Question

Let $f\left(x\right)=\{\begin{array}{cc}1+\frac{2x}{a},& 0\le x<1\\ ax,& 1\le x<2\end{array}$. If $\underset{x\to 1}{lim}f\left(x\right)$ exists, then the value of $a$ could be

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (B), (C)

The correct options are
B $-1$
C $2$

$R.H.L.=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}a(1+h)=a$
$L.H.L.=\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}\{1+\frac{2}{a}(1-h\left)\right\}=1+\frac{2}{a}$
$\underset{x\to 1}{lim}f\left(x\right)$ exists $\Rightarrow$ R.H.L. $=$ L.H.L. Therefore, $a=1+\frac{2}{a}$ or $a=2,-1$
Hence, options 'B' and 'C' are correct.