Key Concepts and Formulas

• Reciprocal Relationship: Two numbers, $a$ and $b$, are reciprocals if their product is 1, i.e., $ab = 1$.
• Difference of Squares: $(a+b)(a-b) = a^2 - b^2$
• Quadratic Formula: The solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

1. Identify Reciprocal Terms We are given the equation $(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10$. We need to check if the bases $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$ are reciprocals. We compute their product: $(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})$ Using the difference of squares formula: $= (\sqrt{3})^2 - (\sqrt{2})^2$ $= 3 - 2$ $= 1$ Since their product is 1, the bases are reciprocals. This is important because it allows us to simplify the equation.

2. Introduce Substitution Let $y = (\sqrt{3}+\sqrt{2})^x$. This substitution will transform the exponential equation into a more manageable algebraic equation. Since $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$ are reciprocals, we have $(\sqrt{3}-\sqrt{2}) = \frac{1}{(\sqrt{3}+\sqrt{2})}$. Therefore: $(\sqrt{3}-\sqrt{2})^x = \left(\frac{1}{(\sqrt{3}+\sqrt{2})}\right)^x = \frac{1}{(\sqrt{3}+\sqrt{2})^x}$ Using our substitution $y = (\sqrt{3}+\sqrt{2})^x$, we get: $(\sqrt{3}-\sqrt{2})^x = \frac{1}{y}$

3. Formulate the Quadratic Equation Substitute $y$ and $\frac{1}{y}$ back into the original equation: $y + \frac{1}{y} = 10$ Now we have an equation solely in terms of $y$, which we can solve. To eliminate the fraction, multiply the entire equation by $y$ (since $y$ is always positive, $y \neq 0$): $y \left(y + \frac{1}{y}\right) = 10y$ $y^2 + 1 = 10y$ Rearrange the terms to form a standard quadratic equation: $y^2 - 10y + 1 = 0$ This is a standard quadratic equation in the form $ay^2 + by + c = 0$, which can be solved using the quadratic formula.

4. Solve the Quadratic Equation for $y$ Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, with $a=1$, $b=-10$, and $c=1$: $y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$ $y = \frac{10 \pm \sqrt{100 - 4}}{2}$ $y = \frac{10 \pm \sqrt{96}}{2}$ Simplify the square root: $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$. Substitute this back into the expression for $y$: $y = \frac{10 \pm 4\sqrt{6}}{2}$ Divide by 2: $y = 5 \pm 2\sqrt{6}$ We have found the two possible values for $y$. These values must be positive because $y = (\sqrt{3}+\sqrt{2})^x$ must be positive for real $x$. Both $5 + 2\sqrt{6}$ and $5 - 2\sqrt{6}$ are positive.

5. Back-Substitute to Find $x$ Now we use the values of $y$ to find the corresponding values of $x$ using our original substitution, $y = (\sqrt{3}+\sqrt{2})^x$.

Case 1: $y = 5 + 2\sqrt{6}$ $(\sqrt{3}+\sqrt{2})^x = 5 + 2\sqrt{6}$ We need to express the right-hand side in terms of the base $(\sqrt{3}+\sqrt{2})$ to solve for $x$. Let's calculate the square of the base: $(\sqrt{3}+\sqrt{2})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$ So, the equation becomes: $(\sqrt{3}+\sqrt{2})^x = (\sqrt{3}+\sqrt{2})^2$ Since the base $(\sqrt{3}+\sqrt{2})$ is greater than 1, the exponential function is one-to-one. Therefore, we can equate the exponents: $x = 2$

Case 2: $y = 5 - 2\sqrt{6}$ $(\sqrt{3}+\sqrt{2})^x = 5 - 2\sqrt{6}$ Similar to Case 1, we need to express $5 - 2\sqrt{6}$ in terms of the base $(\sqrt{3}+\sqrt{2})$. We know that $\sqrt{3}-\sqrt{2} = (\sqrt{3}+\sqrt{2})^{-1}$. Let's square this reciprocal term: $(\sqrt{3}-\sqrt{2})^2 = \left((\sqrt{3}+\sqrt{2})^{-1}\right)^2 = (\sqrt{3}+\sqrt{2})^{-2}$ Let's also calculate $(\sqrt{3}-\sqrt{2})^2$ directly: $(\sqrt{3}-\sqrt{2})^2 = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}$ So, the equation becomes: $(\sqrt{3}+\sqrt{2})^x = (\sqrt{3}+\sqrt{2})^{-2}$ Again, because the base is greater than 1, we can equate the exponents: $x = -2$

6. Identify the Set S and Its Cardinality The set $\mathbf{S}$ is defined as $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. We found two real solutions for $x$: $2$ and $-2$. Therefore, the set $\mathbf{S} = \{2, -2\}$. The number of elements in $\mathbf{S}$ is the count of distinct solutions, which is 2.

Common Mistakes & Tips

• Tip: Always check if terms in exponential equations are reciprocals. Calculating their product is an efficient way to verify this.
• Mistake: Forgetting that the base of an exponential function, when forming a substitution like $y=a^x$, must be positive. Ensure the values obtained for $y$ from the quadratic equation are also positive.
• Tip: When solving $b^x = b^y$ where $b>0$ and $b \neq 1$, you can directly equate the exponents ($x=y$) because the exponential function is one-to-one.

Summary

The given exponential equation can be transformed into a quadratic equation by recognizing the reciprocal nature of its bases and using substitution. Solving the quadratic yields two positive values for the substituted variable. Each of these values, when back-substituted, leads to a unique real solution for $x$ due to the properties of exponential functions. Consequently, the set $\mathbf{S}$ contains two distinct elements.

Final Answer The final answer is $\boxed{2}$, which corresponds to option (C).