If f(x)={3−[cot−1(2x3−3x2)]for x>0{x2}cos(e1/x)for x

R . H . L f ( 0 + ) = L t x → 0 + 3 − [ cot − 1 ( 2 x 3 − 3 x 2 ) ] = 3 − L T x → 0 + [ cot − 1 ( 2 x 3 − 3 x 2 ) ] = 3 − 3 = 0 L . H . L f ( 0 − ) = L t x → 0 − { x 2 } cos ( e 1 x ) = 0

f ( 0 ) = 0 ⇒ continuity of f at x = 0

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If $f (x) = {\begin{matrix} 3 - [\cot^{- 1} (\frac{2 x^{3} - 3}{x^{2}})] & f o r x > 0 \\ {x^{2}} \cos (e^{1 / x}) & f o r x < 0 \end{matrix},$ where ${.}$ and $[.]$ denotes the fractional part and the integral part functions respectively, then which of the following statement does not hold good?

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$f (0^{-}) = 0$

$f (0) = 0 \Rightarrow$ continuity of $f$ at $x = 0$

$f (0^{+}) = 3$

Irremovable discontinuity of $f$ at $x = 0$

answer is B, D.

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Detailed Solution

$R . H . L f (0^{+}) = \underset{x \to 0^{+}}{L t} 3 - [\cot^{- 1} (\frac{2 x^{3} - 3}{x^{2}})] = 3 - \underset{x \to 0^{+}}{L T} [\cot^{- 1} (\frac{2 x^{3} - 3}{x^{2}})] = 3 - 3 = 0 L . H . L f (0^{-}) = \underset{x \to 0^{-}}{L t} {x^{2}} \cos (e^{\frac{1}{x}}) = 0$