Question

If $f\left(x\right)=\frac{x-e^x+cos2x}{x^2},x\ne 0$, is continuous at $x=0$,
where [x] and {x} denotes the greatest integer and fractional part functions, respectively.

Then which of the following is correct?

• A. $f\left(0\right)=5/2$
• B. $\left[f\right(0\left)\right]=-2$
• C. $\left\{f\right(0\left)\right\}=-0.5$
• D. $\left[f\right(0\left)\right]\left\{f\right(0\left)\right\}=-1.5$

Answer 1

The correct option is D $\left[f\right(0\left)\right]\left\{f\right(0\left)\right\}=-1.5$

$\underset{h\to 0}{lim}\frac{x-e^x+1-(1-cos2x)}{x^2}$

$=\underset{h\to 0}{lim}[\frac{x-e^x+1}{x^2}-\frac{(1-cos2x)}{x^2}]$

$=\underset{h\to 0}{lim}[\frac{x+1-(1+x+\frac{x^2}{2})}{x^2}-\frac{2si{n}^{2}x}{x^2}]$ (Using expansion of $e^x$)

$=-\frac{1}{2}-2=-\frac{5}{2}$
Hence, for continuity, $f\left(0\right)=-\frac{5}{2}$

Now, $\left[f\right(0\left)\right]=-3;\left\{f\right(0\left)\right\}=\{-\frac{5}{2}\}=\frac{1}{2}$.
Hence, $\left[f\right(0\left)\right]\left\{f\right(0\left)\right\}=-\frac{3}{2}=-1.5$
Hence, option D is correct.