A) 5 and\[-5\]
B) \[2\sqrt{13}\]and\[-2\sqrt{13}\]
C) 10 and\[-10\]
D) \[\frac{5}{2}\]and\[-\frac{5}{2}\]
Correct Answer: A
Solution :
Let\[f(x)=6\sin x\cos x+4\cos 2x\] \[f(x)=3(2\sin x\cos x)+4\cos 2x\] \[\Rightarrow \]\[f(x)=3\sin 2x+4\cos 2x\] We know that maximum and minimum value of the expression, \[a\sin x+b\cos x\]are\[\sqrt{{{a}^{2}}+{{b}^{2}}}\]and\[-\sqrt{{{a}^{2}}+{{b}^{2}}}\] respectively. \[\therefore \]Maximum value of\[f(x)\] \[=\sqrt{9+16}=\sqrt{25}=5\]and minimum value of \[f(x)=-\sqrt{9+16}=-\sqrt{25}=-5\]
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