Key Concepts and Formulas

• Substitution: Replacing a complex expression with a single variable to simplify the equation.
• Factoring: Expressing a polynomial as a product of simpler polynomials.
• Analyzing Cubic Equations: Using derivatives to determine the number of real roots of a cubic equation.

Step-by-Step Solution

Step 1: Substitution to Simplify the Equation

We are given the equation: $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$ To simplify this equation, we substitute $y = e^x$. Since $x$ is a real number, $y$ must be strictly positive ($y > 0$). Substituting $y$ into the equation yields a polynomial in $y$: $(e^x)^6 - (e^x)^4 - 2(e^x)^3 - 12(e^x)^2 + e^x + 1 = 0$ $y^6 - y^4 - 2y^3 - 12y^2 + y + 1 = 0$

Step 2: Strategic Algebraic Rearrangement

The goal is to rearrange the polynomial to reveal a simpler structure. We rewrite the original exponential equation into a more manageable form. Consider the terms that can form $(e^{3x}-1)^2$: $(e^{3x} - 1)^2 = e^{6x} - 2e^{3x} + 1$ We can rewrite the original equation by isolating these terms: $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$ Rearranging and grouping: $(e^{6x} - 2e^{3x} + 1) - e^{4x} + e^x - 12e^{2x} = 0$ $(e^{3x} - 1)^2 - e^{4x} + e^x - 12e^{2x} = 0$ Now, notice that the terms $-e^{4x} + e^x$ can be obtained by $-e^x(e^{3x} - 1)$: $-e^x(e^{3x} - 1) = -e^{4x} + e^x$ Substituting this into our rearranged equation: $(e^{3x} - 1)^2 - e^x(e^{3x} - 1) - 12e^{2x} = 0$ This transforms the original equation into a more manageable form: $(e^{3x} - 1)^2 - e^x(e^{3x} - 1) = 12e^{2x}$

Step 3: Transformation into a Quadratic Form

We now have the equation: $(e^{3x} - 1)^2 - e^x(e^{3x} - 1) = 12e^{2x}$ Since $e^x > 0$ for all real $x$, we also know $e^{2x} > 0$. We must check if $e^{3x}-1 = 0$ (which implies $x=0$) is a root of the original equation. Substituting $x=0$ into the original equation: $e^0 - e^0 - 2e^0 - 12e^0 + e^0 + 1 = 1 - 1 - 2 - 12 + 1 + 1 = -12 \neq 0$. Thus, $x=0$ is not a root, and $e^{3x} - 1 \neq 0$ when $x=0$. We can safely divide the equation by $e^{2x}$: $\frac{(e^{3x} - 1)^2}{e^{2x}} - \frac{e^x(e^{3x} - 1)}{e^{2x}} = \frac{12e^{2x}}{e^{2x}}$ This can be rewritten as: $\left(\frac{e^{3x} - 1}{e^x}\right)^2 - \frac{e^{3x} - 1}{e^x} = 12$ Let's simplify the term $\frac{e^{3x} - 1}{e^x}$: $\frac{e^{3x} - 1}{e^x} = \frac{e^{3x}}{e^x} - \frac{1}{e^x} = e^{2x} - e^{-x}$ Substituting this back, we get an equation in terms of $(e^{2x} - e^{-x})$: $(e^{2x} - e^{-x})^2 - (e^{2x} - e^{-x}) = 12$ This is a quadratic equation. Let $Z = e^{2x} - e^{-x}$: $Z^2 - Z = 12$ $Z^2 - Z - 12 = 0$ Factoring the quadratic equation: $(Z - 4)(Z + 3) = 0$ This yields two possible values for $Z$: $Z = 4$ or $Z = -3$.

Step 4: Solving for $x$ in Each Case We now solve for $x$ in each of the two cases derived from $Z$.

Case 1: $Z = 4$ Substituting back $Z = e^{2x} - e^{-x}$: $e^{2x} - e^{-x} = 4$ To work with positive exponents, multiply the entire equation by $e^x$ (which is always positive): $(e^{2x} \cdot e^x) - (e^{-x} \cdot e^x) = 4 \cdot e^x$ $e^{3x} - e^0 = 4e^x$ $e^{3x} - 1 = 4e^x$ Rearranging, we get a cubic equation in $e^x$: $e^{3x} - 4e^x - 1 = 0$ Let $u = e^x$. Since $e^x > 0$, we are looking for positive real roots ($u > 0$) of the cubic equation $u^3 - 4u - 1 = 0$. Let $f(u) = u^3 - 4u - 1$. To find the number of positive real roots, we examine its derivative: $f'(u) = 3u^2 - 4$. Setting $f'(u) = 0$ gives $3u^2 = 4$, so $u^2 = 4/3$, and $u = \pm 2/\sqrt{3}$. For $u > 0$, the critical point is $u = 2/\sqrt{3}$.

• For $0 < u < 2/\sqrt{3}$, $f'(u) < 0$, so $f(u)$ is decreasing.
• For $u > 2/\sqrt{3}$, $f'(u) > 0$, so $f(u)$ is increasing. The function $f(u)$ has a local minimum at $u = 2/\sqrt{3}$. The value of this minimum is $f(2/\sqrt{3}) = (2/\sqrt{3})^3 - 4(2/\sqrt{3}) - 1 = \frac{8}{3\sqrt{3}} - \frac{8}{\sqrt{3}} - 1 = \frac{8 - 24}{3\sqrt{3}} - 1 = \frac{-16}{3\sqrt{3}} - 1$, which is negative. As $u \to 0^+$, $f(u) \to -1$. As $u \to \infty$, $f(u) \to \infty$. Since $f(u)$ starts at $-1$, decreases to a negative minimum, and then increases to $\infty$, it must cross the u-axis exactly once for $u > 0$. Therefore, there is exactly one positive real root for $u^3 - 4u - 1 = 0$, which means there is exactly one real solution for $x$ in this case.

Case 2: $Z = -3$ Substituting back $Z = e^{2x} - e^{-x}$: $e^{2x} - e^{-x} = -3$ Multiply by $e^x$: $(e^{2x} \cdot e^x) - (e^{-x} \cdot e^x) = -3 \cdot e^x$ $e^{3x} - 1 = -3e^x$ Rearranging, we get: $e^{3x} + 3e^x - 1 = 0$ Let $u = e^x$. We are looking for positive real roots ($u > 0$) of the cubic equation $u^3 + 3u - 1 = 0$. Let $g(u) = u^3 + 3u - 1$. The derivative is $g'(u) = 3u^2 + 3$. For all real $u$, $g'(u) = 3(u^2 + 1) > 0$, meaning $g(u)$ is strictly increasing for all $u$. As $u \to 0^+$, $g(u) \to -1$. As $u \to \infty$, $g(u) \to \infty$. Since $g(u)$ is strictly increasing, starts at $-1$ for $u=0$, and goes to $\infty$, it must cross the u-axis exactly once for $u > 0$. Therefore, there is exactly one positive real root for $u^3 + 3u - 1 = 0$, which means there is exactly one real solution for $x$ in this case.

Step 5: Conclusion We found one real solution for $x$ from Case 1 and one real solution for $x$ from Case 2. Since $u=e^x$ is a one-to-one function, each unique positive root $u$ corresponds to a unique real root $x$. The total number of distinct real roots for the original equation is $1 + 1 = 2$.

Common Mistakes & Tips:

• Substitution Validity: Always remember that when substituting $y = e^x$, you are only interested in positive values of $y$. Discard any negative or zero roots found for the polynomial in $y$.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with exponents and negative signs. Double-check rearrangements and expansions.
• Division by Zero: Before dividing an equation by an expression, make sure that expression is not equal to zero. Check for the validity of the step for all $x$.

Summary

This problem involves transforming a complex exponential equation into a quadratic form using strategic algebraic manipulation and substitution. By analyzing the resulting cubic polynomials arising from the quadratic equation, we determine the number of positive real roots, which correspond to the number of real solutions for $x$. The equation has two real roots.

Final Answer

The final answer is \boxed{2}, which corresponds to option (A).