Analyzing the Equation $\sin(e^x) = 5x + 5 - x^2$

We need to find the number of real solutions for the equation $\sin(e^x) = 5x + 5 - x^2$. Let's denote the left-hand side as $f(x)$ and the right-hand side as $g(x)$. $f(x) = \sin(e^x)$ $g(x) = 5x + 5 - x^2 = -x^2 + 5x + 5$

Analyzing the Left Hand Side (LHS): $f(x) = \sin(e^x)$

The sine function, $\sin(\cdot)$, always outputs values between -1 and 1. Therefore, the range of $f(x)$ is $[-1, 1]$. $ -1 \le \sin(e^x) \le 1 $ for all real $x$. The term $e^x$ is always positive ($e^x > 0$). As $x$ varies, $e^x$ spans $(0, \infty)$. The function $\sin(e^x)$ will oscillate infinitely often between -1 and 1 as $x$ increases. As $x \to -\infty$, $e^x \to 0^+$, so $f(x) \to \sin(0) = 0$.

Analyzing the Right Hand Side (RHS): $g(x) = -x^2 + 5x + 5$

This is a quadratic function representing a parabola opening downwards. The vertex occurs at $x = -b/(2a) = -5/(2 \times -1) = 5/2 = 2.5$. The maximum value is $g(2.5) = -(2.5)^2 + 5(2.5) + 5 = -6.25 + 12.5 + 5 = 11.25$. The function $g(x)$ is symmetric around $x=2.5$.

Finding Potential Solution Ranges

For a solution to exist, the value of $g(x)$ must be within the range of $f(x)$, i.e., between -1 and 1. So, we must have: $ -1 \le g(x) \le 1 $ $ -1 \le -x^2 + 5x + 5 \le 1 $ We need to solve two inequalities: 1. $ -x^2 + 5x + 5 \le 1 $ $ -x^2 + 5x + 4 \le 0 $ $ x^2 - 5x - 4 \ge 0 $ To find the roots of $x^2 - 5x - 4 = 0$, we use the quadratic formula: $ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-4)}}{2(1)} = \frac{5 \pm \sqrt{25 + 16}}{2} = \frac{5 \pm \sqrt{41}}{2} $ Let $x_1 = \frac{5 - \sqrt{41}}{2} \approx -0.70156$ and $x_2 = \frac{5 + \sqrt{41}}{2} \approx 5.70156$. Since the parabola $x^2 - 5x - 4$ opens upwards, the inequality $x^2 - 5x - 4 \ge 0$ holds when $x \le x_1$ or $x \ge x_2$. $ x \le \frac{5 - \sqrt{41}}{2} $ or $ x \ge \frac{5 + \sqrt{41}}{2} $ 2. $ -x^2 + 5x + 5 \ge -1 $ $ -x^2 + 5x + 6 \ge 0 $ $ x^2 - 5x - 6 \le 0 $ Factoring the quadratic: $ (x-6)(x+1) \le 0 $. This inequality holds when $ -1 \le x \le 6 $. Combining the conditions from both inequalities: We need $x$ such that $ (-1 \le x \le 6) $ AND ($x \le \frac{5 - \sqrt{41}}{2}$ or $x \ge \frac{5 + \sqrt{41}}{2}$). The possible intervals for solutions are: $ I_1 = [-1, \frac{5 - \sqrt{41}}{2}] \approx [-1, -0.70156] $ $ I_2 = [\frac{5 + \sqrt{41}}{2}, 6] \approx [5.70156, 6] $

Counting Solutions in Interval 1 ($I_1$)

In the interval $ I_1 \approx [-1, -0.70156] $: - $g(x)$ increases from $g(-1) = -1$ to $g(\frac{5 - \sqrt{41}}{2}) = 1$. - $f(x) = \sin(e^x)$. At $x = -1$, $e^x = e^{-1} \approx 0.3678$. $f(-1) = \sin(e^{-1}) \approx \sin(0.3678) \approx 0.358$. At $x = \frac{5 - \sqrt{41}}{2} \approx -0.70156$, $e^x = e^{-0.70156} \approx 0.4955$. $f(\frac{5 - \sqrt{41}}{2}) \approx \sin(0.4955) \approx 0.478$. - Let $h(x) = g(x) - f(x)$. We are looking for roots of $h(x)$. $h(-1) = g(-1) - f(-1) = -1 - 0.358 = -1.358$. (Negative) $h(\frac{5 - \sqrt{41}}{2}) = g(\frac{5 - \sqrt{41}}{2}) - f(\frac{5 - \sqrt{41}}{2}) = 1 - 0.478 = 0.522$. (Positive) - The derivative $h'(x) = g'(x) - f'(x)$. $g'(x) = -2x + 5$. For $x \in [-1, -0.7]$, $g'(x)$ is positive (approx 7 to 6.4). $f'(x) = e^x \cos(e^x)$. For $x \in [-1, -0.7]$, $e^x \in [0.3678, 0.4955]$. This range is within $(0, \pi/2)$, so $\cos(e^x)>0$. $f'(x)$ is positive (approx 0.34 to 0.44). - $h'(x) = (\text{positive, decreasing}) - (\text{positive, increasing})$. $h'(x)$ remains positive throughout the interval $I_1$. - Since $h(x)$ is continuous, starts negative ($h(-1)<0$), ends positive ($h(x_1)>0$), and is strictly increasing ($h'(x)>0$), there is exactly one real solution in the interval $I_1$.

Counting Solutions in Interval 2 ($I_2$)

In the interval $ I_2 \approx [5.70156, 6] $: - $g(x)$ decreases from $g(\frac{5 + \sqrt{41}}{2}) = 1$ to $g(6) = -1$. - $f(x) = \sin(e^x)$. At $x \approx 5.70156$, $e^x \approx e^{5.70156} \approx 299.54$. $f(x) \approx \sin(299.54) \approx -0.87$. At $x = 6$, $e^x = e^6 \approx 403.43$. $f(x) \approx \sin(403.43) \approx 0.95$. - Let $h(x) = g(x) - f(x)$. $h(\frac{5 + \sqrt{41}}{2}) = g(x) - f(x) \approx 1 - (-0.87) = 1.87$. (Positive) $h(6) = g(6) - f(6) = -1 - 0.95 = -1.95$. (Negative) - In the interval $I_2$, $e^x$ increases from approx $299.54$ to $403.43$. This range covers approximately $\frac{403.43 - 299.54}{2\pi} \approx 16.5$ full cycles of the sine function. - Since $g(x)$ decreases monotonically from 1 to -1 across $I_2$, and $f(x)$ oscillates rapidly between -1 and 1, $f(x)$ is expected to cross $g(x)$ multiple times. - $h(x)$ starts positive and ends negative. Its derivative $h'(x) = g'(x) - f'(x)$ involves $g'(x)$ (negative) and $f'(x) = e^x \cos(e^x)$ (oscillating rapidly with large magnitude). This wild oscillation of $h'(x)$ implies $h(x)$ also oscillates wildly. - Given that $h(x)$ starts positive and ends negative, and oscillates rapidly, it must cross the x-axis multiple times within $I_2$. It is highly likely that there are many solutions in this interval (likely greater than 2).

Conclusion on the Number of Solutions

Our analysis shows: - There is exactly 1 solution in the interval $I_1 = [-1, \frac{5 - \sqrt{41}}{2}]$. - There are multiple solutions (likely many) in the interval $I_2 = [\frac{5 + \sqrt{41}}{2}, 6]$. Therefore, the total number of real solutions is $1 + (\text{multiple}) \ge 2$. Based on this analysis, the number of solutions is not 0, 1, or 2. The options provided do not seem to perfectly match this analysis, but the existence of solutions in both intervals is clear.