Question

Let$f\left(x\right)=\left\{\begin{array}{c}1+\frac{2x}{a},0\le x<1\\ \frac{x}{a},1\le x\le 2\end{array}\right\}$
If $\underset{x\to 1}{lim}f\left(x\right)$ exists then a=

• A. $1$
• B. $-1$
• C. $2$
• D. None of these

Answer 1

Answer: (B)

$-1$

$f\left(x\right)=\left\{\begin{array}{c}1+\frac{2x}{a},0\le x<1\\ \frac{x}{a},1\le x\le 2\end{array}\right\}$

$\underset{x\to 1}{lim}f\left(x\right)$ exists

$\therefore$ L.H.L(Left hand limit)

$\Rightarrow \underset{x\to -1}{lim}f\left(x\right)=1+\frac{2}{a}$

R.H.L(Right hand limit)

$\Rightarrow \underset{x\to +1}{lim}f\left(x\right)=\frac{1}{a}$

$\therefore L.H.L=R.H.L$

$\therefore 1+\frac{2}{a}=\frac{1}{a}$

$\frac{1}{a}=-1$

$\Rightarrow a=-1$

hence, Option B is correct.