If f ( x ) = s i n x s i n 3 x , x  = nπ then the range of values of f ( x ) for real values of x is

3 − 4 sin 2 x = y ∴ sin 2 x = 4 3 − y ⋅ But 0 < sin 2 x ≤ 1 ∴ ( ∵ sin x = 0 ⇒ x = nπ ) 0 < 4 3 − y ≤ 1 or 0 < 3 − y ≤ 4.

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If $f (x) = \frac{s i n 3 x}{s i n x}, x \neq = nπ$ then the range of values of $f (x)$ for real values of $x$ is

Answer:

Solution:

$3 - 4 sin^{2} x = y$

$∴ sin^{2} x = \frac{3 - y}{4} \cdot But 0 < sin^{2} x \leq 1 ∴ (∵ sin x = 0 \Rightarrow x = nπ) 0 < \frac{3 - y}{4} \leq 1 or 0 < 3 - y \leq 4.$

Stream:JEESubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Equation

⏱ 2mℹ️ Source: QB

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