JEE Main 2026 — Mathematics Algebra

For the function $f(\theta) = \alpha\tan^2\theta + \beta\cot^2\theta$f(θ)=αtan2θ+βcot2θ, applying the AM-GM inequality gives:

$$\alpha\tan^2\theta + \beta\cot^2\theta \ge 2\sqrt{\alpha\tan^2\theta \cdot \beta\cot^2\theta} = 2\sqrt{\alpha\beta}$$αtan2θ+βcot2θ≥2αtan2θ⋅βcot2θ​=2αβ​

Thus, $\min_{0 < \theta < \pi/2} f(\theta) = 2\sqrt{\alpha\beta}$min0<θ<π/2​f(θ)=2αβ​.

For the function $g(\theta) = \alpha\sin^2\theta + \beta\cos^2\theta$g(θ)=αsin2θ+βcos2θ, we can rewrite it as:

$$g(\theta) = \alpha\sin^2\theta + \beta(1 - \sin^2\theta) = \beta + (\alpha - \beta)\sin^2\theta$$g(θ)=αsin2θ+β(1−sin2θ)=β+(α−β)sin2θ

Since $\alpha > \beta > 0$α>β>0, the maximum value of $g(\theta)$g(θ) occurs when $\sin^2\theta$sin2θ is maximum, which is $1$1 (at $\theta = \pi/2$θ=π/2).

Thus, $\max_{0 < \theta < \pi} g(\theta) = \beta + (\alpha - \beta)(1) = \alpha$max0<θ<π​g(θ)=β+(α−β)(1)=α.

Given that $\min f(\theta) = \max g(\theta)$minf(θ)=maxg(θ), we have:

$$2\sqrt{\alpha\beta} = \alpha$$2αβ​=α

Squaring both sides (since $\alpha > 0$α>0), we get:

$$4\alpha\beta = \alpha^2 \Rightarrow \alpha = 4\beta$$4αβ=α2⇒α=4β

For the given Geometric Progression (G.P.):

First term $a = \dfrac{\alpha}{2\beta} = \dfrac{4\beta}{2\beta} = 2$a=2βα​=2β4β​=2

Common ratio $r = \dfrac{2\beta}{\alpha} = \dfrac{2\beta}{4\beta} = \dfrac{1}{2}$r=α2β​=4β2β​=21​

The sum of the first $10$10 terms is:

$$S_{10} = a\dfrac{1 - r^{10}}{1 - r} = 2\dfrac{1 - (1/2)^{10}}{1 - 1/2} = 4\left(1 - \dfrac{1}{1024}\right) = 4\left(\dfrac{1023}{1024}\right) = \dfrac{1023}{256}$$S10​=a1−r1−r10​=21−1/21−(1/2)10​=4(1−10241​)=4(10241023​)=2561023​

We are given $S_{10} = \dfrac{m}{n}$S10​=nm​ with $\gcd(m, n) = 1$gcd(m,n)=1.

Since $256 = 2^8$256=28 and $1023$1023 is odd, they are coprime. Therefore, $m = 1023$m=1023 and $n = 256$n=256.

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

Answer: $1279$1279