Let $f:[0,2] \rightarrow \mathbb{R}$ be the function defined by $$f(x)=(3-\sin(2\pi x))\sin\left(\pi x-\frac{\pi}{4}\right)-\sin\left(3\pi x+\frac{\pi}{4}\right)$$ If $\alpha, \beta \in [0,2]$ are such that $\{x \in [0,2]: f(x) \geq 0\}=[\alpha, \beta]$, then the value of $\beta-\alpha$ is ____

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Trigonometric Equations

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})

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 ____

Answer: 1

Let

θ = π x - \frac{π}{4}

, so

θ \in [- \frac{π}{4}, \frac{7 π}{4}]

.

The inequality becomes:

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

(3 - cos 2 θ) sin θ \geq - sin 3 θ

sin θ (3 - cos 2 θ) + sin 3 θ \geq 0

sin θ (3 - 4 sin^{2} θ + 3 - cos 2 θ) \geq 0

sin θ (6 - 2 (1 - cos 2 θ) - cos 2 θ) \geq 0

sin θ (4 + cos 2 θ) \geq 0

Since

cos 2 θ \geq - 1

, we have

4 + cos 2 θ \geq 3 > 0

.

Therefore,

sin θ \geq 0

, which gives

θ \in [0, π]

.

This means

0 \leq π x - \frac{π}{4} \leq π

, so

\frac{1}{4} \leq x \leq \frac{5}{4}

.

Thus

α = \frac{1}{4}

β = \frac{5}{4}

, and

β - α = 1

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer