Animated Solution for Mathematics - Functions: Let the range of the function $f(x)=\frac{1}{2+\sin 3x+\cos 3x},x\in \mathbb{R}$ be $[a,b]$. If $\alpha$ and $\beta$ are respectively the A.M. and the G.M. of $a$ and $b$, then $\frac{\alpha }{\beta }$ is equal to

Imagine you are standing before a complex-looking function:

At first glance, it looks like a candidate for the quotient rule, a messy derivative, and a long afternoon of algebra. But stop. Take a breath.

In JEE Advanced, the most elegant solutions often hide in plain sight. We are not here to brute-force our way through calculus; we are here to understand the heartbeat of the function.

The numerator is a constant, $1$. This means the entire variation of $f(x)$ is enslaved to the denominator.

If the denominator is large, the function is small. If the denominator is small, the function is large. Our mission is simple: find the range of the denominator, and we find the range of the function.

Now, look at the denominator:

The constant $2$ is just a vertical shift. The real action is in the $\sin 3x+\cos 3x$ part. This is a classic JEE archetype.

Whenever you see $A\sin \theta +B\cos \theta$, your mind should immediately jump to the standard range formula: $[-\sqrt{A^2+B^2},\sqrt{A^2+B^2}]$. Here, $A=1$ and $B=1$.

The angle $3x$ is just a distraction; it oscillates just as freely as $x$. Thus, $\sin 3x+\cos 3x$ must oscillate between $-\sqrt{1^2+1^2}$ and $\sqrt{1^2+1^2}$, which is $[-\sqrt{2},\sqrt{2}]$.

This is the core geometric reality. We have trapped the oscillating part of our function.

Now, let us build the full denominator. We add $2$ to our inequality:

This is the range of our denominator. Now comes the moment of truth. We need the range of $f(x)=\frac{1}{D}$.

When we take the reciprocal of an inequality where all terms are positive, the inequality signs must flip. The minimum of the denominator becomes the maximum of the function, and the maximum of the denominator becomes the minimum.

So, our range $[a,b]$ becomes:

We have successfully isolated $a=\frac{1}{2+\sqrt{2}}$ and $b=\frac{1}{2-\sqrt{2}}$.

We are now in the final stretch. The problem asks for the ratio of the Arithmetic Mean (A.M.) to the Geometric Mean (G.M.). Let $\alpha$ be the A.M. and $\beta$ be the G.M.

For $\alpha$, we calculate $\frac{a+b}{2}$. Substituting our values, we get:

Taking the common denominator, the numerator becomes $(2-\sqrt{2})+(2+\sqrt{2})=4$, and the denominator becomes $(2+\sqrt{2})(2-\sqrt{2})=4-2=2$. Thus, $\alpha =\frac{1}{2}\times \frac{4}{2}=1$.

It is beautiful how the irrational parts simply vanish! Now for $\beta$, the G.M., which is $\sqrt{ab}$:

Finally, the ratio is:

We have arrived at the answer. It wasn't about complex calculus; it was about recognizing the structure, respecting the properties of inequalities, and trusting the algebra. You have mastered the function. The final answer is $\sqrt{2}$.