To find the value of m + n m + n m + n f ( θ ) f(\theta) f ( θ ) g ( θ ) g(\theta) g ( θ )
Step 1: Minimum value of f ( θ ) f(\theta) f ( θ )
The given function is:
f ( θ ) = α tan 2 θ + β cot 2 θ , for 0 < θ < π 2 f(\theta) = \alpha \tan^2 \theta + \beta \cot^2 \theta, \quad \text{for } 0 < \theta < \frac{\pi}{2} f ( θ ) = α tan 2 θ + β cot 2 θ , for 0 < θ < 2 π
Since α > 0 \alpha > 0 α > 0 β > 0 \beta > 0 β > 0 tan 2 θ , cot 2 θ > 0 \tan^2 \theta, \cot^2 \theta > 0 tan 2 θ , cot 2 θ > 0 θ ∈ ( 0 , π 2 ) \theta \in \left(0, \frac{\pi}{2}\right) θ ∈ ( 0 , 2 π ) α tan 2 θ + β cot 2 θ 2 ≥ ( α tan 2 θ ) ( β cot 2 θ ) \frac{\alpha \tan^2 \theta + \beta \cot^2 \theta}{2} \ge \sqrt{(\alpha \tan^2 \theta)(\beta \cot^2 \theta)} 2 α t a n 2 θ + β c o t 2 θ ≥ ( α tan 2 θ ) ( β cot 2 θ ) f ( θ ) ≥ 2 α β f(\theta) \ge 2\sqrt{\alpha \beta} f ( θ ) ≥ 2 α β
Equality holds when α tan 2 θ = β cot 2 θ ⟹ tan 4 θ = β α \alpha \tan^2 \theta = \beta \cot^2 \theta \implies \tan^4 \theta = \frac{\beta}{\alpha} α tan 2 θ = β cot 2 θ ⟹ tan 4 θ = α β θ ∈ ( 0 , π 2 ) \theta \in \left(0, \frac{\pi}{2}\right) θ ∈ ( 0 , 2 π ) min 0 < θ < π 2 f ( θ ) = 2 α β \min_{0 < \theta < \frac{\pi}{2}} f(\theta) = 2\sqrt{\alpha\beta} min 0 < θ < 2 π f ( θ ) = 2 α β
Step 2: Maximum value of g ( θ ) g(\theta) g ( θ )
The given function is:
g ( θ ) = α sin 2 θ + β cos 2 θ , for 0 < θ < π g(\theta) = \alpha \sin^2 \theta + \beta \cos^2 \theta, \quad \text{for } 0 < \theta < \pi g ( θ ) = α sin 2 θ + β cos 2 θ , for 0 < θ < π
Rewriting g ( θ ) g(\theta) g ( θ ) g ( θ ) = α sin 2 θ + β ( 1 − sin 2 θ ) = β + ( α − β ) sin 2 θ g(\theta) = \alpha \sin^2 \theta + \beta (1 - \sin^2 \theta) = \beta + (\alpha - \beta) \sin^2 \theta g ( θ ) = α sin 2 θ + β ( 1 − sin 2 θ ) = β + ( α − β ) sin 2 θ
Given that α > β > 0 \alpha > \beta > 0 α > β > 0 ( α − β ) > 0 (\alpha - \beta) > 0 ( α − β ) > 0 θ ∈ ( 0 , π ) \theta \in (0, \pi) θ ∈ ( 0 , π ) 0 < sin 2 θ ≤ 1 0 < \sin^2 \theta \le 1 0 < sin 2 θ ≤ 1 θ = π 2 \theta = \frac{\pi}{2} θ = 2 π sin 2 ( π 2 ) = 1 \sin^2\left(\frac{\pi}{2}\right) = 1 sin 2 ( 2 π ) = 1 max 0 < θ < π g ( θ ) = β + ( α − β ) ( 1 ) = α \max_{0 < \theta < \pi} g(\theta) = \beta + (\alpha - \beta)(1) = \alpha max 0 < θ < π g ( θ ) = β + ( α − β ) ( 1 ) = α
Step 3: Relationship between α \alpha α β \beta β
We are given that min f ( θ ) = max g ( θ ) \min f(\theta) = \max g(\theta) min f ( θ ) = max g ( θ ) 2 α β = α 2\sqrt{\alpha\beta} = \alpha 2 α β = α
Squaring both sides:
4 α β = α 2 4\alpha\beta = \alpha^2 4 α β = α 2
Since α > 0 \alpha > 0 α > 0 α \alpha α α = 4 β ⟹ α β = 4 \alpha = 4\beta \implies \frac{\alpha}{\beta} = 4 α = 4 β ⟹ β α = 4
Step 4: Sum of the Geometric Progression (G.P.)
The first term a a a r r r a = α 2 β = 4 β 2 β = 2 a = \frac{\alpha}{2\beta} = \frac{4\beta}{2\beta} = 2 a = 2 β α = 2 β 4 β = 2 r = 2 β α = 2 β 4 β = 1 2 r = \frac{2\beta}{\alpha} = \frac{2\beta}{4\beta} = \frac{1}{2} r = α 2 β = 4 β 2 β = 2 1
The sum of the first 10 10 10 S 10 S_{10} S 10 S 10 = a ( 1 − r 10 1 − r ) = 2 ( 1 − ( 1 2 ) 10 1 − 1 2 ) = 4 ( 1 − 1 1024 ) = 4 ( 1023 1024 ) = 1023 256 S_{10} = a \left( \frac{1 - r^{10}}{1 - r} \right) = 2 \left( \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}} \right) = 4 \left( 1 - \frac{1}{1024} \right) = 4 \left( \frac{1023}{1024} \right) = \frac{1023}{256} S 10 = a ( 1 − r 1 − r 10 ) = 2 ( 1 − 2 1 1 − ( 2 1 ) 10 ) = 4 ( 1 − 1024 1 ) = 4 ( 1024 1023 ) = 256 1023
We are given S 10 = m n S_{10} = \frac{m}{n} S 10 = n m gcd ( m , n ) = 1 \gcd(m, n) = 1 g cd( m , n ) = 1 1023 1023 1023 256 = 2 8 256 = 2^8 256 = 2 8 gcd ( 1023 , 256 ) = 1 \gcd(1023, 256) = 1 g cd( 1023 , 256 ) = 1
Therefore:
m = 1023 m = 1023 m = 1023 n = 256 n = 256 n = 256
m + n = 1023 + 256 = 1279 m + n = 1023 + 256 = 1279 m + n = 1023 + 256 = 1279