Let a, b, c be three non-zero real numbers such that the equation 3 acos x + 2bsin x = c, x ∈ [- (π)/(2), (π)/(2)] has two distinct real roots α and β with α + β = (π)/(3) . Then, the value of (b)/(a) is

Correct answer: 0 3 cos x + (2b)/(a)sin x = (c)/(a) Now, 3 cos α + (2b)/(a)sin α = (c)/(a) (i) 3 cos β + (2b)/(a) sin β = (c)/(a) (ii) From equations (i) and (ii) ⇒ 3 [cos α - cos β] + (2b)/(a)(sin α - sin β) = 0 ⇒ 3 [- 2 ((α + β)/(2)) ((α - β)/(2))] + (2b)/(a)[2 ((α + β)/(2)) ((α - β)/(2))] = 0 ⇒ - 3 + 2 3 · (b)/(a) = 0 ⇒ (b)/(a) = (1)/(2) = 0.5

JEE Advanced 2018Paper 1

Exam: JEE Advanced
Year: 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Advanced 2018 · Paper 1 · Mathematics · Trigonometric Ratios & Identities

Let a, b, c be three non-zero real numbers such that the equation 3...

Let

a, b, c

be three non-zero real numbers such that the equation

\sqrt{3}acos x + 2bsin x = c, x \in

\left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]

has two distinct real roots

\alpha and

\beta

with

\alpha + \beta = \dfrac{\pi}{3}

. Then, the value of

\dfrac{b}{a}

Correct answer

Step-by-step solution

\sqrt{3}cos x + \dfrac{2b}{a}sin x = \dfrac{c}{a} $<br />Now,$ \sqrt{3}cos \alpha + \dfrac{2b}{a}sin \alpha = \dfrac{c}{a} \ldots \left(i\right)
$\sqrt{3}cos \beta +$ \dfrac{2b}{a} $sin \beta =$ \dfrac{c}{a} $\ldots \left(ii\right)$

From equations \left(i\right) $and \left(ii\right)$

$⇒ \sqrt{3}\left[cos \alpha - cos \beta\right] + \dfrac{2b}{a}\left(sin \alpha - sin \beta\right) = 0$
$⇒ \sqrt{3}\left[- 2\sin \left(\dfrac{\alpha + \beta}{2}\right)\sin \left(\dfrac{\alpha - \beta}{2}\right)\right] + \dfrac{2b}{a}\left[2\cos \left(\dfrac{\alpha + \beta}{2}\right)\sin \left(\dfrac{\alpha - \beta}{2}\right)\right] = 0$

$⇒ - \sqrt{3} + 2\sqrt{3} \cdot \dfrac{b}{a} = 0$
$⇒ \dfrac{b}{a} = \dfrac{1}{2} = 0.5$

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