Given, F(x)= ⎩⎨⎧x2+3x+2x+2,−1,0,n∈R;n=−2;n=−1 For n∈R we have f(x)=(x+2)(x+1)x+2⇒x+11 F(x)= ⎩⎨⎧x2+3x+2x+2,−1,0,n∈R;n=−2;n=−1∴ Checking continuity at x=−2LHL=x→−2limf(x)=h→0lim2−h+11=h→0limf(x) for x∈R=−2−0+11=−1RHL=x→−2′limf(x)=h→0lim−2+n+11=h→0limf(x) for x∈R=−2+0+11=−1 Hence LHL=RHL=f(−2)=−1 It is continuous at x=−2∴ Checking continuity at x=−1LHL=x→−1−limf(x)=n→0lim−1−h+11=01=∞ and f(−1)=0 Hence It is not continuous at x=−1 from the given option. It is all real number except −1 i.e., R∖{−1}