Question

If $\underset{x\to 1}{lim}(2-x+a[x-1]+b[1+x\left]\right)$ exists, then a and b can take the values (where [.] denotes the greatest integer function)

• A. $a=\frac{1}{3},b=1$
• B. $a=1,b=-1$
• C. $a=9,b=-9$
• D. $a=2,b=\frac{2}{3}$

Answer 1

Answer: (B), (C)

The correct options are
B $a=1,b=-1$
C $a=9,b=-9$

Given $\underset{x\to 1}{lim}(2-x+a[x-1]+b[1+x\left]\right)$ exists

i.e. LHL=RHL

Now $LHL=$ $\underset{x\to 1^{-}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(1-h)$

$=\underset{h\to 0}{lim}2-(1-h)+a[1-h-1]+b[1+1-h]$

$=1+a\left[\right(0-h\left)\right]+b\left[\right(2-h\left)\right]=1+a(-1)+b\left(1\right)=1-a+b$

$RHL=\underset{x\to 1^{+}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(1+h)$

$=\underset{h\to 0}{lim}2-(1+h)+a[1+h-1]+b[1+1+h]$

$=1+a\left[\right(0+h\left)\right]+b\left[\right(2+h\left)\right]$

$=1+2b$

Comparing LHL and RHL , we get $b=-a$

Hence, option B and C satisfies