Answer:
Option A
Explanation:
We have ,
f(x)={√1+ax−√1−axx,−1≤x<0x2+2x−2,0≤x≤1is
Since, f(x) is continuous on [-1,1]
∴ f(x) is continuous at x=0
∴ LHS (at x=0)=RHL(at x=0)
⇒limx→0√1+ax−√1−axx=limx→0x2+2x−2
⇒limx→0(1+ax)−(1−ax)x[√1+ax+√1−ax]=limx→0x2+2x−2
⇒limx→02a√1+ax+√1−ax=limx→0x2+2x−2
⇒ a=-1