Q61 / 160 · TS EAMCET 2012 · Mathematics
Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{ccc}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
where, $[x]$ denotes the integral part of $x$. If $f$ continuous at $x=0$, then $\beta-\alpha$ is equal to
- A$-1$
- Correct answerB$1$
- C$0$
- D$2$
Solution
Given.
$f(x)=\left\{\begin{array}{lll}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
Since, $f$ is continuous at $x=0$.
$\therefore \mathrm{IHI}=f(0)=$ RHI.
Now, I.HI. $=\lim f(x)$
$\begin{aligned} & =\lim _{x \rightarrow 0}\left[\beta+\left(\frac{\sin x-x}{x^3}\right)\right] \\ & =\lim _{h \rightarrow 0}\left[\beta+\left(\frac{\sin h+h}{-h^3}\right)\right] \\ & =\beta+0\end{aligned}$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}}\left[\alpha+\frac{\sin [x]}{x}\right]$
$\begin{aligned} & =\lim _{h \rightarrow 0}\left[\alpha+\frac{\sin [h]}{h}\right] \\ & =\alpha+1\end{aligned}$
and $f(0)=2$
$\therefore$ From Eq. (i), we get
$\begin{array}{ll}\quad \beta+0=2=\alpha+1 \\ \Rightarrow \quad & \beta-\alpha=1\end{array}$
Hence, the correct answer is option: 'B'.