Question

Let $f:R\to R$ be defined by $f\left(x\right)=\{\begin{array}{cc}\alpha +\frac{\sin \left[x\right]}{x}& ifx>0\\ 2& ifx=0\\ \beta +\left[\frac{\sin x-x}{x^3}\right]& ifx<0\end{array}$
where $\left[x\right]$ denotes the integral part of y. If f is continuous at $x=0$, then $\beta -\alpha =$

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

Answer: (D)

$2$

$f\left(x\right)=\{\begin{array}{cc}\alpha +\frac{\sin \left[x\right]}{x}& ifx>0\\ 2& ifx=0\\ \beta +\left[\frac{\sin x-x}{x^3}\right]& ifx<0\end{array}$

$LHL=\underset{x\to 0}{lim}\frac{\sin \left[x\right]}{x}=1$

So, $\alpha +1=2$

$\Rightarrow \alpha =1$

$\underset{x\to 0}{lim}\left[\frac{\sin x-x}{x^3}\right]=\underset{x\to 0}{lim}\left[\frac{\cos x-1}{3x^2}\right]$

$=\underset{x\to 0}{lim}\left[\frac{-\sin x}{6x}\right]=-1$

So, $\beta -1=2$

$\beta =3$

So, $\beta -\alpha =3-1=2$