Explanation
Solution
1. Analyze Non-Differentiability (m):
Check continuity/differentiability at junction points x=2,3,5 and where the max function or [x] changes.
For x≤2, let g(t)=t3−3t. g′(t)=3(t−1)(t+1). The function increases until t=−1 (value 2), decreases until t=1 (value -2), then increases.
maxt≤xg(t) stays at 2 for x∈[−1,2].
Non-differentiable points occur at x=−1,1,3,4,5. However, checking limits at x=2 (f(2)=2 and x2+2x−6∣x=2=2), we find m=4.
2. Calculate I=∫−22f(x)dx:
Divide the interval based on the behavior of max{t3−3t}:
For x∈[−2,−1]: f(x)=x3−3x
For x∈[−1,2]: f(x)=2
I=∫−2−1(x3−3x)dx+∫−122dx
I=[4x4−23x2]−2−1+[2x]−12=((41−23)−(4−6))+(4−(−2))
Correct Option: (C)
