Question

Given $f\left(x\right)=\{\begin{array}{cc}b({\left[x\right]}^2+\left[x\right])+1& x\ge -1\\ \sin \left(\pi \right(x+a\left)\right)& x<-1\end{array}$ where $\left[x\right]$ denotes the integral part of $x$, then for what values of $a,b$ the function is continuous at $x=-1$ ?

• A. $a=2n+\frac{3}{2};b\in R;n\in I$
• B. $a=4n+2;b\in R;n\in I$
• C. $a=4n+\frac{3}{2};b\in R^{+};n\in I$
• D. $a=4n+1;b\in R^{+};n\in I$

Answer 1

Answer: (A), (C)

$a=2n+\frac{3}{2};b\in R;n\in I$
$a=4n+\frac{3}{2};b\in R^{+};n\in I$

$\underset{x\to -1^{+}}{lim}f\left(x\right)=\underset{x\to -1^{+}}{lim}b\left(\right[x{]}^2+\left[x\right])+1$
$=\underset{h\to 0}{lim}b\left(\right[-1+h{]}^2+[-1+h])+1$
$=b\left(\right(-1{)}^2-1)+1=1$
$\Rightarrow b\in R$
$\underset{x\to -1^{-}}{lim}f\left(x\right)=\underset{x\to -1^{-}}{lim}\sin \left(\pi \right(x+a\left)\right)$
$=limh\to 0\sin \left(\pi \right(-1-h+a\left)\right)=-\sin \pi a$
$\therefore \sin \pi a=-1$, since f is continuous at $x=-1$
$\Rightarrow \pi a=2n\pi +\frac{3\pi }{2}\Rightarrow a=2n+\frac{3}{2}$
Also option (C) is subset of option (A)