Application of DerivativesmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Application of Derivatives: Functions Let First Term Common Ratio Sum First (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
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<π/2f(θ)=max⁡0<θ<πg(θ)\displaystyle\min_{0<\theta<\pi/2}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 1010 terms is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to
Solution
Answer: 1279 (± 0.01)
Step 1: tan⁡2θ⋅cot⁡2θ=1\tan^2\theta\cdot\cot^2\theta=1, AM–GM on αtan⁡2θ, βcot⁡2θ\alpha\tan^2\theta,\ \beta\cot^2\theta:
f(θ)=αtan⁡2θ+βcot⁡2θ≥2αtan⁡2θ⋅βcot⁡2θ=2αβ.f(\theta)=\alpha\tan^2\theta+\beta\cot^2\theta\ge2\sqrt{\alpha\tan^2\theta\cdot\beta\cot^2\theta}=2\sqrt{\alpha\beta}.
⇒min⁡0<θ<π/2f=2αβ.\Rightarrow\min_{0<\theta<\pi/2}f=2\sqrt{\alpha\beta}.
 Step 2: cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta:
g(θ)=αsin⁡2θ+β(1−sin⁡2θ)=(α−β)sin⁡2θ+β.g(\theta)=\alpha\sin^2\theta+\beta(1-\sin^2\theta)=(\alpha-\beta)\sin^2\theta+\beta.
 α−β>0\alpha-\beta>0 and sin⁡2θ→1\sin^2\theta\to1 at θ=π2\theta=\tfrac\pi2:
⇒max⁡g=(α−β)(1)+β=α.\Rightarrow\max g=(\alpha-\beta)(1)+\beta=\alpha.
 Step 3: 2αβ=α⇒4αβ=α2⇒4β=α⇒α=4β.2\sqrt{\alpha\beta}=\alpha\Rightarrow4\alpha\beta=\alpha^2\Rightarrow4\beta=\alpha\Rightarrow\alpha=4\beta. Step 4: First term =α2β=4β2β=2=\dfrac{\alpha}{2\beta}=\dfrac{4\beta}{2\beta}=2; ratio =2βα=2β4β=12=\dfrac{2\beta}{\alpha}=\dfrac{2\beta}{4\beta}=\dfrac12. Step 5: Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}, a=2,r=12,n=10a=2,r=\tfrac12,n=10:
S10=2(1−11024)12=4⋅10231024=1023256.S_{10}=\dfrac{2\left(1-\tfrac{1}{1024}\right)}{\tfrac12}=4\cdot\dfrac{1023}{1024}=\dfrac{1023}{256}.
 Step 6: 1023=3⋅11⋅311023=3\cdot11\cdot31, 256=28⇒gcd⁡=1⇒m=1023, n=256256=2^8\Rightarrow\gcd=1\Rightarrow m=1023,\ n=256.
∴m+n=1023+256=1279.\therefore m+n=1023+256=1279.
 Correct answer: 1279
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Application of Derivatives · medium
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a differentiable function such that f(x+y3)=f(x)+f(y)3f\left(\dfrac{x+y}{3}\right)=\dfrac{f(x)+f(y)}{3} for all x,y∈Rx,y\in\mathbb{R} and f′(0)=3f'(0)=3. Then the minimum value of g(x)=3+exf(x)g(x)=3+e^x f(x) is
Application of Derivatives · medium
max⁡0≤x≤π(16sin⁡(x2)cos⁡3(x2))\displaystyle\max_{0\le x\le\pi}\left(16\sin\left(\frac{x}{2}\right)\cos^3\left(\frac{x}{2}\right)\right) is equal to
Application of Derivatives · medium
The number of critical points of the function f(x)={∣sin⁡xx∣,x≠01,x=0f(x)=\begin{cases}\left|\dfrac{\sin x}{x}\right|, & x\ne0\\[4pt] 1, & x=0\end{cases} in the interval (−2π,2π)(-2\pi,2\pi) is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath