Start free

Mathematics · Sequence and Series

JEE Main 2026 — 6 April, Morning Shift — Question 44

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 with α>β>0,α>β>0, let min⁡0<θ<π/2f(θ)=max⁡0<θ<πg(θ)\min_{0<\theta<\pi/2}f(\theta) = \max_{0<\theta<\pi}g(\theta). If the first term of a G.P. is (α2β)(\frac{α}{2β}), its common ratio is (2βα)(\frac{2β}{α}) and the sum of its first 1010 terms is mn,\frac{m}{n}, gcd(m,n)=1,gcd(m,n)=1, then m+nm+n is equal to ______.

Answer: 1279

Numerical answer — enter this value.

Step-by-step solution

f(θ)∣min⁡=2αβ(AM≥GM)\left.f(\theta)\right|_{\min }=2 \sqrt{\alpha \beta}(A M \geq G M) g(θ)=αsin⁡2θ+β(1−sin⁡2θ)=(α−β)sin⁡2θ+βg(\theta)=\alpha \sin ^{2} \theta+\beta\left(1-\sin ^{2} \theta\right)=(\alpha-\beta) \sin ^{2} \theta+\beta ∴g(θ)∣max =α\left.\therefore \mathrm{g}(\theta)\right|_{\text {max }}=\alpha Equation 2αβ=α2 \sqrt{\alpha \beta}=\alpha

4αβ=α2,α=4β(α≠0)\begin{aligned} & 4 \alpha \beta=\alpha^{2} , & \alpha=4 \beta(\alpha \neq 0) \end{aligned}

Now first term (α2β)=2\left(\frac{\alpha}{2 \beta}\right)=2 and common ratio (2βα)=12\left(\frac{2 \beta}{\alpha}\right)=\frac{1}{2} S10=a(1−r10)1−r=1023256=mn\mathrm{S}_{10}=\frac{\mathrm{a}\left(1-\mathrm{r}^{10}\right)}{1-\mathrm{r}}=\frac{1023}{256}=\frac{\mathrm{m}}{\mathrm{n}} m+n=1279\mathrm{m}+\mathrm{n}=1279

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression

More Sequence and Series questions from this paper