Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} \mathrm{a} \cos x+2 b \sin…

Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} \mathrm{a} \cos x+2 b \sin x=c$, $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha+\beta=\frac{\pi}{3}$. Then the value of $\frac{b}{a}$ is
  1. $0 \cdot 1$
  2. 0.5
  3. $-0 \cdot 5$
  4. 1

Solution

$\alpha$ and $\beta$ are roots of $\sqrt{3} \mathrm{a} \cos x+2 \mathrm{~b} \sin x=\mathrm{c}$ $\begin{aligned} & \therefore \quad \sqrt{3} a \cos \alpha+2 b \sin \alpha=c ...(i)\\ & \sqrt{3} a \cos \beta+2 b \sin \beta=c...(ii) \end{aligned}$
Subtracting (ii) from (i), we get $\begin{aligned} & \sqrt{3} a(\cos \alpha-\cos \beta)+2 b(\sin \alpha-\sin \beta)=0 \\ & \begin{aligned} & \Rightarrow \sqrt{3} a[ \left.-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right] \\ &++2 b\left[2 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0 \\ & \Rightarrow-\sqrt{3} a\left[2 \sin \left(\frac{\pi}{6}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right] \\ &++2 b\left[2 \cos \left(\frac{\pi}{6}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0 \end{aligned} \end{aligned}$ $\begin{aligned} & \Rightarrow-\sqrt{3} a \sin \left(\frac{\alpha-\beta}{2}\right)+2 b\left[\sqrt{3} \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0 \\ & \Rightarrow-\sqrt{3} a+2 \sqrt{3} b=0 \Rightarrow \frac{b}{a}=\frac{1}{2}=0.5\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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