Question

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

Answer 1

Hint: Let, $\pi x - \dfrac{\pi }{4} = \theta $ and then solve the question using $\left\{ {x \in \left[ {0,2} \right]:f\left( x \right) \geqslant 0} \right\} = \left[ {\alpha ,\beta } \right]$ where you have to consider $f\left( x \right) \geqslant 0$ . In last find the value of $x$ to know the values of $\alpha ,\beta $and put the required values in $\beta - \alpha $.

Formula Used:
Trigonometric formula –
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
$\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta $

Complete step by step solution:
 Given that,
$f(x) = \left( {3 - \sin \left( {2\pi x} \right)} \right)\sin \left( {\pi x - \dfrac{\pi }{4}} \right) - \sin \left( {3\pi x - \dfrac{\pi }{4}} \right)$
Let, $\pi x - \dfrac{\pi }{4} = \theta $ OR $\pi x = \dfrac{\pi }{4} + \theta $
Now, $\left\{ {x \in \left[ {0,2} \right]:f\left( x \right) \geqslant 0} \right\} = \left[ {\alpha ,\beta } \right] - - - - - - (1)$
$f\left( x \right) \geqslant 0$
$ \Rightarrow \left( {3 - \sin \left( {2\pi x} \right)} \right)\sin \left( {\pi x - \dfrac{\pi }{4}} \right) - \sin \left( {3\pi x - \dfrac{\pi }{4}} \right) \geqslant 0$
$ \Rightarrow \left( {3 - \sin \left( {2\left( {\dfrac{\pi }{4} + \theta } \right)} \right)} \right)\sin \theta - \sin \left( {3\left( {\dfrac{\pi }{4} + \theta } \right) - \dfrac{\pi }{4}} \right) \geqslant 0$
$ \Rightarrow \left( {3 - \sin \left( {\dfrac{\pi }{2} + 2\theta } \right)} \right)\sin \theta - \sin \left( {\dfrac{{3\pi }}{4} + \theta - \dfrac{\pi }{4}} \right) \geqslant 0$
$ \Rightarrow 3\sin \theta - \sin \theta \sin \left( {\dfrac{\pi }{2} + 2\theta } \right) - \sin \left( {\pi + 3\theta } \right) \geqslant 0$
$ \Rightarrow 3\sin \theta - \sin \theta \cos 2\theta + \sin 3\theta \geqslant 0$
$ \Rightarrow 3\sin \theta - \sin \theta \left( {1 - 2{{\sin }^2}\theta } \right) + \left( {3\sin \theta - 4{{\sin }^3}\theta } \right) \geqslant 0$
$ \Rightarrow \sin \theta \left[ {3 - \left( {1 - 2{{\sin }^2}\theta } \right) + 3 - 4{{\sin }^2}\theta } \right] \geqslant 0$
$ \Rightarrow \sin \theta \left[ {5 - 2{{\sin }^2}\theta } \right] \geqslant 0$
$ \Rightarrow \sin \theta \left[ {5 - 2\left( {\dfrac{{1 - \cos 2\theta }}{2}} \right)} \right] \geqslant 0$
$ \Rightarrow \sin \theta \left[ {4 + \cos 2\theta } \right] \geqslant 0$
$ \Rightarrow \sin \theta \geqslant 0$
$\therefore \theta \in \left[ {0,\pi } \right]$
$0 \leqslant \theta \leqslant \pi $
$0 \leqslant \pi x - \dfrac{\pi }{4} \leqslant \pi $
$\dfrac{\pi }{4} \leqslant \pi x \leqslant \pi + \dfrac{\pi }{4}$
$\dfrac{\pi }{4} \leqslant \pi x \leqslant \dfrac{{5\pi }}{4}$
$\dfrac{1}{4} \leqslant x \leqslant \dfrac{5}{4}$
$x \in \left[ {\dfrac{1}{4},\dfrac{5}{4}} \right] - - - - - (2)$
From equation (1) and (2)
$\alpha = \dfrac{1}{4},\beta = \dfrac{5}{4}$
Hence, the value of $\beta - \alpha $ is $1$.

Note: The key concept involved in solving this problem is the good knowledge of Trigonometry formula, ratio, and identities. Students must remember that in which quadrant the sign will change and in which the trigonometric function will change.