Let the range of the function f ( x ) = 1 2 + sin 3 x + cos 3 x , x be [ a , b ] . If and are respectively the A.M. and the G.M. of a and b, then is equal to [2024]

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Solution

Q.
Let the range of the function $f (x) = \frac{1}{2 + \sin 3 x + \cos 3 x},$ $x \in ℝ$ be $[a, b]$ . If $α$ and $β$ are respectively the A.M. and the G.M. of $a$ and $b,$ then $\frac{α}{β}$ is equal to [2024]

1 $π$

2 $2$

3 $\sqrt{π}$

4 $\sqrt{2}$

Ans.
(4)

$f (x) = \frac{1}{2 + \sin 3 x + \cos 3 x}$

Now, $- \sqrt{2} \leq \sin 3 x + \cos 3 x \leq \sqrt{2}$

$\frac{1}{2 + \sqrt{2}} \leq \frac{1}{2 + \sin 3 x + \cos 3 x} \leq \frac{1}{2 - \sqrt{2}}$

$[a, b] = [\frac{1}{2 + \sqrt{2}}, \frac{1}{2 - \sqrt{2}}]$

Now, A.M. of $a$ and $b$ i.e., $α = \frac{1}{2} [\frac{1}{2 + \sqrt{2}} + \frac{1}{2 - \sqrt{2}}]$

$\frac{1}{2} [\frac{4}{4 - 2}] = 1$

$β = G.M. of a and b = \sqrt{\frac{1}{2 + \sqrt{2}} \cdot \frac{1}{2 - \sqrt{2}}} = \sqrt{\frac{1}{4 - 2}} = \frac{1}{\sqrt{2}}$

So, $\frac{α}{β} = \frac{\frac{1}{1}}{\sqrt{2}} = \sqrt{2}$

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