JEE Advanced Model Paper 19 with solutions for online practice

If x ∈ [–1, 1], f(g(x)) = sin x – 1 for –1 ≤ x < 0
= sin2 x for 0 ≤ x ≤ 1
|g(x)| = –sin x for –1 ≤ x < 0
= sin x for 0 ≤ x ≤ 1
f(|g(x)|) = sin2 x ∀ x ∈ [–1, 1]. Further
|(g(x))| = 1 – sin x for –1 ≤ x < 0
= sin2 x for 0 ≤ x ≤ 1.
∴ h(x) = sin2 x – sin x + 1 for –1 ≤ x < 0
= 2sin2 x for 0 ≤ x ≤ 1
h′(x) = sin 2x – cos x for –1 < x < 0
= 4sin x cos x for 0 < x < 1. The value of h(x) at x =0 is less than the neighboring values.
So local minimum at x = 0