Ta có: \(y = \dfrac{1}{{\sin x + \cos x}}\)\( = \dfrac{1}{{\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)}}\).
Có \(0 < x < \dfrac{\pi }{4} \Rightarrow \dfrac{\pi }{4} < x + \dfrac{\pi }{4} < \dfrac{{3\pi }}{4}\) nên \(\dfrac{{\sqrt 2 }}{2} < \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1,\forall x \in \left( {0;\dfrac{\pi }{2}} \right)\).
Do đó \(1 < \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) \le \sqrt 2 \) \( \Rightarrow 1 > \dfrac{1}{{\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)}} \ge \dfrac{1}{{\sqrt 2 }}\) hay \(\dfrac{1}{{\sqrt 2 }} \le y < 1\).
Vậy \(\mathop {\min }\limits_{\left( {0;\dfrac{\pi }{2}} \right)} y = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\), dấu “=” xảy ra khi \(x = \dfrac{\pi }{4}\).
Chọn D.
