Let $f (x) = {[x]}^{2} + \sqrt{\{x\}},$ , where [] & {} respectively denotes the greatest integer and fractional part of functions, then

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f(x) is continuous at all integral points

f(x) is not differentiable $\forall$ x $\in$ Z

f(x) is discontinuous as x $\in$ Z -{1}

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

answer is C.

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Detailed Solution

$f (x) = {[x]}^{2} + \sqrt{{x}} ∴ f (x) = {[x]}^{2} + \sqrt{x - [x]} \Rightarrow f (1) = 1 + 0 = 1 a n d L . H . L = \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} {[x]}^{2} + \sqrt{x - [x]} = \lim_{x \to 1^{-}} 0 + \sqrt{x - 0} = \sqrt{1} = 1 R . H . L = \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} {[x]}^{2} + \sqrt{x - [x]} = \lim_{x \to 1^{+}} 1 + \sqrt{x - 1} = 1 + 0 = 1 \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) ∴ f is continuous at x = 1$

$f (2) = 4 + 0 = 4 \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} {[x]}^{2} + \sqrt{x - 1} = 1 + \sqrt{1} = 2 \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} {[x]}^{2} + \sqrt{x - [x]} = \lim_{x \to 2^{+}} 4 + \sqrt{x - 2} = 4 ∴ f is discontinuous at x = 2 Similarly f is dicontinuous at all integers (\neq 1) f is dicontinuous as x \in Z - {1} Option C is correct$