For the functions f( )= ^2 + ^2 , and g( )= ^2 + ^2 , 0, let 0

Question

For the functions $f( heta )=\alpha { an }^{2} heta +\beta {\cot }^{2} heta $, and $g( heta )=\alpha {\sin }^{2} heta +\beta {\cos }^{2} heta ,\alpha >\beta >0$, let $\underset{0< heta <\frac{\pi }{2}}{min}f( heta )=\underset{0< heta <\pi }{max}g( heta )$. If the first term of a G.P. is $(\frac{\alpha }{2\beta })$, its common ratio is $(\frac{2\beta }{\alpha })$ and the sum of its first 10 terms is $\frac{m}{n},gcd(m,n)=1$, then $m+n$ is equal to $____$ . Answer: 1279

Accepted Answer

For

$f(	heta )=\alpha {	an }^{2}	heta +\beta {\cot }^{2}	heta $

and

$g(	heta )=\alpha {\sin }^{2}	heta +\beta {\cos }^{2}	heta ,\, \alpha >\beta >0$

we are given that

$\underset0We will first find these two values.

1. Minimum value of $f(	heta )$
Let

$x={	an }\^{2}	heta .$

Since $00$.

Then

${\cot }\^{2}	heta =\frac{1}{{	an }\^{2}	heta }=\frac{1}{x}.$

So

$f(	heta )=\alpha x+\frac{\beta }{x}.$

Now use the standard result for $x>0$:

$ax+\frac{b}{x}\ge 2\sqrt{ab}.$

Hence,

$\alpha x+\frac{\beta }{x}\ge 2\sqrt{\alpha \beta }.$

Therefore,

$minf(	heta )=2\sqrt{\alpha \beta }.$

2. Maximum value of $g(	heta )$
We have

$g(	heta )=\alpha {\sin }\^{2}	heta +\beta {\cos }\^{2}	heta .$

Using

${\cos }\^{2}	heta =1-{\sin }\^{2}	heta ,$

we get

$g(	heta )=\alpha {\sin }\^{2}	heta +\beta (1-{\sin }\^{2}	heta )$

$=\beta +(\alpha -\beta ){\sin }\^{2}	heta .$

Since $\alpha >\beta $, the coefficient of ${\sin }\^{2}	heta $ is positive. So $g(	heta )$ is maximum when

${\sin }\^{2}	heta =1.$

For $0Thus,

$maxg(	heta )=\beta +(\alpha -\beta )=\alpha .$

3. Use the given condition
Given

$minf(	heta )=maxg(	heta ),$

so

$2\sqrt{\alpha \beta }=\alpha .$

Now square both sides:

$4\alpha \beta ={\alpha }^{2}.$

Since $\alpha >0$, divide by $\alpha $:

$4\beta =\alpha .$

So,

$\alpha =4\beta .$

4. G.P. data
First term:

$a=\frac{\alpha }{2\beta }=\frac{4\beta }{2\beta }=2.$

Common ratio:

$r=\frac{2\beta }{\alpha }=\frac{2\beta }{4\beta }=\frac{1}{2}.$

So the G.P. is

$2,1,\frac{1}{2},\frac{1}{4},\ldots$

5. Sum of first 10 terms
For a G.P.,

${S}_{n}=\frac{a(1-{r}^{n})}{1-r},\, r
e 1.$

Thus,

${S}_{10}=\frac{2(1-{(\frac{1}{2})}^{10})}{1-\frac{1}{2}}.$

Now simplify:

${S}_{10}=2\cdot \frac{1-\frac{1}{1024}}{\frac{1}{2}}$

$=2\cdot 2(1-\frac{1}{1024})$

$=4\cdot \frac{1023}{1024}$

$=\frac{4092}{1024}$

$=\frac{1023}{256}.$

So,

$\frac{m}{n}=\frac{1023}{256}.$

Hence,

$m+n=1023+256=1279.$

Therefore, the answer is

$1279.$

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