If $2\sin^{3}x+\sin 2x\cos x+4\sin x-4=0$ has exactly 3 solutions in the interval $\left[0,\frac{n\pi}{2}\right]$, $n\in\mathrm{N}$, then the roots of the equation $x^{2}+nx+(n-3)=0$ belong to:

(C) $\left(-\frac{\sqrt{17}}{2},\frac{\sqrt{17}}{2}\right)$

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Trigonometric Equations

2 sin^{3} x + sin 2 x cos x + 4 sin x - 4 = 0

has exactly 3 solutions in the interval

[0, \frac{nπ}{2}]

n \in N

, then the roots of the equation

x^{2} + n x + (n - 3) = 0

belong to:

Answer: (B)

2 sin^{3} x + 2 sin x cos^{2} x + 4 sin x - 4 = 0

2 sin^{3} x + 2 sin x (1 - sin^{2} x) + 4 sin x - 4 = 0 \Rightarrow 6 sin x - 4 = 0 \Rightarrow sin x = \frac{2}{3}

.

In

[0, \frac{nπ}{2}]

sin x = \frac{2}{3}

has exactly 3 solutions when

n = 5

(two solutions in each full period up to

2 π

, plus one more in the extra half-period up to

\frac{5 π}{2}

).

With

n = 5

x^{2} + 5 x + 2 = 0 \Rightarrow x = \frac{- 5 \pm 17}{2}

, and both roots are negative since

17 < 5

— the roots lie in

(- \infty, 0)

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option