График функции \(f\left( x \right) = a\sin x + b\) проходит через точки \(\left( {0;2} \right)\) и \(\left( {\frac{\pi }{2};\frac{1}{2}} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{2 = a\sin 0 + b}\\{\frac{1}{2} = a\sin \frac{\pi }{2} + b}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2 = a \cdot 0 + b}\\{\frac{1}{2} = a + b\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = 2\,\,\,\,\,\,\,}\\{a + b = \frac{1}{2}}\end{array}} \right.} \right.} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a = — \frac{3}{2}.\)
Ответ: – 1,5.
Задача 11. На рисунке изображён график функции \(f\left( x \right) = a\,\,\sin x + b.\) Найдите a.