Find \[\alpha \] and \[\beta \] so that the function \[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {2\sin x,}&{{\text{for}}\, - \pi \leqslant x \leqslant - \dfrac{\pi }{2}} \\
  {\alpha \sin x + \beta ,}&{{\text{for}}\, - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}} \\
  {\cos x,}&{{\text{for}}\,\dfrac{\pi }{2} \leqslant x \leqslant \pi }
\end{array}} \right\}\]

Hint: We are asked to find the values of \[\alpha \] and \[\beta \] . The value of the function is given for different intervals, using suitable limits on the function form two equations having the terms \[\alpha \] and \[\beta \] . Then solve those equations to find the value of \[\alpha \] and \[\beta \] .

Complete step-by-step answer:
Given, the function \[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {2\sin x,}&{{\text{for}}\, - \pi \leqslant x \leqslant - \dfrac{\pi }{2}} \\
  {\alpha \sin x + \beta ,}&{{\text{for}}\, - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}} \\
  {\cos x,}&{{\text{for}}\,\dfrac{\pi }{2} \leqslant x \leqslant \pi }
\end{array}} \right\}\]
Let us first form the equations to find the value of \[\alpha \] and \[\beta \] .
Taking the limit \[x = - \dfrac{\pi }{2}\] we find the value of \[f(x)\] for \[2\sin x\] and \[\alpha \sin x + \beta \] .
 \[f\left( { - \dfrac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to - \dfrac{\pi }{2}} 2\sin x = 2\sin \left( { - \dfrac{\pi }{2}} \right) = - 2\] (i)
 \[f\left( { - \dfrac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to - \dfrac{\pi }{2}} \alpha \sin x + \beta = \alpha \sin \left( { - \dfrac{\pi }{2}} \right) + \beta = - \alpha + \beta \] (ii)
Equating (i) and (ii), we have
 \[ - \alpha + \beta = - 2\] (iii)
Taking the limit \[x = \dfrac{\pi }{2}\] we find the value of \[f(x)\] for \[\alpha \sin x + \beta \] and \[\cos x\] .
 \[f\left( {\dfrac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \alpha \sin x + \beta = \alpha \sin \left( {\dfrac{\pi }{2}} \right) + \beta = \alpha + \beta \] (iv)
 \[f\left( {\dfrac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \cos x = \cos \left( {\dfrac{\pi }{2}} \right) = 0\] (v)
Equating (iv) and (v), we get
 \[\alpha + \beta = 0\] (v)
Now, adding equations (iii) and (v), we get
 \[- \alpha + \beta + \alpha + \beta = - 2 + 0 \\
   \Rightarrow 2\beta = - 2 \\
   \Rightarrow \beta = - 1 \]
Now, putting the value of \[\beta \] in equation (v), we get
 \[\alpha + \left( { - 1} \right) = 0 \\
   \Rightarrow \alpha = 1 \]
Therefore, the values of \[\alpha \] and \[\beta \] are \[1\] and \[ - 1\] respectively.
So, the correct answer is “ \[\alpha \] and \[\beta \] are \[1\] and \[ - 1\] ”.

Note: In such types of questions, where the function is given in terms of intervals use suitable limits to find the value of the function at that point. For solving equations always check the number of variables you need to find out, when the number of equations is equal to the number of variables only then you can find out the variables. Suppose here there were two variables you need to find out so we form two equations to find them. If there are three variables then you need at least three equations to find the value of those variables. So, when you are asked to find the values of some variables check whether you can form as many equations as the number of variables.