f(−3+h)=6⇒f(−3)=6 ⇒f(x) is continuous at x=−3 At x=3 LHL =
lim
h→0
f(3−h)=6 RHL=
lim
h→0
f(3+h)=
lim
h→0
(3+(3+h))=6 f(3)=6 ⇒f(x) is continuous at x=3 ∴f(x) is continuous ∀x∈R. ∴α=0 Again, f(x)={
3−x
x<−3
6
−3≤x≤3
3+x
x>9
. f′(x)={
−1
x<−3
0
−3<x<3
1
x>3
. ∵f′(−3−)=−1 and f′(−3+)=0 f′(−3−)≠f′(−3+) ⇒f(x) is not differentiable at x=−3 and f′(3−)=0 and f′(3+)=1≠f′(3+) ⇒f(x) is not differentiable at x=3 ∴β=2 Hence, α+β=0+2=2