Solution :
Here x = 2t – |t-1| and y = 2t2 + t|t|.
Now when t < 0;
x = 2t – {-(t-1)} = 3t – 1 and y = 2t2 – t2 = t2 ⟹ = y = 19(x+1)2
= when 0 ≤ t < 1
x = 2t – {-(t-1)} = 3t – 1 and y = 2t2 – t2 = 3t2 ⟹ = y = 13(x+1)2
when t ≥ 1;
x = 2t – (t-1) = t + 1 and y = 2t2 + t2 = 3t2 ⟹ = y = 3(x+1)2
Thus, y = f(x) = 19(x+1)2, x < -1
y = f(x) = 13(x+1)2, -1≤x < 2
y = f(x) = 3(x+1)2, x≥ 2
We have to check differentiability at x = -1 and 2.
Differentiabilty at x = -1;
LHD = f'(−1)− = limh→0f(−1−h)–f(−1)−h = limh→0 19(−1−h+1)2–0−h = 0
RHD = f'(−1)+ = limh→0f(−1+h)–f(−1)h = limh→0 13(−1+h+1)2–0h = 0
Hence f(x) is differentiable at x = -1
⟹ continuous at x = -1.
To check differentiability at x = 2;
LHD = f'(2)− = limh→013(2−h+1)2–3−h = 2
RHD = f'(2)+ = limh→03(2+h−1)2–3h = 6
Hence f(x) is not differentiable at x = 2.
But continuous at x = 2, because LHD and RHD both are finite.
f(x) is continuous for all x and differentiable for all x, except x = 2.