Discussion Forum

1)

 If  $f(x)=\begin{cases}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} ,& -1\leq x <0\\ \frac{x^{2}+2}{x-2}, & 0\leq x \leq 1\end{cases} is$

 continuous  on [-1,1] , than a =

Answer:

Option A

Explanation:

We have ,

$f(x)=\begin{cases}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} ,& -1\leq x <0\\ \frac{x^{2}+2}{x-2}, & 0\leq x \leq 1\end{cases} is$

 Since, f(x)  is continuous on [-1,1]

 $\therefore$  f(x) is continuous  at x=0

 $\therefore$  LHS  (at x=0)=RHL(at x=0)

 $\Rightarrow \lim_{x \rightarrow 0}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

 $\Rightarrow \lim_{x \rightarrow 0}\frac{(1+ax)-{(1-ax)}}{x[\sqrt{1+ax}+\sqrt{1-ax}]}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

$\Rightarrow \lim_{x \rightarrow 0}\frac{2a}{\sqrt{1+ax}+\sqrt{1-ax}}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

$\Rightarrow$      a=-1