For the function and , , let . If the first term of a G.P. is , its common ratio is and the sum of its first 10 terms is , gcd (m, n) = 1, then m + n is equal to: [2026]

Question

For the function $f (θ) = α \tan^{2} θ + β \cot^{2} θ$ and $g (θ) = α \sin^{2} θ + β \cos^{2} θ$ , $α > β > 0$ , let $\min_{0 < θ < \frac{π}{2}} f (θ) = \max_{0 < θ < π} g (θ)$ . If the first term of a G.P. is $(\frac{α}{2 β})$ , its common ratio is $(\frac{2 β}{a})$ and the sum of its first 10 terms is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to: [2026]

Smart Achievers · Accepted Answer

(1279)

We have, $f (θ) = α \tan^{2} θ + β \cot^{2} θ$ and $g (θ) = α \sin^{2} θ + β \cos^{2} θ$

Using A.M. $\geq$ G.M.

$\min_{0 < θ < \frac{π}{2}} f (θ) = \max_{0 < θ < π} g (θ)$

$\Rightarrow f (θ) \geq 2 \sqrt{α β}$

$\Rightarrow f {(θ)}_{\min} = 2 \sqrt{α β}$

Now, $g (θ) = α \sin^{2} θ + β \cos^{2} θ$

$= α \sin^{2} θ + β (1 - \sin^{2} θ)$

$\Rightarrow g (θ) = β + (α - β) \sin^{2} θ$

$\Rightarrow g {(θ)}_{\max} = β + α - β = α$

$∴ 2 \sqrt{α β} = α$

Squaring on both sides, we get

$4 α β = α^{2} \Rightarrow α^{2} - 4 α β = 0$

$α (α - 4 β) = 0$

$\Rightarrow 4 β = α$ [ $∵ α > 0$ ]

$\frac{α}{β} = 4$

$∴ T_{1} = \frac{α}{2 β} = 2$ and common ratio

$(r) = \frac{2 β}{α} = 2 \times \frac{1}{4} = \frac{1}{2}$

Sum of first 10 terms of G.P.

$= 2 [\frac{1 - {(\frac{1}{2})}^{10}}{1 - \frac{1}{2}}]$

$= 2^{2} [\frac{2^{10} - 1}{2^{10}}] = \frac{1024 - 1}{256} = \frac{1023}{256}$

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

$∴$ m + n = 1279

Solution

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Solution

Q. For the function f(θ)=αtan2θ+βcot2θ and g(θ)= αsin2θ+βcos2θ, α>β>0, let min0<θ<π2f(θ)=max0<θ<πg(θ). If the first term of a G.P. is (α2β), its common ratio is (2βa) and the sum of its first 10 terms is mn, gcd (m, n) = 1, then m + n is equal to: [2026]

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