График функции \(f\left( x \right) = a\cos x + b\) проходит через точки \(\left( {0; — 1} \right)\) и \(\left( {\frac{\pi }{2};1} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 1 = a\cos 0 + b}\\{ 1 = a\cos \frac{\pi }{2} + b}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{ — 1 = a \cdot 1 + b}\\{1 = a \cdot 0 + b}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a + b = — 1}\\{b = 1\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a = — 2.\)
Ответ: – 2.