- Home
- Standard 11
- Mathematics
माना $f:[0,2] \rightarrow R$ एक फलन है जो
$f(x)=(3-\sin (2 \pi x)) \sin \left(\pi x-\frac{\pi}{4}\right)-\sin \left(3 \pi x+\frac{\pi}{4}\right)$
द्वारा परिभाषित है। यदि $\alpha, \beta \in[0,2]$ इस प्रकार है कि $\{ x \in[0,2]: f( x ) \geq 0\}=[\alpha, \beta]$ हो, तो $\beta-\alpha$ का मान होगा
$0$
$1$
$5$
$6$
Solution
Let $\pi x-\frac{\pi}{4}=\theta \in\left[\frac{-\pi}{4}, \frac{7 \pi}{4}\right]$
$\text { So, } \quad\left(3-\sin \left(\frac{\pi}{2}+2 \theta\right)\right) \sin \theta \geq \sin (\pi+3 \theta)$
$\Rightarrow \quad(3-\cos 2 \theta) \sin \theta \geq-\sin 3 \theta$
$\Rightarrow \quad \sin \theta\left[3-4 \sin ^2 \theta+3-\cos 2 \theta\right] \geq 0$
$\Rightarrow \quad \sin \theta(6-2(1-\cos 2 \theta)-\cos 2 \theta) \geq 0$
$\Rightarrow \quad \sin \theta(4+\cos 2 \theta) \geq 0$
$\Rightarrow \quad \sin \theta \geq 0$
$\Rightarrow \quad \theta \in[0, \pi] \Rightarrow 0 \leq \pi x-\frac{\pi}{4} \leq \pi$
$\Rightarrow \quad x \in\left[\frac{1}{4}, \frac{5}{4}\right]$
$\Rightarrow \quad \beta-\alpha=1$
