Given a function f(x)=[x]2+{x}

Let k∈ILtx→K+ f(x) = k2 – 0Ltx→K− f(x) = ( k -1)2 +1It will be continuous at x = k if k2 = ( k -1)2 +1⇒ 2k = 2 ⇒ k = 1∴ continuous only at k = 1 otherwise discontinuous for all integers.

Given a function f(x)=[x]2+{x}

Let k∈ILtx→K+ f(x) = k2 – 0Ltx→K− f(x) = ( k -1)2 +1It will be continuous at x = k if k2 = ( k -1)2 +1⇒ 2k = 2 ⇒ k = 1∴ continuous only at k = 1 otherwise discontinuous for all integers.

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Given a function $f (x) = {[x]}^{2} + \sqrt{{x}}$

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f(x) is continuous at all integers

f(x) is continuous & differentiable at all integers.

f(x) is discontinuous at all integers except x = $\pm 1$

f(x) is discontinuous at all integers except at $x = 1$

answer is D.

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Detailed Solution

Let k $\in$ I
$\underset{x \to K^{+}}{L t}$ f(x) = k² – 0
$\underset{x \to K^{-}}{L t}$ f(x) = ( k -1)² +1
It will be continuous at x = k if k² = ( k -1)² +1 $\Rightarrow$ 2k = 2 $\Rightarrow$ k = 1
$∴$ continuous only at k = 1 otherwise discontinuous for all integers.