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Mathematics · Quadratic Equations

JEE Main 2026 — 5 April, Morning Shift — Question 29

Let tan A, tan B, where A,B∈(-π/2,π/2) be the roots of the quadratic equation x² - 2x - 5 = 0. Then 20 sin²((A+B)/2) is equal to:

  1. Option A:

    10+1010+\sqrt{10}

  2. Option B:

    10−21010-2\sqrt{10}

  3. Option C:

    10−31010-3\sqrt{10}

    Correct
  4. Option D:

    10−1010-\sqrt{10}

Answer: C

Step-by-step solution

x2−2x−5=0x^{2}-2 x-5=0 tan⁡A+tan⁡B=2;tan⁡Atan⁡B=−5\tan \mathrm{A}+\tan \mathrm{B}=2 ; \tan \mathrm{A} \tan \mathrm{B}=-5 ∴tan⁡(A+B)=21−(−5)=13\therefore \tan (\mathrm{A}+\mathrm{B})=\frac{2}{1-(-5)}=\frac{1}{3} ⇒cos⁡(A+B)=310\Rightarrow \cos (\mathrm{A}+\mathrm{B})=\frac{3}{\sqrt{10}} ∴20(sin⁡2( A+B2))=102(1−cos⁡(A+B))\therefore 20\left(\sin ^{2}\left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right)\right)=\frac{10}{2}(1-\cos (\mathrm{A}+\mathrm{B})) =10(1−310)=(10−310)=10\left(1-\frac{3}{\sqrt{10}}\right)=(10-3 \sqrt{10})

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Mathematics
Chapter
Quadratic Equations
Topic
Nature of Roots of Quadratic Equation

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