Question

Let $f:\left[0,2\right]\to R$ be the function defined by

$f\left(x\right)=\left(3-\sin \left(2\mathrm{\pi x}\right)\right)\sin \left(\mathrm{\pi x}-\frac{\mathrm{\pi }}{4}\right)-\sin \left(3\mathrm{\pi x}+\frac{\mathrm{\pi }}{4}\right)$

If , $\alpha ,\beta \in \left[0,2\right]$ are such that $\left\{x\in \left[0,2\right]:f\left(x\right)\ge 0\right\}=\left[\alpha ,\beta \right]$, then the value of $\beta -\alpha$ is _____

Answer 1

Finding the value of $\beta -\alpha$

Given, the function is:

$f\left(x\right)=\left(3-\sin \left(2\mathrm{\pi x}\right)\right)\sin \left(\mathrm{\pi x}-\frac{\mathrm{\pi }}{4}\right)-\sin \left(3\mathrm{\pi x}+\frac{\mathrm{\pi }}{4}\right)$

Let,

$$\begin{gathered} \Rightarrow \mathrm{\pi x}-\frac{\mathrm{\pi }}{4}=\mathrm{\theta } \\ \Rightarrow f\left(x\right)\ge 0 \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow \left(3-\sin \left(2\mathrm{\theta }+\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{sin\theta }-\sin \left(3\left(\mathrm{\theta }+\frac{\mathrm{\pi }}{4}\right)+\mathrm{\pi }\right)+\mathrm{\pi }\ge 0 \\ \Rightarrow \left(3-\sin \left(2\mathrm{\theta }+\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{sin\theta }-\sin \left(3\mathrm{\theta }+\mathrm{\pi }\right)+\mathrm{\pi }\ge 0 \\ \Rightarrow 3\mathrm{sin\theta }-\cos 2\mathrm{\theta }\mathrm{sin\theta }+\sin 3\mathrm{\theta }\ge 0......\left(1\right) \end{gathered}$$

we know that,

$$\begin{gathered} \sin 3\theta =3\sin \theta -4{\sin }^3\theta \\ \cos 2\theta =1-2{\sin }^2\theta \end{gathered}$$

Substituting the above values in the Equation $\left(1\right)$:

$$\begin{gathered} \Rightarrow 3\sin \theta -\left(1-2{\sin }^2\theta \right)\sin \theta +\left(3\sin \theta -4{\sin }^3\theta \right)\ge 0 \\ \Rightarrow 3\sin \theta -\sin \theta +2{\sin }^3\theta +3\sin \theta -4{\sin }^3\theta \ge 0 \\ \Rightarrow 5\sin \theta -2{\sin }^3\theta \ge 0 \\ \Rightarrow \sin \theta \left(5-2{\sin }^2\theta \right)\ge 0 \end{gathered}$$

Therefore,

$$\begin{gathered} \Rightarrow 5-2{\sin }^2\mathrm{\theta }\ge 0 \\ \Rightarrow 2{\sin }^2\mathrm{\theta }=5 \\ \Rightarrow {\sin }^2\mathrm{\theta }=\frac{5}{2} \end{gathered}$$

So that we consider

$$\begin{gathered} \Rightarrow \sin \theta \ge 0\to x\in \left[0,2\right] \\ \Rightarrow \mathrm{\pi x}-\frac{\mathrm{\pi }}{4}\in \left[-\frac{\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right] \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow \theta \in \left[0,\mathrm{\pi }\right] \\ \Rightarrow \mathrm{\pi x}-\frac{\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right] \\ \Rightarrow \mathrm{x}\in \left[\frac{1}{4},\frac{5}{4}\right] \end{gathered}$$

Considering the above value as,

$$\begin{gathered} \alpha =\frac{1}{4} \\ \beta =\frac{5}{4} \end{gathered}$$

Then the value of $\beta -\alpha$ will be,

$$\begin{gathered} \beta -\alpha =\frac{5}{4}-\frac{1}{4} \\ \beta -\alpha =\frac{5-1}{4} \\ \beta -\alpha =\frac{4}{4} \\ \beta -\alpha =1 \end{gathered}$$

Hence, the correct value of $\beta -\alpha$ is equal to $1$.