Question

The function $f\left(x\right)=a[x+1]+b[x-1],(a\ne 0,b\ne 0)$ where $\left[x\right]$ is the greatest integer function is continuous at $x=1$ if

• A. $a=2b$
• B. $a=b$
• C. $a+b=0$
• D. $a+2b=0$

Answer 1

Answer: (C)

$a+b=0$

$f\left(x\right)=a[x+1]+b[x-1]$

for

$f\left(0\right)=a[0+1]+b[0-b]$

$a-b$

$l{t}_{x\to 0^{+}}a[x+1]+b[x-1]$

$a+b$

$l{t}_{x\to 0^{-}}a[x+1]+b[x-1]$

$0+(-2b)$

$l{t}_{x\to 0^{+}}=l{t}_{x\to 0^{-}}$

$\therefore a-b=-2b$

$a+b=0$

$l{t}_{x\to 0^{+}}a[x+1]+b[x-1]$

$2a+b\left(0\right)=2a$

$l{t}_{x\to 0^{-}}a[x+1]+b[x-1]$

$a-b$

$2a=a-b$

$a+b=0$.