Let f(x) =
-x - π/2, x ≤ -π/2
-cos x, -π/2 < x ≤ 0
x², 0 < x ≤ 1
ln x, x > 1
Then
f(x) is continuous at x = -π/2
f(x) is differentiable at x = 1
f(x) is differentiable at x = ∞
f(x) is not differentiable at x = 0

Given:
f(x) =
-x - π/2, x ≤ -π/2
-cos x, -π/2 < x ≤ 0
x², 0 < x ≤ 1
ln x, x > 1
At x = -π/2,
LHL = limx→-π/2 f(x) = limh→0 -(-π/2 - h) - π/2 = 0
RHL = limx→-π/2+ f(x) = limh→0 -cos(-π/2 + h) = -cos(-π/2) = 0
f(-π/2) = -(-π/2) - π/2 = 0
Since LHL = RHL = f(-π/2) = 0, f(x) is continuous at x = -π/2.
At x = 0,
LHL = limx→0- f(x) = limh→0 -cos(-h) = -1
RHL = limx→0+ f(x) = limh→0 h² = 0
Since LHL ≠ RHL, f(x) is not continuous at x = 0, hence not differentiable at x = 0.
At x = 1,
LHL = limx→1- f(x) = limh→0 (1 - h)² = 1
RHL = limx→1+ f(x) = limh→0 ln(1 + h) = 0
Since LHL ≠ RHL, f(x) is not continuous at x = 1, hence not differentiable at x = 1.