A) \[-4\]
B) \[-8\]
C) \[-2\]
D) \[-2\sqrt{2}\]
Correct Answer: A
Solution :
| \[\int\limits_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\]\[=\int\limits_{0}^{\pi /2}{\frac{{{\sec }^{2}}xdx}{{{b}^{2}}{{\tan }^{2}}x+{{a}^{2}}}}\] |
| \[=\frac{1}{{{b}^{2}}}\int\limits_{0}^{\pi /2}{\frac{{{\sec }^{2}}xdx}{{{\tan }^{2}}x+{{\left( \frac{a}{b} \right)}^{2}}}}\] |
| \[=\frac{1}{{{b}^{2}}}\int\limits_{0}^{\infty }{\frac{dt}{{{t}^{2}}+{{\left( \frac{a}{b} \right)}^{2}}}}\] (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 \[a=b\] |
| \[=2\sqrt{2}\Rightarrow -\sqrt{{{a}^{2}}+{{b}^{2}}}\ge -\sqrt{8+8}=-4\] |
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