
 

\lim_{x\rightarrow a^{-}}\:f(x)=\lim_{x\rightarrow a^{+}}\:f(x)=f(a)

- wherein

   

 

 

\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1^{+}}\left ( [x^{2}]+\sqrt{\left \{ x \right \}^{2}} \right )\Rightarrow \lim_{x\rightarrow 1^{+}}(1+0)=1

\because \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{-}}(0+1)=1\; \; \; \; \; andf(1)=1\Rightarrow \lim_{x\rightarrow 1}f(x)=f(1)

\therefore continuous at x=1

Similarly we check for another integers

 

 

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Gunjita

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