If the range of f ( θ ) = sin ⁡ 4 θ + 3 cos ⁡ 2 θ sin ⁡ 4 θ + cos ⁡ 2 θ , θ ∈ R f(θ)= sin 4 θ+cos 2 θ sin 4 θ+3cos 2 θ ​ ,θ∈R is [ α , β ] [α,β], then the sum of the infinite G.P., whose first term is 64 64 and the common ratio is α β β α ​ , is equal to ____.

1. Understanding the given function

The function is

Key observations:

1. (\sin^{2}\theta) and (\cos^{2}\theta) always lie between $0$ and $1$ and they add up to $1$.
2. By rewriting everything in terms of just one of them, the messy trigonometric look disappears and the problem turns into an ordinary algebraic range question.

2. Converting to a single variable

Put $s = \sin^{2}\theta \quad(0 \le s \le 1) \implies \cos^{2}\theta = 1-s$ Then

Therefore

So

3. Finding where the fraction is biggest and smallest

Because both numerator and denominator are smooth polynomials we can use calculus:

1. Compute derivative:
   $f'(s)=\frac{2s(s-2)}{\bigl(s^{2}-s+1\bigr)^{2}}$
2. On $0<s<1$, the factor $s$ is positive and $(s-2)$ is negative, so $f'(s)<0$.
3. Hence $f(s)$ decreases steadily from $s=0$ to $s=1$.

Edge values give the range:

Therefore $\boxed{\alpha=1,\;\beta=3}\qquad \text{and the range is }[1,3].$

4. The infinite GP

Given:

• first term (a=64)
• common ratio (r=\dfrac{\alpha}{\beta}=\dfrac{1}{3}) (chosen because it is <1, so the GP converges)

The sum of an infinite GP is $S=\frac{a}{1-r}.$ Plugging in the numbers:

Hence the required value is 96.

What is happening here?

Think of $\theta$ as a knob you can turn any amount you like.
When you turn it, two numbers – (\sin\theta) and (\cos\theta) – wiggle up and down between $-1$ and $1$.
The question builds a fancy fraction with powers of those wiggles and asks “How high and how low can that fraction go?”

Once we know the lowest value (call it (\alpha)) and the highest value (call it (\beta)), we use them to make a new money-box called a geometric progression (GP):

• first coin we drop in: 64
• every next coin is only a fixed part of the previous coin (the common ratio).
  Here that fraction is (\dfrac{\alpha}{\beta}) – a small number because (\alpha<\beta).

Since we keep dropping smaller and smaller coins forever, the total money settles to a nice finite sum. The problem finally asks for that total.

Step-by-step calculation

1. Rewrite in one variable
   $s=\sin^{2}\theta,\;0\le s\le1,\;\cos^{2}\theta=1-s$ $f(s)=\frac{s^{2}-3s+3}{s^{2}-s+1}$

2. Find derivative
   $f'(s)=\frac{2s(s-2)}{(s^{2}-s+1)^{2}}$ On $(0,1]$, $f'(s)<0\;\Rightarrow\;f$ is decreasing.

3. Evaluate at endpoints
   $f(0)=3,\qquad f(1)=1$ $\therefore \text{Range}=\bigl[\alpha,\,\beta\bigr]=[1,3] \;(\alpha=1,\,\beta=3).$

4. Infinite GP $a=64,\quad r=\frac{\alpha}{\beta}=\frac{1}{3}\;(\lvert r\rvert<1)$ Sum: $S_{\infty}=\frac{a}{1-r}=\frac{64}{1-\tfrac{1}{3}}=\frac{64}{\tfrac{2}{3}}=96.$

[\boxed{96}]