Let a, b, c be three non-zero real numbers such that the equation 3acos⁡x+2bsin⁡x=c, x∈-π2, π2 has two distinct real roots α and β with α+β=π3. Then, the value of ba is

JEE Advanced · Mathematics · 7. Trigonometry

Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} a \cos x + 2 b \sin x = c, x \in [- \frac{π}{2}, \frac{π}{2}]$ has two distinct real roots $α$ and $β$ with $α + β = \frac{π}{3}$ . Then, the value of $\frac{b}{a}$ is

A 0.52
B 0.5
C 0.14
D 0.1

Verified Solution

Answer & Solution

Correct Answer

(B) 0.5

Step-by-step Solution

Detailed explanation

\sqrt{3} \cos x + \frac{2 b}{a} \sin x = \frac{c}{a}

Now,

\sqrt{3} \cos α + \frac{2 b}{a} \sin α = \frac{c}{a} \dots (i)

\sqrt{3} \cos β + \frac{2 b}{a} \sin β = \frac{c}{a} \dots (ii)

From equations

(i)

and

(ii)

$\Rightarrow \sqrt{3}[\cos \alpha-\cos \beta]+\frac{2 b}{a}(\sin \alpha-\sin \beta)=0 $
$ \Rightarrow \sqrt{3}\left[-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right]+\frac{2 b}{a}[2 \cos$ $\left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)]=0 $
$ \Rightarrow-\sqrt{3}+2 \sqrt{3} \cdot \frac{b}{a}=0 $
$ \Rightarrow \frac{b}{a}=\frac{1}{2}=0.5$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right) ; k=1,2, \ldots, 9$

List - I	List - II
(A) For each $z_k$ there exists a $z_j$ such $z_k \cdot z_j=1$	(P) True
(B) There exists a $k \in\{1,2, \ldots, 9\}$ such that $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers	(Q) False
(C) $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}$ equals	(R) 1
(D) $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	(S) 2

JEE Advanced 2014 Hard

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From JEE Advanced

Match the compounds in LIST-

I

with the observations in LIST-

II

, and choose the correct option.

	LIST- $I$		LIST- $II$
$(I)$	Aniline	$(P)$	Sodium fusion extract of the compound on boiling with ${FeSO}_{4}$ , followed by acidification with conc. $H_{2} {SO}_{4}$ , gives Prussian blue color.
$(II)$	$o -$ Cresol	$(Q)$	Sodium fusion extract of the compound on treatment with sodium nitroprusside gives blood red color.
$(III)$	Cysteine	$(R)$	Addition of the compound to a saturated solution of ${NaHCO}_{3}$ results in effervescence.
$(IV)$	Caprolactam	$(S)$	The compound reacts with bromine water to give a white precipitate.
		$(T)$	Treating the compound with neutral ${FeCl}_{3}$ solution produces violet color.

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List - I	List - II
(A) For each \(z_k\) there exists a \(z_j\) such \(z_k \cdot z_j=1\)	(P) True
(B) There exists a \(k \in\{1,2, \ldots, 9\}\) such that \(z_1 \cdot z=z_k\) has no solution z in the set of complex numbers	(Q) False
(C) \(\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}\) equals	(R) 1
(D) \(1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)\) equals	(S) 2