График функции \(f\left( x \right) = a\sin x + b\) проходит через точки \(\left( {0;1} \right)\) и \(\left( {\dfrac{\pi }{2}; — \dfrac{3}{2}} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{1 = a\sin 0 + b}\\{ — \dfrac{3}{2} = a\sin \dfrac{\pi }{2} + b}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{1 = a \cdot 0 + b}\\{ — \dfrac{3}{2} = a \cdot 1 + b\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = 1\,\,\,\,\,\,\,}\\{a + b = — \dfrac{3}{2}}\end{array}} \right.} \right.} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a = — \dfrac{5}{2}.\)
Ответ: – 2,5.
Задача 12. На рисунке изображён график функции \(f\left( x \right) = a\,\,\sin x + b.\) Найдите a.