First slide

Continuity

Question

$If f (x) = \frac{x - e^{x} + \cos 2 x}{x^{2}}, x \neq 0, is contitnuous at x = 0, then$ where [x] and {r} denote the greatest integer and fractional part functions, respectively.

Moderate

A

$f (0) = \frac{5}{2}$
B

$[f (0)] = - 2$
C

$\{f (0)\} = 0$
D

$[f (0)] \{f (0)\} = - 1.5$

Solution

$\begin{array}{l} \lim_{x \to 0} \frac{x - e^{x} + 1 - (1 - \cos 2 x)}{x^{2}} \\ = \lim_{x \to 0} [\frac{x - - e^{x} + 1}{x^{2}} - \frac{(1 - \cos 2 x)}{x^{2}}] \\ = \lim_{x \to 0} [\frac{x + 1 - (1 + x + \frac{x^{2}}{2})}{x^{2}} - \frac{2 \sin^{2} x}{x^{2}}] (Using expansion of e^{x}) \\ = - \frac{1}{2} - 2 \\ = - \frac{5}{2} \\ Hence, for continuity, f (0) = - \frac{5}{2} . \\ Now, [f (0)] = - 3; \{f (0)\} = \{- \frac{5}{2}\} = \frac{1}{2} . \\ Hencee, [f (0)] \{f (0)\} = - \frac{3}{2} = - 1.5 \end{array}$

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