Let f : [ 0 , 2 ] be the function defined by f ( x ) = ( 3 - sin ( 2 x ) ) sin ( x - 4 ) - sin ( 3 x + 4 ) . If [ 0 , 2 ] are such that { x [ 0 , 2 ] : f ( x ) 0 } = , then the value of is _______. [2020]

Question

Let $f : [0, 2] \to ℝ$ be the function defined by $f (x) = (3 - \sin (2 π x)) \sin (π x - \frac{π}{4}) - \sin (3 π x + \frac{π}{4}) .$

If $α, β \in [0, 2]$ are such that ${x \in [0, 2] : f (x) \geq 0} = [α, β],$ then the value of $β - α$ is _______. [2020]

Smart Achievers · Accepted Answer

(1)

Let $π x - \frac{π}{4} = θ \in [\frac{- π}{4}, \frac{7 π}{4}]$

$∵$ $f (x) \geq 0$

$So, (3 - \sin (\frac{π}{2} + 2 θ)) \sin θ \geq \sin (π + 3 θ)$

$\Rightarrow (3 - \cos 2 θ) \sin θ \geq - \sin 3 θ$

$\Rightarrow \sin θ [3 - 4 \sin^{2} θ + 3 - \cos 2 θ] \geq 0$

$\Rightarrow \sin θ [6 - 2 (1 - \cos 2 θ) - \cos 2 θ] \geq 0$

$\Rightarrow \sin θ (4 + \cos 2 θ) \geq 0$ $\Rightarrow \sin θ \geq 0$

$\Rightarrow θ \in [0, π]$ $\Rightarrow 0 \leq π x - \frac{π}{4} \leq π \Rightarrow x \in [\frac{1}{4}, \frac{5}{4}]$

$\Rightarrow [α, β] = [\frac{1}{4}, \frac{5}{4}]$ $∴$ $β - α = \frac{5}{4} - \frac{1}{4} = 1$

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