已知函数\(f\left ( {x} \right )=2x+a{x}^{2}+b\mathrm{cos}x\)在点 \((\frac{\pi }{2},f(\frac{\pi }{2}))\) 处的切线方程为 \(y = \frac{{3\pi }}{4}\) ．
（Ⅰ）求 \(a\) ， \(b\) 的值，并讨论 \(f(x)\) 在 \([0,\frac{\pi }{2}]\) 上的增减性；
（Ⅱ）若 \(f({x_1}) = f({x_2})\) ，且 \(0 < {x_1} < {x_2} < \pi \) ，求证： \(f'(\frac{{{x_1} + {x_2}}}{2}) < 0\) ．
（参考公式：\(\mathrm{cos}\theta -\mathrm{cos}φ=-2\mathrm{sin}\frac {\theta +φ} {2}\mathrm{sin}\frac {\theta -φ} {2}\) ）
（Ⅰ）由题意知\({f}^{\, '}\left ( {x} \right )=2+2ax-b\mathrm{sin}x\) ， \(\therefore \) \(\left\{ {\begin{array}{*{20}{l}} {f'(\frac{\pi }{2}) = 0} \\\ {f(\frac{\pi }{2}) = \frac{{3\pi }}{4}} \end{array}} \right.\) 解得 \(\left\{ {\begin{array}{*{20}{l}} {a = - \frac{1}{\pi }} \\\ {b = 1} \end{array}} \right.\)
故\(f\left ( {x} \right )=2x-\frac {1} {\pi }{x}^{2}+\mathrm{cos}x\)， \({f}^{\, '}\left ( {x} \right )=2-\frac {2} {\pi }x-\mathrm{sin}x\) ．当 \(0⩽x⩽\frac{\pi }{2}\) 时， \(f'(x)\) 为减函数，且 \(f'(\frac{\pi }{2}) = 0\) ，
\(\therefore f'(x) > 0\) ， \(f(x)\) 为增函数．
（Ⅱ）证明：由 \(f({x_1}) = f({x_2})\) ，得\( 2{x}_{1}-\frac{{x}_{1}^{2}}{\pi }+\mathrm{cos}{x}_{1}=2{x}_{2}-\frac{{x}_{2}^{2}}{\pi }+\mathrm{cos}{x}_{2}\mathrm{ }\)，
所以\( 2\left({x}_{1}-{x}_{2}\right)-\frac{1}{\pi }\left({x}_{1}+{x}_{2}\right)\left({x}_{1}-{x}_{2}\right)+\mathrm{cos}{x}_{1}-\mathrm{cos}{x}_{2}=0\)，两边同除以 \({x_1} - {x_2}\) ，得\( 2-\frac{1}{\pi }\left({x}_{1}+{x}_{2}\right)+\frac{\mathrm{cos}{x}_{1}-\mathrm{cos}{x}_{2}}{{x}_{1}-{x}_{2}}=0\) ，所以\( 2-\frac{1}{\pi }\left({x}_{1}+{x}_{2}\right)+\frac{-2\mathrm{sin}\frac{{x}_{1}+{x}_{2}}{2}\mathrm{sin}\frac{{x}_{1}-{x}_{2}}{2}}{{x}_{1}-{x}_{2}}=0\)，
令\({x_0} = \frac{{{x_1} + {x_2}}}{2}\) ，得\( 2-\frac{2}{\pi }{x}_{0}-\frac{2\mathrm{sin}{x}_{0}\mathrm{sin}\frac{{x}_{1}-{x}_{2}}{2}}{{x}_{1}-{x}_{2}}=0\)，得\( 2-\frac{2}{\pi }{x}_{0}=\frac{2\mathrm{sin}{x}_{0}\mathrm{sin}\frac{{x}_{1}-{x}_{2}}{2}}{{x}_{1}-{x}_{2}}\)．因为 \({f}^{\, '}\left ( {x} \right )=2-\frac {2x} {\pi }-\mathrm{sin}x\)，
所以\({f}^{\, '}\left ( {x} \right )=2-\frac {2} {\pi }{x}_{0}-\mathrm{sin}{x}_{0}=\frac {2\mathrm{sin}{x}_{0}\mathrm{sin}\frac {{x}_{1}-{x}_{2}} {2}} {{x}_{1}-{x}_{2}}-\mathrm{sin}{x}_{0}=\mathrm{sin}{x}_{0}\left ( {\frac {\mathrm{sin}\frac {{x}_{1}-{x}_{2}} {2}} {\frac {{x}_{1}-{x}_{2}} {2}}-1} \right )\) ，
因为\( \frac{\mathrm{sin}\frac{{x}_{1}-{x}_{2}}{2}}{\frac{{x}_{1}-{x}_{2}}{2}}=\frac{\mathrm{sin}\frac{{x}_{2}-{x}_{1}}{2}}{\frac{{x}_{2}-{x}_{1}}{2}}\)，又 \(\frac{{{x_2} - {x_1}}}{2} \in (0,\frac{\pi }{2})\) ，易知\( 0<\mathrm{sin}\frac{{x}_{2}-{x}_{1}}{2}<\frac{{x}_{2}-{x}_{1}}{2}\)，所以 \( \frac{\mathrm{sin}\frac{{x}_{1}-{x}_{2}}{2}}{\frac{{x}_{1}-{x}_{2}}{2}}-1<0\)，又 \({x_0} \in (0,\pi )\) ，所以\( \mathrm{sin}{x}_{0}>0\)，故 \(f'({x_0}) < 0\) ，得 \(f'(\frac{{{x_1} + {x_2}}}{2}) < 0\) ．
