• # question_answer If $\int\limits_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}=\frac{\pi }{16}.}$ Then minimum value of $a\cos x+b\sin \text{ }x$ is A) $-4$ B) $-8$ C) $-2$ D) $-2\sqrt{2}$

 $\int\limits_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}$$=\int\limits_{0}^{\pi /2}{\frac{{{\sec }^{2}}xdx}{{{b}^{2}}{{\tan }^{2}}x+{{a}^{2}}}}$ $=\frac{1}{{{b}^{2}}}\int\limits_{0}^{\pi /2}{\frac{{{\sec }^{2}}xdx}{{{\tan }^{2}}x+{{\left( \frac{a}{b} \right)}^{2}}}}$ $=\frac{1}{{{b}^{2}}}\int\limits_{0}^{\infty }{\frac{dt}{{{t}^{2}}+{{\left( \frac{a}{b} \right)}^{2}}}}$                   (Putting tan$x=t$) $=\left. \frac{1}{{{b}^{2}}}\times \frac{b}{a}{{\tan }^{-1}}\frac{bt}{a} \right|_{0}^{\infty }\,\,\,\,=\frac{1}{ab}\left( \frac{\pi }{2} \right)=\frac{\pi }{2ab}$Thus $a\frac{\pi }{2ab}=\frac{\pi }{16}\Rightarrow ab=8$Now minimum value of $a\cos x+b\sin x$is $-\sqrt{{{a}^{2}}+{{b}^{2}}},$which is least if $a=b$ $=2\sqrt{2}\Rightarrow -\sqrt{{{a}^{2}}+{{b}^{2}}}\ge -\sqrt{8+8}=-4$