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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Circle

Let the arc A C of a circle subtend a right angle at the centre O ....

Let the arc A Cofacirclesubtendarightangleatthecentreof a circle subtend a right angle at the centreO.Ifthepoint. If the pointBonthearcACon the arc A C, divides the arc A Csuchthat length of arc⁡AB length of arc⁡BC=15such that \frac{\text{ length of } \operatorname{arc} A B}{\text{ length of } \operatorname{arc} B C}=\frac{1}{5}, and OC→=αOA→+βOB→\overrightarrow{O C}=\alpha \overrightarrow{O A}+\beta \overrightarrow{O B}, then α+2(3−1)β\alpha+\sqrt{2}(\sqrt{3}-1) \beta is equal toLet the arc A C of a circle subtend a right angle at the centre O . If the point B on the arc A C , divides the arc A C

Options

  1. A232 \sqrt{3}
  2. B2−32-\sqrt{3}
  3. C535 \sqrt{3}
  4. D2+32+\sqrt{3}

Correct answer

B. 2−32-\sqrt{3}

Step-by-step solution

Let the arc A C of a circle subtend a right angle at the centre O . If the point B on the arc A C , divides the arc A C
c→=αa→+βb→…..(1)a→⋅c→=αa→⋅a→+βb→⋅a→0=α+βcos⁡15∘….(2)(1)⇒b→⋅c→=αa→⋅b→+βb→⋅b→⇒cos⁡75∘=αcos⁡15∘+β….(3)(2)(3)⇒cos⁡75∘=−βcos⁡215∘+β\begin{aligned} & \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}} \ldots . .(1) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}} \\ & 0=\alpha+\beta \cos 15^{\circ} \ldots .(2) \\ & (1) \Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}} \\ & \Rightarrow \cos 75^{\circ}=\alpha \cos 15^{\circ}+\beta \ldots .(3) \\ & (2) &(3) \Rightarrow \cos 75^{\circ}=-\beta \cos ^2 15^{\circ}+\beta\end{aligned}\beta=\frac{\cos 75^{\circ}}{\sin ^2 15^{\circ}}=\frac{1}{\sin 15^{\circ}}=\frac{2 \sqrt{2}}{\sqrt{3}-1}(2)(2)\Rightarrow \alpha=\frac{-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \therefore \overrightarrow{\mathrm{c}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \overrightarrow{\mathrm{a}}+\left(\frac{2 \sqrt{2}}{\sqrt{3}-1}\right) \overrightarrow{\mathrm{b}}NowNowα+2(3−1)β=−(3+1)(3−1)+2(3−1)⋅223−1=−(3+1)22+4=−3−1−23+82=2−3\begin{aligned} & \alpha+\sqrt{2}(\sqrt{3}-1) \beta=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}+\frac{\sqrt{2}(\sqrt{3}-1) \cdot 2 \sqrt{2}}{\sqrt{3}-1} \\ & =\frac{-(\sqrt{3}+1)^2}{2}+4 \\ & =\frac{-3-1-2 \sqrt{3}+8}{2} \\ & =2-\sqrt{3} \end{aligned}$
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