JEE Advanced Model Paper 20 with solutions for online practice

f(x) = 0 for 0 ≤ x < 1
= x for 1 ≤ x < 2
= 2(x – 1) for 2 ≤ x < 3
∴ For 0 ≤ x < 1, f o f(x) = f[f(x)] = f(0) = 0.
Now let 1 ≤ x < 2 then f o f(x) = f[f(x)] = f(x) = x
Now let 2 ≤ x < 5/2
then 2 ≤ 2x – 2 < 3
f o f(x) = f[f(x)] = f(2x – 2)
= 2(2x – 2) – 2 = 4x – 6.
and if x ≥ 5/2
, 2x – 2 ≥ 3, so f(x) is undefined.
Hence g(x) = f o f(x) = 0 for 0 ≤ x < 1
= x for 1 ≤ x < 2
= 4x – 6 for 2 ≤ x < 5/2
.Domain of f o f(x) is [0, 5/2). Observe that g(x) is continuous but not differentiable at x = 2.
Now g′(x) = 1 > 0 for 1 ≤ x < 2
= 4 > 0 for 2 ≤ x < 5/2
g(2) = 2
∴ g(x) has no max or min at 2