[GK2015,National I,12]Integers and derivatives

Let $f(x,a) = e^x(2x-1) - ax+a$ . Consider $a=1$ , then $f(x,1) = e^x(2x-1) - x+1$ and $f'(x,1) = e^x(2x-1) + 2e^x - 1$ . $f'(x,1) = 0$ when $x=0$ . Since $f''(0,1) > 0$ , $\min (f(x,1)) = f(0,1) = 0$ , implying $f(x,1) \ge 0$ for all $x$ . Since if we reduce the value of $a$ , we are shifting the curve $f(x,a)$ downward, for $f(x,a) < 0,$ , $a$ must be less than 1 or $a < 1$ and the integer $x_0$ must be 0 or $x_0 = 0$ .

Now consider $f(x,0) = e^x(2x-1)$ . We note that $f(x,0) = 0$ , when $x \to -\infty$ and $x = \frac 12$ . This means that $f(x,0) < 0$ for all $x < \frac 12$ . For $f(x_0, a)$ to have only one integer solution, we need $f(-1,a) \ge 0$ or $e^{-1}(-2-1) +a + a \ge 0$ $\implies a \ge \dfrac 3{2e}$ . Note that $f \left(0, \frac 3{2e}\right) < 0$ but $f \left(1, \frac 3{2e}\right) > 0$ , therefore there is only one $x_0$ which is 0.

Therefore, $a \in \boxed{\left[\dfrac 3{2e}, 1\right)}$ .

I only had to determine which answer best fit the situation.

$\text{Plot3D}\left[\left\{-a x+a+e^x (2 x-1),0\right\},\left\{a,\frac{3}{2 e},1\right\},\left\{x,-1,\frac{1}{2}\right\}, \\ \text{AxesLabel}\to \{\text{a},\text{x},\text{f}\},\text{PlotStyle}\to \{\text{Orange},\{\text{Opacity}[0.3],\text{Blue}\}\},\text{ClippingStyle}\to \left( \begin{array}{cc} \text{Opacity}[0.2] & \text{Magenta} \\ \end{array} \right)\right]$

The orange surface is the function for various values of x and a. The blue plane marks the function = 0 values. The magenta areas are where the function values are outside the plot volume.