Question

If $f\left(x\right)=\{\begin{array}{c}2x+b(x<\alpha )\\ x+d(x\ge \alpha )\end{array}$is such that
$\underset{x\to \alpha }{lim}f\left(x\right)=L$, then $L=$

• A. $2d-b$
• B. $b-d$
• C. $d+b$
• D. $b-2d$

Answer 1

Answer: (A)

$2d-b$

$\begin{array}{c}lim\\ h\to 0^{-}\end{array}f\left(x\right)=2(\alpha -h)+b=2\alpha +b=L.............\left(1\right)$

$\begin{array}{c}lim\\ h\to 0^{+}\end{array}f\left(x\right)=(\alpha +h)+d=L\alpha =L-d................\left(2\right)$

Substituting value of euation (2)in (1), we get
$2(L-d)+b=L$
$L=2d-b$

Hence, option 'A' is correct.