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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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