Limits Continuity And Differentiability Question 22
Question: Given $ f(x)=b({{[x]}^{2}}+[x])+1 $ for $ x\ge -1 $ $ =\sin (\pi (x+a)) $ for $ x<-1 $ where [x] denotes the integral part of x, then for what values of a, b, the function is continuous at x = -1?
Options:
A) $ a=2n+(3/2);b\in R;n\in I $
B) $ a=4n+2;b\in R;n\in I $
C) $ a=4n+(3/2);b\in {R^{+1}};n\in I $
D) $ a=4n+1;b\in {R^{+}};n\in I $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(-1)=b(1-1)+1=1;\underset{h\to 0}{\mathop{\lim }}f(-1+h)=1 $
$ \underset{h\to 0}{\mathop{\lim }}f(-1-h)=\underset{h\to 0}{\mathop{\lim }}\sin (\pi (-1-h)+\pi a) $ $ =\sin (-\pi +\pi a)=-sin\pi a $
For continuous $ \sin \pi a=-1=\sin ( 2n\pi +\frac{3\pi }{2} ) $ $ \Rightarrow \pi a=2n\pi +\frac{3\pi }{2}\Rightarrow a=2n+\frac{3}{2} $
Hence, $ a=2n+\frac{3}{2},n\in I $ and $ b\in R $ .
BETA
