\(y = \left| {4{\rm{f}}\left( {\sin x} \right) + \cos 2x - a} \right| = \left| {4f\left( {\sin x} \right) - 2{{\sin }^2}x + 1 - a} \right|.\)

Đặt \(t = \sin x\,,\,\,t \in \left( {0\,;\,\,1} \right)\) do \(x \in \left( {0\,;\,\,\frac{\pi }{2}} \right)\).

Bài toán trở thành: Có bao nhiêu số nguyên dương \[a\] để hàm số \({\rm{y}} = \left| {4{\rm{f}}\left( {\rm{t}} \right) - 2{{\rm{t}}^2} + 1 - {\rm{a}}} \right|\) nghịch biến trên khoảng \(\left( {0\,;\,\,1} \right)\).

Ta có: \[{\rm{y'}} = \frac{{\left[ {4{\rm{f'}}\left( {\rm{t}} \right) - 4{\rm{t}}} \right]\left[ {4{\rm{f}}\left( {\rm{t}} \right) - 2{{\rm{t}}^2} + 1 - {\rm{a}}} \right]}}{{\left| {4{\rm{f}}\left( {\rm{t}} \right) - 2{{\rm{t}}^2} + 1 - {\rm{a}}} \right|}} \le 0\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right)\,\,(*).\]

Với \({\rm{t}} \in \left( {0\,;\,\,1} \right)\) thì đồ thị hàm số \({\rm{y}} = {\rm{f'}}\left( {\rm{t}} \right)\) nằm phía dưới trục \({\rm{Ox}}\)

\( \Rightarrow {\rm{f'}}\left( {\rm{t}} \right) < 0\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right) \Rightarrow {\rm{f'}}\left( {\rm{t}} \right) - {\rm{t}} < 0\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right)\)

Khi đó: \((*) \Leftrightarrow 4f\left( {\rm{t}} \right) - 2{t^2} + 1 - a \ge 0\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right) \Leftrightarrow a \le 4f\left( {\rm{t}} \right) - 2{t^2} + 1\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right)\).

Xét hàm số \({\rm{g}}\left( {\rm{t}} \right) = 4{\rm{f}}\left( {\rm{t}} \right) - 2{{\rm{t}}^2} + 1\) trên \(\left( {0\,;\,\,1} \right)\).

Ta có \(g'\left( {\rm{t}} \right) = 4f'\left( {\rm{t}} \right) - 4t < 0 \Rightarrow g\left( {\rm{t}} \right) > g\left( 1 \right) = 4f\left( 1 \right) - 2 \cdot 1 + 1 = 3\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right)\).

Do đó \(a \le 3 < g\left( {\rm{t}} \right)\,,\,\,\forall {\rm{t}} \in \left( {0\,;\,\,1} \right)\). Vậy \(0 < a \le 3 \Rightarrow a \in \left\{ {1\,;\,\,2\,;\,\,3} \right\}\). Chọn B.