If the range of f( )= ^4 +3 ^2 ^4 + ^2 , R is [ , ], then the sum of the infinite G.P., wh…

Question

If the range of $f(	heta )=\frac{{\sin }^{4}	heta +3{\cos }^{2}	heta }{{\sin }^{4}	heta +{\cos }^{2}	heta },	heta \in R$ is $[\alpha ,\beta ]$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha }{\beta }$, is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 96

Accepted Answer

To determine the range of the function $f(	heta )=\frac{{\sin }^{4}	heta +3{\cos }^{2}	heta }{{\sin }^{4}	heta +{\cos }^{2}	heta }$, let's start by simplifying the expression. Let ${\sin }^{2}	heta =x$, so ${\cos }^{2}	heta =1-x$. The function then transforms into:

$f(x)=\frac{{x}^{2}+3(1-x)}{{x}^{2}+(1-x)}$

Simplify the numerator and denominator separately:

Numerator: ${x}^{2}+3-3x$

Denominator: ${x}^{2}+1-x$

Thus, the function becomes:

$f(x)=\frac{{x}^{2}+3-3x}{{x}^{2}+1-x}=\frac{{x}^{2}-3x+3}{{x}^{2}-x+1}$

Next, we need to find the range of this function. Let's analyze the function by testing specific values of $x$ in the interval $[0,1]$ (since ${\sin }^{2}	heta $ ranges from 0 to 1):

When $x=0$:

$f(0)=\frac{{0}^{2}-3(0)+3}{{0}^{2}-0+1}=\frac{3}{1}=3$

When $x=1$:

$f(1)=\frac{{1}^{2}-3(1)+3}{{1}^{2}-1+1}=\frac{1-3+3}{1-1+1}=\frac{1}{1}=1$

It appears that $f(x)$ achieves values within $[1,3]$. To confirm this, we need to solve the quadratic inequality:

$1\le \frac{{x}^{2}-3x+3}{{x}^{2}-x+1}\le 3$

By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:

$\alpha =1$

$\beta =3$

The common ratio of the infinite geometric progression is:

$\frac{\alpha }{\beta }=\frac{1}{3}$

Given the first term $a=64$, the sum $S$ of the infinite geometric progression can be given as:

$S=\frac{a}{1-r}$

Substituting the values $a=64$ and $r=\frac{1}{3}$, we get:

$S=\frac{64}{1-\frac{1}{3}}=\frac{64}{\frac{2}{3}}=64	imes \frac{3}{2}=96$

Therefore, the sum of the infinite geometric progression is 96.

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