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MathematicsInteger+4 / -1Question 24 of 75

JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 24

For the functions f(θ)=αtan⁡2θ+βcot⁡2θf(\theta)=\alpha\tan^2\theta+\beta\cot^2\theta, and g(θ)=αsin⁡2θ+βcos⁡2θg(\theta)=\alpha\sin^2\theta+\beta\cos^2\theta, α>β>0\alpha>\beta>0, let min⁡0<θ<π2f(θ)=max⁡0<θ<πg(θ)\displaystyle\min_{0<\theta<\frac\pi2}f(\theta)=\max_{0<\theta<\pi}g(\theta). If the first term of a G.P. is (α2β)\left(\dfrac\alpha{2\beta}\right), its common ratio is (2βα)\left(\dfrac{2\beta}\alpha\right) and the sum of its first 10 terms is mn\dfrac mn, gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to ________.

Answer and worked solution

Official answer: 1279

Extremes.

  • By AM-GM, min⁡f=2αβ\min f=2\sqrt{\alpha\beta}.
  • Since α>β\alpha>\beta, max⁡g=α\max g=\alpha.

So 2αβ=α2\sqrt{\alpha\beta}=\alpha, and α=4β\alpha=4\beta.

G.P. First term 2, ratio 12\tfrac12:
S10=4(1−2−10)=1023256.S_{10}=4\left(1-2^{-10}\right)=\frac{1023}{256}.

Answer: m+n=1279m+n=1279