Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is an algebra problem, maybe there's a way to apply inequalities like Cauchy-Schwarz or AM-HM or AM-GM. Let me recall those.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can use that here. But the problem is that the expression we need to minimize has different coefficients in the denominators. So maybe I need to adjust the variables or split them in a way that allows me to apply AM-GM effectively.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that first. Let me set up the Lagrangian. Let me denote the function to minimize as $f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ and the constraint is $g(a, b, c, d) = a + b + c + d - 1 = 0$.

The Lagrangian is $ \mathcal{L} = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} - \lambda(a + b + c + d - 1) $.

Taking partial derivatives with respect to each variable and setting them equal to zero:

For $a$:
$$
\frac{\partial \mathcal{L}}{\partial a} = -\frac{1}{a^2} - \lambda = 0 \implies \lambda = -\frac{1}{a^2}
$$

For $b$:
$$
\frac{\partial \mathcal{L}}{\partial b} = -\frac{1}{b^2} - \lambda = 0 \implies \lambda = -\frac{1}{b^2}
$$

For $c$:
$$
\frac{\partial \mathcal{L}}{\partial c} = -\frac{4}{c^2} - \lambda = 0 \implies \lambda = -\frac{4}{c^2}
$$

For $d$:
$$
\frac{\partial \mathcal{L}}{\partial d} = -\frac{16}{d^2} - \lambda = 0 \implies \lambda = -\frac{16}{d^2}
$$

So from the first two equations, since both equal to $\lambda$, we have:
$$
-\frac{1}{a^2} = -\frac{1}{b^2} \implies \frac{1}{a^2} = \frac{1}{b^2} \implies a = b
$$
Similarly, from the first and third equations:
$$
-\frac{1}{a^2} = -\frac{4}{c^2} \implies \frac{1}{a^2} = \frac{4}{c^2} \implies c^2 = 4a^2 \implies c = 2a
$$
Since all variables are positive, we can ignore the negative root.

Similarly, from the first and fourth equations:
$$
-\frac{1}{a^2} = -\frac{16}{d^2} \implies \frac{1}{a^2} = \frac{16}{d^2} \implies d^2 = 16a^2 \implies d = 4a
$$

So now, we have relations between variables: $a = b$, $c = 2a$, $d = 4a$. Let me substitute these into the constraint equation $a + b + c + d = 1$:

Substituting:
$a + a + 2a + 4a = 1 \implies 8a = 1 \implies a = \frac{1}{8}$

Therefore, $a = b = \frac{1}{8}$, $c = 2a = \frac{1}{4}$, $d = 4a = \frac{1}{2}$

Now, let me compute the value of the original expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
$$

Calculating each term:

- $\frac{1}{1/8} = 8$
- Similarly, the second term is also 8
- $\frac{4}{1/4} = 16$
- $\frac{16}{1/2} = 32$

Adding them up: 8 + 8 + 16 + 32 = 64.

So according to this method, the minimum value is 64. But wait, let me check if this is indeed the minimum. Maybe I can verify using another method, like Cauchy-Schwarz or AM-GM.

Alternatively, maybe I can use the Cauchy-Schwarz inequality. Let me think. The Cauchy-Schwarz inequality in the form $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I need to relate the given expression to the constraint. Alternatively, maybe using the method of weighted AM-HM inequality?

Alternatively, let me think of splitting the variables. Since the coefficients in the denominators are different, maybe I can adjust variables to make them similar. For example, if I consider variables in such a way that the terms in the expression can be paired with the constraint terms.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me consider vectors. Suppose I have two vectors such that their dot product relates to the given expression. But I need to think more carefully.

Alternatively, maybe I can use the method of Lagrange multipliers again but check if there are other critical points. However, since the problem is convex? Wait, the function to minimize is convex? Let me check. The function $f(a,b,c,d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ is convex in each variable individually, but the sum of convex functions is convex. However, the constraint is linear. So the problem is a convex optimization problem, so the critical point found via Lagrange multipliers should be the global minimum. Therefore, 64 is the minimum.

But let me try another approach to confirm. Let me try using AM-GM inequality. Let me think about how to apply AM-GM here. The problem is that the terms in the expression have different coefficients. Maybe I can split the variables into parts so that I can apply AM-GM. For example, if I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$, perhaps I can think of them as multiple terms.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me recall that for positive reals, $(\sum \frac{x_i^2}{a_i}) \geq \frac{(\sum x_i)^2}{\sum a_i}$. But I need to relate this to the given expression. Let me think. Suppose I set up something like:

Let me consider the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me think of this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then, by Cauchy-Schwarz inequality:

$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2
$$

Wait, that's the Cauchy-Schwarz in the form $(\sum \frac{x_i^2}{a_i})(\sum a_i) \geq (\sum x_i)^2$. So here, if I take $x_1 = 1$, $x_2 = 1$, $x_3 = 2$, $x_4 = 4$, and $a_1 = a$, $a_2 = b$, $a_3 = c$, $a_4 = d$. Then:

$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2 = 8^2 = 64
$$

Therefore, since $a + b + c + d = 1$, we have:

$$
\left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \times 1 \geq 64
$$

Therefore, the expression is at least 64. And equality holds when $\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}$, that is, $\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$.

Let me check this condition. If equality holds, then:

From $\frac{1}{a} = \frac{1}{b}$, so $a = b$.

From $\frac{1}{a} = \frac{2}{c}$, so $c = 2a$.

From $\frac{1}{a} = \frac{4}{d}$, so $d = 4a$.

Which is exactly the same relations as found via Lagrange multipliers. Therefore, substituting back into the constraint $a + b + c + d = 1$ gives $a + a + 2a + 4a = 8a = 1$, so $a = 1/8$, and so on. Therefore, equality is achieved at these values, so the minimum value is indeed 64.

Therefore, both methods lead to the same result, so I can be confident that 64 is the correct minimum value.

But just to be thorough, let me test with some other values. Suppose I take different values for a, b, c, d that satisfy the constraint and see if the expression is larger than 64.

For example, let me take a = b = c = d = 1/4. Then the expression becomes:

$\frac{1}{1/4} + \frac{1}{1/4} + \frac{4}{1/4} + \frac{16}{1/4} = 4 + 4 + 16 + 64 = 88$, which is larger than 64.

Another test case: Let me take a = 1/8, b = 1/8, c = 1/4, d = 1/2 as found earlier. Then the expression is 8 + 8 + 16 + 32 = 64.

What if I take a different set? Let me try a = 1/10, b = 1/10, then c and d would need to satisfy 1/10 + 1/10 + c + d = 1 => c + d = 7/10. Let me choose c = 2a = 1/5, then d = 7/10 - 1/5 = 7/10 - 2/10 = 5/10 = 1/2. Then compute the expression:

1/a + 1/b + 4/c + 16/d = 10 + 10 + 4/(1/5) + 16/(1/2) = 20 + 20 + 32 = 72, which is still larger than 64.

Alternatively, suppose I take a = 1/9, b = 1/9, then c + d = 1 - 2/9 = 7/9. Let me take c = 2a = 2/9, then d = 7/9 - 2/9 = 5/9. Then compute the expression:

1/(1/9) + 1/(1/9) + 4/(2/9) + 16/(5/9) = 9 + 9 + 18 + (16*9)/5 = 27 + 18 + 144/5 = 45 + 28.8 = 73.8, which is still higher than 64.

Alternatively, let me try to take a different approach where maybe c or d are different. Suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. That gives 64. If I perturb slightly, say a = 0.12, b = 0.12, then c + d = 1 - 0.24 = 0.76. Let me take c = 2a = 0.24, then d = 0.76 - 0.24 = 0.52. Then compute the expression:

1/0.12 ≈ 8.333, same for 1/b, so total for a and b is ≈ 16.666. Then 4/c = 4 / 0.24 ≈ 16.666, and 16/d ≈ 16 / 0.52 ≈ 30.769. Total ≈ 16.666 + 16.666 + 30.769 ≈ 64.101, which is slightly above 64. So that seems to confirm that 64 is indeed the minimum.

Alternatively, if I take a = 1/8 + ε, b = 1/8 + ε, then c = 2a = 2*(1/8 + ε) = 1/4 + 2ε, d = 4a = 1/2 + 4ε. Then the constraint becomes:

a + b + c + d = (1/8 + ε) + (1/8 + ε) + (1/4 + 2ε) + (1/2 + 4ε) = 1/8 + 1/8 + 1/4 + 1/2 + ε + ε + 2ε + 4ε = (1/8 + 1/8 + 2/8 + 4/8) + 8ε = (8/8) + 8ε = 1 + 8ε. But the constraint is supposed to be 1, so we need to adjust. Therefore, if I take a = 1/8 + ε, then to maintain the constraint, I need to adjust other variables. Maybe this is getting too complicated, but the previous test case shows that perturbing slightly increases the value, so it's likely that 64 is indeed the minimum.

Alternatively, let me think if there's a way to use AM-GM directly. Let me consider splitting the variables. For example, since the coefficients in the denominators are 1, 1, 4, 16, which are squares of 1, 1, 2, 4. Maybe I can think of the terms as multiple variables. For example, if I consider the term $\frac{1}{a}$, maybe I can think of it as $\frac{1^2}{a}$, similarly for others. Then, using Cauchy-Schwarz as before. Which we already did, and that gives the same result.

Alternatively, maybe I can use the AM-GM inequality on the terms. Let me think. Suppose I want to relate the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$ with the constraint $a + b + c + d = 1$.

But since the terms are reciprocals, it's tricky. However, if I can write the expression as a sum of terms that can be related to the variables in the constraint. For example, if I consider variables multiplied by coefficients.

Alternatively, maybe I can use the method of substitution. Since a + b + c + d = 1, maybe express one variable in terms of others. But that might complicate things.

Alternatively, think of the problem as optimizing with weights. For example, the expression is similar to weighted harmonic mean. But I need to think of how to apply that.

Alternatively, think of the problem as minimizing $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ with a + b + c + d = 1. Let me consider using the method of Lagrange multipliers again, but maybe I can check if there are other critical points. However, since the Lagrangian method gave a unique critical point, and the function tends to infinity as any variable approaches 0, that critical point should be the global minimum.

Therefore, combining all these methods, I can be confident that the minimum value is 64.

But just to make sure, let me check another case where variables are not following the relations found earlier. For example, suppose I take a = 0.2, b = 0.2, c = 0.3, d = 0.3. Then a + b + c + d = 0.2 + 0.2 + 0.3 + 0.3 = 1.0. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.3 + 16/0.3 = 5 + 5 + 13.333... + 53.333... ≈ 5 + 5 + 13.33 + 53.33 ≈ 76.66, which is still higher than 64.

Another case: Let me take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Then the sum is 0.1 + 0.1 + 0.2 + 0.6 = 1.0. The expression is 1/0.1 + 1/0.1 + 4/0.2 + 16/0.6 = 10 + 10 + 20 + 26.666... ≈ 66.666..., which is still higher than 64.

Alternatively, take a = 0.125, b = 0.125, c = 0.25, d = 0.5, which is the critical point. Then the expression is 8 + 8 + 16 + 32 = 64.

So all these tests confirm that 64 is indeed the minimum.

Therefore, I think the answer is 64.

**Final Answer**
\boxed{64}

To find the minimum value of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1
$$
with $ a, b, c, d > 0 $, we can use the **Cauchy-Schwarz inequality** in a strategic way.

---

### Step 1: Apply Cauchy-Schwarz Inequality

We rewrite the expression as:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}
$$
This allows us to apply the **Cauchy-Schwarz inequality** in the form:
$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2
$$
Since $ a + b + c + d = 1 $, the inequality becomes:
$$
\left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \cdot 1 \geq (8)^2 = 64
$$
Thus, the expression is **at least** 64.

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when:
$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}
$$
Let’s denote this common value as $ k $. Then:
- $ a = \frac{1}{k} $
- $ b = \frac{1}{k} $
- $ c = \frac{2}{k} $
- $ d = \frac{4}{k} $

Substitute into the constraint $ a + b + c + d = 1 $:
$$
\frac{1}{k} + \frac{1}{k} + \frac{2}{k} + \frac{4}{k} = \frac{8}{k} = 1 \Rightarrow k = 8
$$

So:
- $ a = b = \frac{1}{8} $
- $ c = \frac{2}{8} = \frac{1}{4} $
- $ d = \frac{4}{8} = \frac{1}{2} $

---

### Step 3: Compute the Expression

Now compute the original expression:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = 8 + 8 + 16 + 32 = 64
$$

---

### Step 4: Confirm Minimality

We have shown that the expression is **at least** 64 by Cauchy-Schwarz, and that the value **64** is achieved when the equality condition is satisfied. Therefore, the **minimum value** of the expression is:

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is an algebra problem, maybe there's a way to apply inequalities like Cauchy-Schwarz or AM-HM or AM-GM. Let me recall those.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can use that here. But the problem is that the expression we need to minimize has different coefficients in the denominators. So maybe I need to adjust the variables or split them in a way that allows me to apply AM-GM effectively.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that first. Let me set up the Lagrangian. Let me denote the function to minimize as $f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ and the constraint is $g(a, b, c, d) = a + b + c + d - 1 = 0$.

The Lagrangian is $ \mathcal{L} = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} - \lambda(a + b + c + d - 1) $.

Taking partial derivatives with respect to each variable and setting them equal to zero:

For $a$:
$$
\frac{\partial \mathcal{L}}{\partial a} = -\frac{1}{a^2} - \lambda = 0 \implies \lambda = -\frac{1}{a^2}
$$

For $b$:
$$
\frac{\partial \mathcal{L}}{\partial b} = -\frac{1}{b^2} - \lambda = 0 \implies \lambda = -\frac{1}{b^2}
$$

For $c$:
$$
\frac{\partial \mathcal{L}}{\partial c} = -\frac{4}{c^2} - \lambda = 0 \implies \lambda = -\frac{4}{c^2}
$$

For $d$:
$$
\frac{\partial \mathcal{L}}{\partial d} = -\frac{16}{d^2} - \lambda = 0 \implies \lambda = -\frac{16}{d^2}
$$

So from the first two equations, since both equal to $\lambda$, we have:
$$
-\frac{1}{a^2} = -\frac{1}{b^2} \implies \frac{1}{a^2} = \frac{1}{b^2} \implies a = b
$$
Similarly, from the first and third equations:
$$
-\frac{1}{a^2} = -\frac{4}{c^2} \implies \frac{1}{a^2} = \frac{4}{c^2} \implies c^2 = 4a^2 \implies c = 2a
$$
Since all variables are positive, we can ignore the negative root.

Similarly, from the first and fourth equations:
$$
-\frac{1}{a^2} = -\frac{16}{d^2} \implies \frac{1}{a^2} = \frac{16}{d^2} \implies d^2 = 16a^2 \implies d = 4a
$$

So now, we have relations between variables: $a = b$, $c = 2a$, $d = 4a$. Let me substitute these into the constraint equation $a + b + c + d = 1$:

Substituting:
$a + a + 2a + 4a = 1 \implies 8a = 1 \implies a = \frac{1}{8}$

Therefore, $a = b = \frac{1}{8}$, $c = 2a = \frac{1}{4}$, $d = 4a = \frac{1}{2}$

Now, let me compute the value of the original expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
$$

Calculating each term:

- $\frac{1}{1/8} = 8$
- Similarly, the second term is also 8
- $\frac{4}{1/4} = 16$
- $\frac{16}{1/2} = 32$

Adding them up: 8 + 8 + 16 + 32 = 64.

So according to this method, the minimum value is 64. But wait, let me check if this is indeed the minimum. Maybe I can verify using another method, like Cauchy-Schwarz or AM-GM.

Alternatively, maybe I can use the Cauchy-Schwarz inequality. Let me think. The Cauchy-Schwarz inequality in the form $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I need to relate the given expression to the constraint. Alternatively, maybe using the method of weighted AM-HM inequality?

Alternatively, let me think of splitting the variables. Since the coefficients in the denominators are different, maybe I can adjust variables to make them similar. For example, if I consider variables in such a way that the terms in the expression can be paired with the constraint terms.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me consider vectors. Suppose I have two vectors such that their dot product relates to the given expression. But I need to think more carefully.

Alternatively, maybe I can use the method of Lagrange multipliers again but check if there are other critical points. However, since the problem is convex? Wait, the function to minimize is convex? Let me check. The function $f(a,b,c,d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ is convex in each variable individually, but the sum of convex functions is convex. However, the constraint is linear. So the problem is a convex optimization problem, so the critical point found via Lagrange multipliers should be the global minimum. Therefore, 64 is the minimum.

But let me try another approach to confirm. Let me try using AM-GM inequality. Let me think about how to apply AM-GM here. The problem is that the terms in the expression have different coefficients. Maybe I can split the variables into parts so that I can apply AM-GM. For example, if I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$, perhaps I can think of them as multiple terms.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me recall that for positive reals, $(\sum \frac{x_i^2}{a_i}) \geq \frac{(\sum x_i)^2}{\sum a_i}$. But I need to relate this to the given expression. Let me think. Suppose I set up something like:

Let me consider the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me think of this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then, by Cauchy-Schwarz inequality:

$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2
$$

Wait, that's the Cauchy-Schwarz in the form $(\sum \frac{x_i^2}{a_i})(\sum a_i) \geq (\sum x_i)^2$. So here, if I take $x_1 = 1$, $x_2 = 1$, $x_3 = 2$, $x_4 = 4$, and $a_1 = a$, $a_2 = b$, $a_3 = c$, $a_4 = d$. Then:

$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2 = 8^2 = 64
$$

Therefore, since $a + b + c + d = 1$, we have:

$$
\left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \times 1 \geq 64
$$

Therefore, the expression is at least 64. And equality holds when $\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}$, that is, $\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$.

Let me check this condition. If equality holds, then:

From $\frac{1}{a} = \frac{1}{b}$, so $a = b$.

From $\frac{1}{a} = \frac{2}{c}$, so $c = 2a$.

From $\frac{1}{a} = \frac{4}{d}$, so $d = 4a$.

Which is exactly the same relations as found via Lagrange multipliers. Therefore, substituting back into the constraint $a + b + c + d = 1$ gives $a + a + 2a + 4a = 8a = 1$, so $a = 1/8$, and so on. Therefore, equality is achieved at these values, so the minimum value is indeed 64.

Therefore, both methods lead to the same result, so I can be confident that 64 is the correct minimum value.

But just to be thorough, let me test with some other values. Suppose I take different values for a, b, c, d that satisfy the constraint and see if the expression is larger than 64.

For example, let me take a = b = c = d = 1/4. Then the expression becomes:

$\frac{1}{1/4} + \frac{1}{1/4} + \frac{4}{1/4} + \frac{16}{1/4} = 4 + 4 + 16 + 64 = 88$, which is larger than 64.

Another test case: Let me take a = 1/8, b = 1/8, c = 1/4, d = 1/2 as found earlier. Then the expression is 8 + 8 + 16 + 32 = 64.

What if I take a different set? Let me try a = 1/10, b = 1/10, then c and d would need to satisfy 1/10 + 1/10 + c + d = 1 => c + d = 7/10. Let me choose c = 2a = 1/5, then d = 7/10 - 1/5 = 7/10 - 2/10 = 5/10 = 1/2. Then compute the expression:

1/a + 1/b + 4/c + 16/d = 10 + 10 + 4/(1/5) + 16/(1/2) = 20 + 20 + 32 = 72, which is still larger than 64.

Alternatively, suppose I take a = 1/9, b = 1/9, then c + d = 1 - 2/9 = 7/9. Let me take c = 2a = 2/9, then d = 7/9 - 2/9 = 5/9. Then compute the expression:

1/(1/9) + 1/(1/9) + 4/(2/9) + 16/(5/9) = 9 + 9 + 18 + (16*9)/5 = 27 + 18 + 144/5 = 45 + 28.8 = 73.8, which is still higher than 64.

Alternatively, let me try to take a different approach where maybe c or d are different. Suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. That gives 64. If I perturb slightly, say a = 0.12, b = 0.12, then c + d = 1 - 0.24 = 0.76. Let me take c = 2a = 0.24, then d = 0.76 - 0.24 = 0.52. Then compute the expression:

1/0.12 ≈ 8.333, same for 1/b, so total for a and b is ≈ 16.666. Then 4/c = 4 / 0.24 ≈ 16.666, and 16/d ≈ 16 / 0.52 ≈ 30.769. Total ≈ 16.666 + 16.666 + 30.769 ≈ 64.101, which is slightly above 64. So that seems to confirm that 64 is indeed the minimum.

Alternatively, if I take a = 1/8 + ε, b = 1/8 + ε, then c = 2a = 2*(1/8 + ε) = 1/4 + 2ε, d = 4a = 1/2 + 4ε. Then the constraint becomes:

a + b + c + d = (1/8 + ε) + (1/8 + ε) + (1/4 + 2ε) + (1/2 + 4ε) = 1/8 + 1/8 + 1/4 + 1/2 + ε + ε + 2ε + 4ε = (1/8 + 1/8 + 2/8 + 4/8) + 8ε = (8/8) + 8ε = 1 + 8ε. But the constraint is supposed to be 1, so we need to adjust. Therefore, if I take a = 1/8 + ε, then to maintain the constraint, I need to adjust other variables. Maybe this is getting too complicated, but the previous test case shows that perturbing slightly increases the value, so it's likely that 64 is indeed the minimum.

Alternatively, let me think if there's a way to use AM-GM directly. Let me consider splitting the variables. For example, since the coefficients in the denominators are 1, 1, 4, 16, which are squares of 1, 1, 2, 4. Maybe I can think of the terms as multiple variables. For example, if I consider the term $\frac{1}{a}$, maybe I can think of it as $\frac{1^2}{a}$, similarly for others. Then, using Cauchy-Schwarz as before. Which we already did, and that gives the same result.

Alternatively, maybe I can use the AM-GM inequality on the terms. Let me think. Suppose I want to relate the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$ with the constraint $a + b + c + d = 1$.

But since the terms are reciprocals, it's tricky. However, if I can write the expression as a sum of terms that can be related to the variables in the constraint. For example, if I consider variables multiplied by coefficients.

Alternatively, maybe I can use the method of substitution. Since a + b + c + d = 1, maybe express one variable in terms of others. But that might complicate things.

Alternatively, think of the problem as optimizing with weights. For example, the expression is similar to weighted harmonic mean. But I need to think of how to apply that.

Alternatively, think of the problem as minimizing $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ with a + b + c + d = 1. Let me consider using the method of Lagrange multipliers again, but maybe I can check if there are other critical points. However, since the Lagrangian method gave a unique critical point, and the function tends to infinity as any variable approaches 0, that critical point should be the global minimum.

Therefore, combining all these methods, I can be confident that the minimum value is 64.

But just to make sure, let me check another case where variables are not following the relations found earlier. For example, suppose I take a = 0.2, b = 0.2, c = 0.3, d = 0.3. Then a + b + c + d = 0.2 + 0.2 + 0.3 + 0.3 = 1.0. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.3 + 16/0.3 = 5 + 5 + 13.333... + 53.333... ≈ 5 + 5 + 13.33 + 53.33 ≈ 76.66, which is still higher than 64.

Another case: Let me take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Then the sum is 0.1 + 0.1 + 0.2 + 0.6 = 1.0. The expression is 1/0.1 + 1/0.1 + 4/0.2 + 16/0.6 = 10 + 10 + 20 + 26.666... ≈ 66.666..., which is still higher than 64.

Alternatively, take a = 0.125, b = 0.125, c = 0.25, d = 0.5, which is the critical point. Then the expression is 8 + 8 + 16 + 32 = 64.

So all these tests confirm that 64 is indeed the minimum.

Therefore, I think the answer is 64.

**Final Answer**
\boxed{64}

To find the minimum value of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1
$$
with $ a, b, c, d > 0 $, we can use the **Cauchy-Schwarz inequality** in a strategic way.

---

### Step 1: Apply Cauchy-Schwarz Inequality

We rewrite the expression as:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}
$$
This allows us to apply the **Cauchy-Schwarz inequality** in the form:
$$
\left( \frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \right)(a + b + c + d) \geq (1 + 1 + 2 + 4)^2
$$
Since $ a + b + c + d = 1 $, the inequality becomes:
$$
\left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \cdot 1 \geq (8)^2 = 64
$$
Thus, the expression is **at least** 64.

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when:
$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}
$$
Let’s denote this common value as $ k $. Then:
- $ a = \frac{1}{k} $
- $ b = \frac{1}{k} $
- $ c = \frac{2}{k} $
- $ d = \frac{4}{k} $

Substitute into the constraint $ a + b + c + d = 1 $:
$$
\frac{1}{k} + \frac{1}{k} + \frac{2}{k} + \frac{4}{k} = \frac{8}{k} = 1 \Rightarrow k = 8
$$

So:
- $ a = b = \frac{1}{8} $
- $ c = \frac{2}{8} = \frac{1}{4} $
- $ d = \frac{4}{8} = \frac{1}{2} $

---

### Step 3: Compute the Expression

Now compute the original expression:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = 8 + 8 + 16 + 32 = 64
$$

---

### Step 4: Confirm Minimality

We have shown that the expression is **at least** 64 by Cauchy-Schwarz, and that the value **64** is achieved when the equality condition is satisfied. Therefore, the **minimum value** of the expression is:

$$
\boxed{64}
$$

Solution:
By Cauchy-Schwarz,
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \ge (1 + 1 + 2 + 4)^2 = 64.\]Equality occurs when $a = b = \frac{c}{2} = \frac{d}{4}$ and $a + b + c + d = 1.$  We can solve to get $a = \frac{1}{8},$ $b = \frac{1}{8},$ $c = \frac{1}{4},$ and $d = \frac{1}{2},$ so the minimum value is $\boxed{64}.$


Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basic properties of inverse trigonometric functions.

First, I know that the domain of both $ \arccos x $ and $ \arcsin x $ is $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, the question is about the range, so I need to figure out all the possible values that $ f(x) $ can take as $ x $ varies from -1 to 1.

Let me think about the relationship between $ \arccos x $ and $ \arcsin x $. I remember there's an identity that connects them. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Is that right? Let me verify with a quick example. If $ x = 0 $, then $ \arccos 0 = \frac{\pi}{2} $ and $ \arcsin 0 = 0 $, so their sum is $ \frac{\pi}{2} $. If $ x = 1 $, $ \arccos 1 = 0 $ and $ \arcsin 1 = \frac{\pi}{2} $, so again the sum is $ \frac{\pi}{2} $. Similarly for $ x = -1 $, $ \arccos(-1) = \pi $ and $ \arcsin(-1) = -\frac{\pi}{2} $, so their sum is $ \pi - \frac{\pi}{2} = \frac{\pi}{2} $. Okay, so that identity seems to hold.

So, if I let $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Therefore, substituting into $ f(x) $, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2
$$

Let me expand that:

$$
\left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = \frac{\pi^2}{4} - \pi \theta + 2\theta^2
$$

So, $ f(x) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. But since $ \theta = \arcsin x $, and $ x \in [-1, 1] $, $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $. Wait, but $ \arcsin x $ actually gives values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, right? So $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

But maybe I can express $ f(x) $ in terms of $ \theta $, and then find its range over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Alternatively, since $ \theta = \arcsin x $, and $ x = \sin \theta $, but maybe that's complicating things. Let me stick with the expression $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $, where $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Alternatively, maybe I can think of this as a quadratic function in terms of $ \theta $. Let me write it again:

$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}
$$

So, this is a quadratic function in $ \theta $, opening upwards because the coefficient of $ \theta^2 $ is positive (2). Therefore, its minimum occurs at the vertex, and the maximum values will be at the endpoints of the interval $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Let me recall that for a quadratic function $ ax^2 + bx + c $, the vertex is at $ x = -\frac{b}{2a} $. So here, $ a = 2 $, $ b = -\pi $, so the vertex is at $ \theta = -\frac{-\pi}{2 \times 2} = \frac{\pi}{4} $.

So, the minimum of $ f(\theta) $ occurs at $ \theta = \frac{\pi}{4} $, and the maximum occurs at one of the endpoints $ \theta = -\frac{\pi}{2} $ or $ \theta = \frac{\pi}{2} $.

Therefore, to find the range of $ f(x) $, I need to compute $ f(\theta) $ at $ \theta = \frac{\pi}{4} $, $ \theta = -\frac{\pi}{2} $, and $ \theta = \frac{\pi}{2} $, and then determine the minimum and maximum values.

Let me compute each of these:

First, at $ \theta = \frac{\pi}{4} $:

$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{8} $
- $ -\pi \times \frac{\pi}{4} = -\frac{\pi^2}{4} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}
$$

So, the minimum value is $ \frac{\pi^2}{8} $.

Now, let's check the endpoints. First, $ \theta = \frac{\pi}{2} $:

$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $
- $ -\pi \times \frac{\pi}{2} = -\frac{\pi^2}{2} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}
$$

Similarly, for $ \theta = -\frac{\pi}{2} $:

$$
f\left( -\frac{\pi}{2} \right) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $
- $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4}
$$

Wait, hold on. Wait, that seems like a problem. Wait, let me check again. Wait, the original function is $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. So when $ \theta = -\frac{\pi}{2} $, substituting:

First term: $ 2 \times (-\frac{\pi}{2})^2 = 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

So total is $ \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $. Hmm, okay. But wait, when $ \theta = \frac{\pi}{2} $, we had $ \frac{\pi^2}{4} $, and when $ \theta = -\frac{\pi}{2} $, we have $ \frac{5\pi^2}{4} $. Therefore, the maximum value is $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $. Therefore, the range of $ f(x) $ is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

Wait, but hold on. Let me check if there are any other critical points or if I made a mistake here. Because the quadratic function is opening upwards, so the minimum is at the vertex, which we found at $ \theta = \frac{\pi}{4} $, and the maximum occurs at the endpoint with the larger value. Since $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, and the function is quadratic, so the maximum is at one of the endpoints. But when we plug in $ \theta = -\frac{\pi}{2} $, we get $ \frac{5\pi^2}{4} $, and at $ \theta = \frac{\pi}{2} $, we get $ \frac{\pi^2}{4} $. Therefore, the maximum is indeed $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $. Therefore, the range is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But let me verify this with another approach to make sure. Let me think about the original function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Since $ \arccos x + \arcsin x = \frac{\pi}{2} $, maybe I can use this identity to express $ f(x) $ in terms of a single trigonometric function.

Let me denote $ \alpha = \arccos x $, so $ \arcsin x = \frac{\pi}{2} - \alpha $. Therefore, $ f(x) = \alpha^2 + \left( \frac{\pi}{2} - \alpha \right)^2 $. Expanding this:

$$
\alpha^2 + \left( \frac{\pi^2}{4} - \pi \alpha + \alpha^2 \right) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4}
$$

Which is the same expression as before, but now $ \alpha = \arccos x $, so $ \alpha \in [0, \pi] $, since $ \arccos x $ ranges from 0 to $ \pi $ as $ x $ goes from 1 to -1. Wait, so here $ \alpha \in [0, \pi] $, whereas earlier I considered $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. But since $ \alpha = \frac{\pi}{2} - \theta $, then if $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, then $ \alpha \in [0, \pi] $. So, in this case, if I consider $ \alpha \in [0, \pi] $, then the function $ f(\alpha) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $, and we can analyze this quadratic function over $ \alpha \in [0, \pi] $.

So, again, the quadratic function $ f(\alpha) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $. The vertex is at $ \alpha = \frac{\pi}{4} $, as before. Since the coefficient of $ \alpha^2 $ is positive, the minimum is at $ \alpha = \frac{\pi}{4} $, and the maximum occurs at one of the endpoints $ \alpha = 0 $ or $ \alpha = \pi $.

Calculating $ f(0) $:

$$
2(0)^2 - \pi(0) + \frac{\pi^2}{4} = \frac{\pi^2}{4}
$$

Calculating $ f(\pi) $:

$$
2(\pi)^2 - \pi(\pi) + \frac{\pi^2}{4} = 2\pi^2 - \pi^2 + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4}
$$

So, same results as before. Therefore, the minimum value is $ \frac{\pi^2}{8} $ when $ \alpha = \frac{\pi}{4} $, and the maximum is $ \frac{5\pi^2}{4} $ when $ \alpha = \pi $. Therefore, the range of $ f(x) $ is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check if there are any other critical points or if I need to consider something else. For example, maybe there's a mistake in assuming that the maximum occurs at the endpoints? Let me think. Since the quadratic function is convex (opening upwards), the maximum on a closed interval will indeed be at one of the endpoints. Therefore, since $ \alpha \in [0, \pi] $, the maximum is at $ \alpha = \pi $ or $ \alpha = 0 $, which we have already calculated. Therefore, the maximum is $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $.

But let me check with specific values of $ x $ to see if this makes sense. For example, when $ x = 0 $, $ \arccos 0 = \frac{\pi}{2} $, $ \arcsin 0 = 0 $, so $ f(0) = (\frac{\pi}{2})^2 + 0 = \frac{\pi^2}{4} $, which is the value we found at $ \alpha = 0 $, which is one of the endpoints.

When $ x = 1 $, $ \arccos 1 = 0 $, $ \arcsin 1 = \frac{\pi}{2} $, so $ f(1) = 0 + (\frac{\pi}{2})^2 = \frac{\pi^2}{4} $, same as when $ x = 0 $.

When $ x = -1 $, $ \arccos(-1) = \pi $, $ \arcsin(-1) = -\frac{\pi}{2} $, so $ f(-1) = \pi^2 + (-\frac{\pi}{2})^2 = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $, which matches our previous calculation.

Now, let's check the minimum value. When does $ f(x) $ reach its minimum? At $ \alpha = \frac{\pi}{4} $, which corresponds to $ x = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $. Let me compute $ f(\frac{\sqrt{2}}{2}) $:

$ \arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} $, and $ \arcsin(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} $, so $ f(\frac{\sqrt{2}}{2}) = (\frac{\pi}{4})^2 + (\frac{\pi}{4})^2 = 2 \times \frac{\pi^2}{16} = \frac{\pi^2}{8} $, which matches our earlier result.

So, that seems correct. Therefore, the minimum value is $ \frac{\pi^2}{8} $, and the maximum is $ \frac{5\pi^2}{4} $. Therefore, the range of $ f(x) $ is all real numbers from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But let me just think again if there's a possibility that the function could attain values outside of this interval. For example, suppose I take some other value of $ x $, say $ x = \frac{1}{2} $. Then $ \arccos(\frac{1}{2}) = \frac{\pi}{3} $, and $ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $. Therefore, $ f(\frac{1}{2}) = (\frac{\pi}{3})^2 + (\frac{\pi}{6})^2 = \frac{\pi^2}{9} + \frac{\pi^2}{36} = \frac{4\pi^2 + \pi^2}{36} = \frac{5\pi^2}{36} \approx 1.37 $, whereas $ \frac{\pi^2}{8} \approx 1.23 $, so this is larger than the minimum. Similarly, if I take $ x = \frac{\sqrt{2}}{2} $, we get the minimum.

Another test: take $ x = \frac{\sqrt{3}}{2} $. Then $ \arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} $, and $ \arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} $. Then $ f(x) = (\frac{\pi}{6})^2 + (\frac{\pi}{3})^2 = \frac{\pi^2}{36} + \frac{\pi^2}{9} = \frac{5\pi^2}{36} $, same as before. So, same value.

So, seems like the minimum is indeed $ \frac{\pi^2}{8} $, and the maximum is $ \frac{5\pi^2}{4} $. Therefore, the range is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check another point. Let me take $ x = \frac{\sqrt{2}}{2} $, which we already did, and got $ \frac{\pi^2}{8} $. What about $ x = \frac{\sqrt{3}}{2} $, which we saw gives $ \frac{5\pi^2}{36} \approx 1.37 $, which is greater than $ \frac{\pi^2}{8} \approx 1.23 $. So, that seems okay.

Alternatively, let me check the value of $ f(x) $ at $ x = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $, which we already did. So, that's the minimum.

Therefore, I think my conclusion is correct. The range is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But just to be thorough, let me consider the function $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $ over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Wait, earlier I considered $ \theta = \arcsin x $, which ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, but when I considered $ \alpha = \arccos x $, which ranges from 0 to $ \pi $. However, since $ \theta = \frac{\pi}{2} - \alpha $, then when $ \alpha $ goes from 0 to $ \pi $, $ \theta $ goes from $ \frac{\pi}{2} $ to $ -\frac{\pi}{2} $. Therefore, the interval for $ \theta $ is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, which is consistent.

But in this case, when I considered $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, the function $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $, and we found that the minimum is at $ \theta = \frac{\pi}{4} $, and the maximum at $ \theta = -\frac{\pi}{2} $. However, when I considered $ \alpha \in [0, \pi] $, the maximum was at $ \alpha = \pi $, which corresponds to $ \theta = -\frac{\pi}{2} $. So, same result.

Therefore, regardless of the substitution, the conclusion is the same. Therefore, I think the range is indeed $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check if there's a possibility that the function could have a higher value somewhere else. For example, suppose I take $ \theta $ approaching $ -\frac{\pi}{2} $, then $ f(\theta) $ approaches $ \frac{5\pi^2}{4} $, which is the maximum. Similarly, as $ \theta $ approaches $ \frac{\pi}{2} $, $ f(\theta) $ approaches $ \frac{\pi^2}{4} $. Therefore, since the function is continuous on the closed interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, by the Extreme Value Theorem, it must attain its maximum and minimum, which we have found.

Therefore, I think I can be confident that the range is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But just to make sure, let me compute the derivative of $ f(\theta) $ with respect to $ \theta $, and check for critical points.

So, $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. The derivative is $ f'(\theta) = 4\theta - \pi $. Setting this equal to zero:

$ 4\theta - \pi = 0 \implies \theta = \frac{\pi}{4} $, which is the critical point we found earlier. Since the second derivative is $ f''(\theta) = 4 > 0 $, this is indeed a minimum. Therefore, there are no other critical points, so the maximum must be at the endpoints.

Therefore, the conclusion is correct.

Therefore, the range of $ f(x) $ is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To determine the range of the function
$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$
we begin by leveraging a key identity involving inverse trigonometric functions:

$$
\arccos x + \arcsin x = \frac{\pi}{2}, \quad \text{for all } x \in [-1, 1].
$$

Let us define $ \theta = \arcsin x $, so that $ \arccos x = \frac{\pi}{2} - \theta $. Substituting into the function, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2.
$$

Expanding this expression:

$$
f(x) = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}.
$$

Now, since $ \theta = \arcsin x $, and the domain of $ \arcsin x $ is $ [-1, 1] $, the range of $ \theta $ is:

$$
\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

Thus, we are now analyzing the function:

$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}, \quad \text{for } \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

---

### Step 1: Identify the nature of the function

This is a quadratic function in $ \theta $, with a positive leading coefficient (2), so it opens **upwards**. Therefore, it has a **minimum** at its vertex and **maximum** values at the endpoints of the interval.

---

### Step 2: Find the vertex

The vertex of a quadratic $ a\theta^2 + b\theta + c $ occurs at:

$$
\theta = -\frac{b}{2a} = -\frac{-\pi}{2 \cdot 2} = \frac{\pi}{4}.
$$

So, the minimum value of $ f(\theta) $ occurs at $ \theta = \frac{\pi}{4} $.

---

### Step 3: Evaluate the function at the vertex and endpoints

- **At $ \theta = \frac{\pi}{4} $:**

$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

- **At $ \theta = -\frac{\pi}{2} $:**

$$
f\left( -\frac{\pi}{2} \right) = 2\left( \frac{\pi^2}{4} \right) + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

- **At $ \theta = \frac{\pi}{2} $:**

$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi^2}{4} \right) - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

---

### Step 4: Determine the range

From the above evaluations:

- The **minimum** value of $ f(x) $ is $ \frac{\pi^2}{8} $, achieved at $ \theta = \frac{\pi}{4} $.
- The **maximum** value of $ f(x) $ is $ \frac{5\pi^2}{4} $, achieved at $ \theta = -\frac{\pi}{2} $.

Since the function is continuous on the closed interval $ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] $, and the function is quadratic (hence smooth), the range is the closed interval between these two values.

---

### Final Answer

$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}
$$

[Assistant]
Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basic properties of inverse trigonometric functions.

First, I know that the domain of both $ \arccos x $ and $ \arcsin x $ is $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, the question is about the range, so I need to figure out all the possible values that $ f(x) $ can take as $ x $ varies from -1 to 1.

Let me think about the relationship between $ \arccos x $ and $ \arcsin x $. I remember there's an identity that connects them. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Is that right? Let me verify with a quick example. If $ x = 0 $, then $ \arccos 0 = \frac{\pi}{2} $ and $ \arcsin 0 = 0 $, so their sum is $ \frac{\pi}{2} $. If $ x = 1 $, $ \arccos 1 = 0 $ and $ \arcsin 1 = \frac{\pi}{2} $, so again the sum is $ \frac{\pi}{2} $. Similarly for $ x = -1 $, $ \arccos(-1) = \pi $ and $ \arcsin(-1) = -\frac{\pi}{2} $, so their sum is $ \pi - \frac{\pi}{2} = \frac{\pi}{2} $. Okay, so that identity seems to hold.

So, if I let $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Therefore, substituting into $ f(x) $, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2
$$

Let me expand that:

$$
\left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = \frac{\pi^2}{4} - \pi \theta + 2\theta^2
$$

So, $ f(x) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. But since $ \theta = \arcsin x $, and $ x \in [-1, 1] $, $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $. Wait, but $ \arcsin x $ actually gives values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, right? So $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

But maybe I can express $ f(x) $ in terms of $ \theta $, and then find its range over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Alternatively, since $ \theta = \arcsin x $, and $ x = \sin \theta $, but maybe that's complicating things. Let me stick with the expression $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $, where $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Alternatively, maybe I can think of this as a quadratic function in terms of $ \theta $. Let me write it again:

$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}
$$

So, this is a quadratic function in $ \theta $, opening upwards because the coefficient of $ \theta^2 $ is positive (2). Therefore, its minimum occurs at the vertex, and the maximum values will be at the endpoints of the interval $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Let me recall that for a quadratic function $ ax^2 + bx + c $, the vertex is at $ x = -\frac{b}{2a} $. So here, $ a = 2 $, $ b = -\pi $, so the vertex is at $ \theta = -\frac{-\pi}{2 \times 2} = \frac{\pi}{4} $.

So, the minimum of $ f(\theta) $ occurs at $ \theta = \frac{\pi}{4} $, and the maximum occurs at one of the endpoints $ \theta = -\frac{\pi}{2} $ or $ \theta = \frac{\pi}{2} $.

Therefore, to find the range of $ f(x) $, I need to compute $ f(\theta) $ at $ \theta = \frac{\pi}{4} $, $ \theta = -\frac{\pi}{2} $, and $ \theta = \frac{\pi}{2} $, and then determine the minimum and maximum values.

Let me compute each of these:

First, at $ \theta = \frac{\pi}{4} $:

$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{8} $
- $ -\pi \times \frac{\pi}{4} = -\frac{\pi^2}{4} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}
$$

So, the minimum value is $ \frac{\pi^2}{8} $.

Now, let's check the endpoints. First, $ \theta = \frac{\pi}{2} $:

$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $
- $ -\pi \times \frac{\pi}{2} = -\frac{\pi^2}{2} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}
$$

Similarly, for $ \theta = -\frac{\pi}{2} $:

$$
f\left( -\frac{\pi}{2} \right) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating term by term:

- $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $
- $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $
- $ \frac{\pi^2}{4} $

Adding them up:

$$
\frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4}
$$

Wait, hold on. Wait, that seems like a problem. Wait, let me check again. Wait, the original function is $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. So when $ \theta = -\frac{\pi}{2} $, substituting:

First term: $ 2 \times (-\frac{\pi}{2})^2 = 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

So total is $ \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $. Hmm, okay. But wait, when $ \theta = \frac{\pi}{2} $, we had $ \frac{\pi^2}{4} $, and when $ \theta = -\frac{\pi}{2} $, we have $ \frac{5\pi^2}{4} $. Therefore, the maximum value is $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $. Therefore, the range of $ f(x) $ is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

Wait, but hold on. Let me check if there are any other critical points or if I made a mistake here. Because the quadratic function is opening upwards, so the minimum is at the vertex, which we found at $ \theta = \frac{\pi}{4} $, and the maximum occurs at the endpoint with the larger value. Since $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, and the function is quadratic, so the maximum is at one of the endpoints. But when we plug in $ \theta = -\frac{\pi}{2} $, we get $ \frac{5\pi^2}{4} $, and at $ \theta = \frac{\pi}{2} $, we get $ \frac{\pi^2}{4} $. Therefore, the maximum is indeed $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $. Therefore, the range is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But let me verify this with another approach to make sure. Let me think about the original function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Since $ \arccos x + \arcsin x = \frac{\pi}{2} $, maybe I can use this identity to express $ f(x) $ in terms of a single trigonometric function.

Let me denote $ \alpha = \arccos x $, so $ \arcsin x = \frac{\pi}{2} - \alpha $. Therefore, $ f(x) = \alpha^2 + \left( \frac{\pi}{2} - \alpha \right)^2 $. Expanding this:

$$
\alpha^2 + \left( \frac{\pi^2}{4} - \pi \alpha + \alpha^2 \right) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4}
$$

Which is the same expression as before, but now $ \alpha = \arccos x $, so $ \alpha \in [0, \pi] $, since $ \arccos x $ ranges from 0 to $ \pi $ as $ x $ goes from 1 to -1. Wait, so here $ \alpha \in [0, \pi] $, whereas earlier I considered $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. But since $ \alpha = \frac{\pi}{2} - \theta $, then if $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, then $ \alpha \in [0, \pi] $. So, in this case, if I consider $ \alpha \in [0, \pi] $, then the function $ f(\alpha) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $, and we can analyze this quadratic function over $ \alpha \in [0, \pi] $.

So, again, the quadratic function $ f(\alpha) = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $. The vertex is at $ \alpha = \frac{\pi}{4} $, as before. Since the coefficient of $ \alpha^2 $ is positive, the minimum is at $ \alpha = \frac{\pi}{4} $, and the maximum occurs at one of the endpoints $ \alpha = 0 $ or $ \alpha = \pi $.

Calculating $ f(0) $:

$$
2(0)^2 - \pi(0) + \frac{\pi^2}{4} = \frac{\pi^2}{4}
$$

Calculating $ f(\pi) $:

$$
2(\pi)^2 - \pi(\pi) + \frac{\pi^2}{4} = 2\pi^2 - \pi^2 + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4}
$$

So, same results as before. Therefore, the minimum value is $ \frac{\pi^2}{8} $ when $ \alpha = \frac{\pi}{4} $, and the maximum is $ \frac{5\pi^2}{4} $ when $ \alpha = \pi $. Therefore, the range of $ f(x) $ is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check if there are any other critical points or if I need to consider something else. For example, maybe there's a mistake in assuming that the maximum occurs at the endpoints? Let me think. Since the quadratic function is convex (opening upwards), the maximum on a closed interval will indeed be at one of the endpoints. Therefore, since $ \alpha \in [0, \pi] $, the maximum is at $ \alpha = \pi $ or $ \alpha = 0 $, which we have already calculated. Therefore, the maximum is $ \frac{5\pi^2}{4} $, and the minimum is $ \frac{\pi^2}{8} $.

But let me check with specific values of $ x $ to see if this makes sense. For example, when $ x = 0 $, $ \arccos 0 = \frac{\pi}{2} $, $ \arcsin 0 = 0 $, so $ f(0) = (\frac{\pi}{2})^2 + 0 = \frac{\pi^2}{4} $, which is the value we found at $ \alpha = 0 $, which is one of the endpoints.

When $ x = 1 $, $ \arccos 1 = 0 $, $ \arcsin 1 = \frac{\pi}{2} $, so $ f(1) = 0 + (\frac{\pi}{2})^2 = \frac{\pi^2}{4} $, same as when $ x = 0 $.

When $ x = -1 $, $ \arccos(-1) = \pi $, $ \arcsin(-1) = -\frac{\pi}{2} $, so $ f(-1) = \pi^2 + (-\frac{\pi}{2})^2 = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $, which matches our previous calculation.

Now, let's check the minimum value. When does $ f(x) $ reach its minimum? At $ \alpha = \frac{\pi}{4} $, which corresponds to $ x = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $. Let me compute $ f(\frac{\sqrt{2}}{2}) $:

$ \arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} $, and $ \arcsin(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} $, so $ f(\frac{\sqrt{2}}{2}) = (\frac{\pi}{4})^2 + (\frac{\pi}{4})^2 = 2 \times \frac{\pi^2}{16} = \frac{\pi^2}{8} $, which matches our earlier result.

So, that seems correct. Therefore, the minimum value is $ \frac{\pi^2}{8} $, and the maximum is $ \frac{5\pi^2}{4} $. Therefore, the range of $ f(x) $ is all real numbers from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But let me just think again if there's a possibility that the function could attain values outside of this interval. For example, suppose I take some other value of $ x $, say $ x = \frac{1}{2} $. Then $ \arccos(\frac{1}{2}) = \frac{\pi}{3} $, and $ \arcsin(\frac{1}{2}) = \frac{\pi}{6} $. Therefore, $ f(\frac{1}{2}) = (\frac{\pi}{3})^2 + (\frac{\pi}{6})^2 = \frac{\pi^2}{9} + \frac{\pi^2}{36} = \frac{4\pi^2 + \pi^2}{36} = \frac{5\pi^2}{36} \approx 1.37 $, whereas $ \frac{\pi^2}{8} \approx 1.23 $, so this is larger than the minimum. Similarly, if I take $ x = \frac{\sqrt{2}}{2} $, we get the minimum.

Another test: take $ x = \frac{\sqrt{3}}{2} $. Then $ \arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} $, and $ \arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} $. Then $ f(x) = (\frac{\pi}{6})^2 + (\frac{\pi}{3})^2 = \frac{\pi^2}{36} + \frac{\pi^2}{9} = \frac{5\pi^2}{36} $, same as before. So, same value.

So, seems like the minimum is indeed $ \frac{\pi^2}{8} $, and the maximum is $ \frac{5\pi^2}{4} $. Therefore, the range is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check another point. Let me take $ x = \frac{\sqrt{2}}{2} $, which we already did, and got $ \frac{\pi^2}{8} $. What about $ x = \frac{\sqrt{3}}{2} $, which we saw gives $ \frac{5\pi^2}{36} \approx 1.37 $, which is greater than $ \frac{\pi^2}{8} \approx 1.23 $. So, that seems okay.

Alternatively, let me check the value of $ f(x) $ at $ x = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $, which we already did. So, that's the minimum.

Therefore, I think my conclusion is correct. The range is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

But just to be thorough, let me consider the function $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $ over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Wait, earlier I considered $ \theta = \arcsin x $, which ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, but when I considered $ \alpha = \arccos x $, which ranges from 0 to $ \pi $. However, since $ \theta = \frac{\pi}{2} - \alpha $, then when $ \alpha $ goes from 0 to $ \pi $, $ \theta $ goes from $ \frac{\pi}{2} $ to $ -\frac{\pi}{2} $. Therefore, the interval for $ \theta $ is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, which is consistent.

But in this case, when I considered $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, the function $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $, and we found that the minimum is at $ \theta = \frac{\pi}{4} $, and the maximum at $ \theta = -\frac{\pi}{2} $. However, when I considered $ \alpha \in [0, \pi] $, the maximum was at $ \alpha = \pi $, which corresponds to $ \theta = -\frac{\pi}{2} $. So, same result.

Therefore, regardless of the substitution, the conclusion is the same. Therefore, I think the range is indeed $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But let me check if there's a possibility that the function could have a higher value somewhere else. For example, suppose I take $ \theta $ approaching $ -\frac{\pi}{2} $, then $ f(\theta) $ approaches $ \frac{5\pi^2}{4} $, which is the maximum. Similarly, as $ \theta $ approaches $ \frac{\pi}{2} $, $ f(\theta) $ approaches $ \frac{\pi^2}{4} $. Therefore, since the function is continuous on the closed interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, by the Extreme Value Theorem, it must attain its maximum and minimum, which we have found.

Therefore, I think I can be confident that the range is $ [\frac{\pi^2}{8}, \frac{5\pi^2}{4}] $.

But just to make sure, let me compute the derivative of $ f(\theta) $ with respect to $ \theta $, and check for critical points.

So, $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. The derivative is $ f'(\theta) = 4\theta - \pi $. Setting this equal to zero:

$ 4\theta - \pi = 0 \implies \theta = \frac{\pi}{4} $, which is the critical point we found earlier. Since the second derivative is $ f''(\theta) = 4 > 0 $, this is indeed a minimum. Therefore, there are no other critical points, so the maximum must be at the endpoints.

Therefore, the conclusion is correct.

Therefore, the range of $ f(x) $ is from $ \frac{\pi^2}{8} $ to $ \frac{5\pi^2}{4} $.

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To determine the range of the function
$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$
we begin by leveraging a key identity involving inverse trigonometric functions:

$$
\arccos x + \arcsin x = \frac{\pi}{2}, \quad \text{for all } x \in [-1, 1].
$$

Let us define $ \theta = \arcsin x $, so that $ \arccos x = \frac{\pi}{2} - \theta $. Substituting into the function, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2.
$$

Expanding this expression:

$$
f(x) = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}.
$$

Now, since $ \theta = \arcsin x $, and the domain of $ \arcsin x $ is $ [-1, 1] $, the range of $ \theta $ is:

$$
\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

Thus, we are now analyzing the function:

$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}, \quad \text{for } \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

---

### Step 1: Identify the nature of the function

This is a quadratic function in $ \theta $, with a positive leading coefficient (2), so it opens **upwards**. Therefore, it has a **minimum** at its vertex and **maximum** values at the endpoints of the interval.

---

### Step 2: Find the vertex

The vertex of a quadratic $ a\theta^2 + b\theta + c $ occurs at:

$$
\theta = -\frac{b}{2a} = -\frac{-\pi}{2 \cdot 2} = \frac{\pi}{4}.
$$

So, the minimum value of $ f(\theta) $ occurs at $ \theta = \frac{\pi}{4} $.

---

### Step 3: Evaluate the function at the vertex and endpoints

- **At $ \theta = \frac{\pi}{4} $:**

$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

- **At $ \theta = -\frac{\pi}{2} $:**

$$
f\left( -\frac{\pi}{2} \right) = 2\left( \frac{\pi^2}{4} \right) + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

- **At $ \theta = \frac{\pi}{2} $:**

$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi^2}{4} \right) - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

---

### Step 4: Determine the range

From the above evaluations:

- The **minimum** value of $ f(x) $ is $ \frac{\pi^2}{8} $, achieved at $ \theta = \frac{\pi}{4} $.
- The **maximum** value of $ f(x) $ is $ \frac{5\pi^2}{4} $, achieved at $ \theta = -\frac{\pi}{2} $.

Since the function is continuous on the closed interval $ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] $, and the function is quadratic (hence smooth), the range is the closed interval between these two values.

---

### Final Answer

$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}
$$

Solution:
First, we claim that $\arccos x + \arcsin x = \frac{\pi}{2}$ for all $x \in [-1,1].$

Note that
\[\cos \left( \frac{\pi}{2} - \arcsin x \right) = \cos (\arccos x) = x.\]Furthermore, $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},$ so $0 \le \frac{\pi}{2} - \arcsin x \le \pi.$  Therefore,
\[\frac{\pi}{2} - \arcsin x = \arccos x,\]so $\arccos x + \arcsin x = \frac{\pi}{2}.$

Let $\alpha = \arccos x$ and $\beta = \arcsin x,$ so $\alpha + \beta = \frac{\pi}{2}.$  Then
\begin{align*}
f(x) &= (\arccos x)^2 + (\arcsin x)^2 \\
&= \alpha^2 + \beta^2 \\
&= \left( \frac{\pi}{2} - \beta \right)^2 + \beta^2 \\
&= 2 \beta^2 - \pi \beta + \frac{\pi^2}{4} \\
&= 2 \left( \beta - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}.
\end{align*}Since $-\frac{\pi}{2} \le \beta \le \frac{\pi}{2},$ the range of $f(x)$ is $\boxed{\left[ \frac{\pi^2}{8}, \frac{5 \pi^2}{4} \right]}.$


Explanation:
Okay, so I need to find the number of dice, n, such that when you roll n fair 6-sided dice, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me think about how to approach this.

First, let me recall some probability concepts. When dealing with probabilities involving multiple dice, especially when looking for specific numbers of successes (in this case, dice showing numbers other than 1), the binomial probability formula is usually useful. The binomial formula is:

P(k) = C(n, k) * p^k * (1 - p)^(n - k)

Where:
- P(k) is the probability of exactly k successes,
- C(n, k) is the combination of n things taken k at a time,
- p is the probability of success on a single trial,
- (1 - p) is the probability of failure.

In this problem, a "success" would be rolling a number other than 1. Let me confirm that. So each die has 6 sides, numbers 1 through 6. The probability of rolling a number other than 1 is 5/6, since there are 5 favorable outcomes (2, 3, 4, 5, 6) out of 6. Conversely, the probability of rolling a 1 is 1/6.

But the problem says the probability that exactly two of them show a number other than 1 is 25/216. Wait, so exactly two dice show numbers other than 1, which would mean that the other (n - 2) dice show 1. So, using the binomial formula, the probability should be:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

So, I need to solve this equation for n. Let me write that down:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

First, let me simplify the left-hand side. Let me recall that C(n, 2) is n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Hmm, okay. Let me see if I can simplify this equation. Let me first note that 25/36 is (5/6)^2, which is already there. Let me also note that 216 is 6^3. Let me see if I can write all terms with base 6 or something.

Alternatively, maybe I can divide both sides by 25 to make it simpler. Let me try that.

Dividing both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Wait, because 25/216 divided by 25 is 1/216. Let me check:

Original equation:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Divide both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Yes, that's correct. Now, let me simplify the left-hand side. Let me combine the constants:

(1/36) * (1/6)^{n - 2} = (1/6^2) * (1/6)^{n - 2} = (1/6)^{n}

Because 6^2 * 6^{n - 2} = 6^n. Therefore, the left-hand side becomes:

[n(n - 1)/2] * (1/6)^n = 1/216

So:

[n(n - 1)/2] * (1/6)^n = 1/216

Let me write that as:

[n(n - 1)/2] / 6^n = 1/216

Which can be rewritten as:

[n(n - 1)/2] = 6^n / 216

But 6^n / 216 is equal to 6^n / 6^3 = 6^{n - 3}

Therefore:

n(n - 1)/2 = 6^{n - 3}

So now, the equation simplifies to:

n(n - 1)/2 = 6^{n - 3}

Hmm, okay. Now, I need to solve for n here. Since n is a positive integer (number of dice), I can try plugging in small integer values for n and see which one satisfies the equation.

But first, let me check if I did the algebra correctly. Let me go back step by step.

Original equation:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

Which is:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Divide both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Then, combining the exponents:

(1/36) * (1/6)^{n - 2} = (1/6^2) * (1/6)^{n - 2} = (1/6)^{n}

Yes, that's correct. Therefore:

[n(n - 1)/2] * (1/6)^n = 1/216

Multiply both sides by 6^n:

n(n - 1)/2 = 6^n / 216

Which is:

n(n - 1)/2 = 6^{n - 3}

Yes, because 216 is 6^3. So 6^n / 6^3 = 6^{n - 3}

So, the equation is n(n - 1)/2 = 6^{n - 3}

Now, let me try plugging in values for n. Since n has to be at least 2, because we are talking about exactly two dice showing numbers other than 1. So n must be ≥ 2.

Let me try n = 3:

Left-hand side: 3*2/2 = 3

Right-hand side: 6^{3 - 3} = 6^0 = 1

So 3 ≠ 1. Not equal.

n = 4:

Left-hand side: 4*3/2 = 6

Right-hand side: 6^{4 - 3} = 6^1 = 6

Oh! 6 = 6. So n = 4 is a solution. Let me check if there are other solutions.

n = 5:

Left-hand side: 5*4/2 = 10

Right-hand side: 6^{5 - 3} = 6^2 = 36

10 ≠ 36. Not equal.

n = 2:

Left-hand side: 2*1/2 = 1

Right-hand side: 6^{2 - 3} = 6^{-1} = 1/6 ≈ 0.166...

Not equal.

n = 1:

But n must be at least 2, since we need exactly two dice showing numbers other than 1. So n=1 is invalid.

n=0? Also invalid.

What about n=5? As above, LHS=10, RHS=36. Not equal.

n=6:

LHS: 6*5/2 = 15

RHS: 6^{6 - 3} = 6^3 = 216

15 ≠ 216.

n=7:

LHS: 7*6/2 = 21

RHS: 6^4 = 1296

Still not equal.

So, seems like n=4 is the only solution here. Let me check n=3 again. Wait, n=3 gives LHS=3, RHS=1. Not equal. So n=4 is the only integer solution here.

But let me check if there's a possibility of another solution for higher n? For example, maybe n= something else?

But as n increases, the RHS is 6^{n - 3}, which grows exponentially, while the LHS is quadratic in n. So for n ≥ 5, RHS is growing much faster than LHS. Therefore, after n=4, RHS is way larger than LHS. For n=4, they are equal. For n=3, LHS is larger than RHS. For n=2, LHS is 1, RHS is 1/6. So, seems like n=4 is the only solution.

But let me check n=4 in the original probability equation to make sure.

Original equation: probability is C(4, 2)*(5/6)^2*(1/6)^{4 - 2} = ?

Compute that:

C(4, 2) is 6.

(5/6)^2 is 25/36.

(1/6)^{2} is 1/36.

Multiply them all together: 6 * (25/36) * (1/36) = 6 * 25 / (36 * 36) = 150 / 1296.

Simplify 150/1296: divide numerator and denominator by 6: 25 / 216. Which is exactly the given probability. So yes, n=4 is correct.

Therefore, the answer is 4. Wait, but let me check if there's another possible n? For example, maybe negative numbers? But n has to be positive integer. So no. So n=4 is the only solution.

But let me think again. Maybe I made a mistake in simplifying the equation? Let me check again.

Original equation:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216

For n=4:

C(4,2)=6, (5/6)^2=25/36, (1/6)^{2}=1/36. Multiply them: 6*(25/36)*(1/36)= 6*25/(36*36)= 150/1296=25/216. Correct.

So that works. So n=4 is correct. Therefore, the answer is 4.

But let me check n=5 again. If n=5, then the probability would be C(5,2)*(5/6)^2*(1/6)^{3} = 10*(25/36)*(1/216) = 10*25/(36*216). Let me compute that:

Denominator: 36*216 = 7776. Numerator: 250. So 250/7776. Simplify: divide numerator and denominator by 2: 125/3888 ≈ 0.03215... whereas 25/216 ≈ 0.1157... So not equal. So n=5 is not.

Similarly, n=3: C(3,2)=3, (5/6)^2*(1/6)^1= 25/36 * 1/6 = 25/216. Multiply by 3: 75/216 = 25/72 ≈ 0.347... which is bigger than 25/216 ≈ 0.1157. So not equal.

So n=4 is the only solution. Therefore, the answer is 4.

But just to make sure, let me check if there's another way to approach this problem, maybe using combinations?

Alternatively, think about the total number of possible outcomes when rolling n dice: 6^n.

The number of favorable outcomes is the number of ways to choose 2 dice to show numbers other than 1, and the remaining (n - 2) dice to show 1. So:

Number of favorable outcomes = C(n, 2) * 5^2 * 1^{n - 2}

Because for the two chosen dice, each has 5 possibilities (numbers 2-6), and the rest have 1 possibility (number 1).

Therefore, the probability is [C(n, 2) * 5^2 * 1^{n - 2}] / 6^n = [C(n, 2) * 25] / 6^n

Set equal to 25/216:

[C(n, 2) * 25] / 6^n = 25/216

Divide both sides by 25:

C(n, 2)/6^n = 1/216

Multiply both sides by 6^n:

C(n, 2) = 6^n / 216 = 6^{n - 3}

Which is the same equation as before: C(n, 2) = 6^{n - 3}

Which again leads to n(n - 1)/2 = 6^{n - 3}

So same equation. Therefore, same result. So n=4 is the solution.

Therefore, I think the answer is 4. Let me check once more if there's any other possible n. For example, n=0? Not possible. n=1? No. n=2? Let's check:

For n=2, the probability would be C(2,2)*(5/6)^2*(1/6)^0 = 1*(25/36)*1 = 25/36 ≈ 0.694, which is not 25/216. So no. So n=4 is the only solution.

Therefore, the answer is 4. So I think that's it.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that the probability of **exactly two out of $ n $ fair 6-sided dice** showing a number **other than 1** is $ \frac{25}{216} $. We are to find the value of $ n $.

---

### Step 1: Understanding the Probability Model

Each die has 6 faces, and the probability of rolling a number **other than 1** is:

$$
p = \frac{5}{6}
$$

The probability of rolling a **1** is:

$$
1 - p = \frac{1}{6}
$$

We are looking for the probability that **exactly two** of the $ n $ dice show a number other than 1. This is a classic **binomial probability** scenario, where:

- $ k = 2 $ (number of successes),
- $ p = \frac{5}{6} $ (probability of success),
- $ n $ is the total number of trials (dice rolled),
- $ 1 - p = \frac{1}{6} $ (probability of failure).

The binomial probability formula is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

Substituting the known values:

$$
P(2) = \binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2}
$$

We are told that this equals $ \frac{25}{216} $, so:

$$
\binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

---

### Step 2: Simplify the Equation

Recall that:

$$
\binom{n}{2} = \frac{n(n - 1)}{2}
$$

So the equation becomes:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{25}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by 25:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{216}
$$

Now simplify the constants:

$$
\left( \frac{1}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6^2} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6} \right)^n
$$

So the equation becomes:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{6} \right)^n = \frac{1}{216}
$$

Multiply both sides by $ 6^n $:

$$
\frac{n(n - 1)}{2} = \frac{6^n}{216}
$$

Note that $ 216 = 6^3 $, so:

$$
\frac{n(n - 1)}{2} = 6^{n - 3}
$$

---

### Step 3: Solve the Equation

We now solve:

$$
\frac{n(n - 1)}{2} = 6^{n - 3}
$$

Try small integer values of $ n $:

- $ n = 2 $: LHS = $ \frac{2 \cdot 1}{2} = 1 $, RHS = $ 6^{-1} = \frac{1}{6} $ → Not equal
- $ n = 3 $: LHS = $ \frac{3 \cdot 2}{2} = 3 $, RHS = $ 6^0 = 1 $ → Not equal
- $ n = 4 $: LHS = $ \frac{4 \cdot 3}{2} = 6 $, RHS = $ 6^1 = 6 $ → **Equal**
- $ n = 5 $: LHS = $ \frac{5 \cdot 4}{2} = 10 $, RHS = $ 6^2 = 36 $ → Not equal

As $ n $ increases, the RHS grows exponentially while the LHS grows quadratically, so no further solutions exist.

---

### Step 4: Verify the Solution

Let’s verify $ n = 4 $ in the original probability expression:

$$
P(2) = \binom{4}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^2 = 6 \cdot \frac{25}{36} \cdot \frac{1}{36} = \frac{150}{1296} = \frac{25}{216}
$$

This matches the given probability.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of dice, n, such that when you roll n fair 6-sided dice, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me think about how to approach this.

First, let me recall some probability concepts. When dealing with probabilities involving multiple dice, especially when looking for specific numbers of successes (in this case, dice showing numbers other than 1), the binomial probability formula is usually useful. The binomial formula is:

P(k) = C(n, k) * p^k * (1 - p)^(n - k)

Where:
- P(k) is the probability of exactly k successes,
- C(n, k) is the combination of n things taken k at a time,
- p is the probability of success on a single trial,
- (1 - p) is the probability of failure.

In this problem, a "success" would be rolling a number other than 1. Let me confirm that. So each die has 6 sides, numbers 1 through 6. The probability of rolling a number other than 1 is 5/6, since there are 5 favorable outcomes (2, 3, 4, 5, 6) out of 6. Conversely, the probability of rolling a 1 is 1/6.

But the problem says the probability that exactly two of them show a number other than 1 is 25/216. Wait, so exactly two dice show numbers other than 1, which would mean that the other (n - 2) dice show 1. So, using the binomial formula, the probability should be:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

So, I need to solve this equation for n. Let me write that down:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

First, let me simplify the left-hand side. Let me recall that C(n, 2) is n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Hmm, okay. Let me see if I can simplify this equation. Let me first note that 25/36 is (5/6)^2, which is already there. Let me also note that 216 is 6^3. Let me see if I can write all terms with base 6 or something.

Alternatively, maybe I can divide both sides by 25 to make it simpler. Let me try that.

Dividing both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Wait, because 25/216 divided by 25 is 1/216. Let me check:

Original equation:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Divide both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Yes, that's correct. Now, let me simplify the left-hand side. Let me combine the constants:

(1/36) * (1/6)^{n - 2} = (1/6^2) * (1/6)^{n - 2} = (1/6)^{n}

Because 6^2 * 6^{n - 2} = 6^n. Therefore, the left-hand side becomes:

[n(n - 1)/2] * (1/6)^n = 1/216

So:

[n(n - 1)/2] * (1/6)^n = 1/216

Let me write that as:

[n(n - 1)/2] / 6^n = 1/216

Which can be rewritten as:

[n(n - 1)/2] = 6^n / 216

But 6^n / 216 is equal to 6^n / 6^3 = 6^{n - 3}

Therefore:

n(n - 1)/2 = 6^{n - 3}

So now, the equation simplifies to:

n(n - 1)/2 = 6^{n - 3}

Hmm, okay. Now, I need to solve for n here. Since n is a positive integer (number of dice), I can try plugging in small integer values for n and see which one satisfies the equation.

But first, let me check if I did the algebra correctly. Let me go back step by step.

Original equation:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

Which is:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Divide both sides by 25:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Then, combining the exponents:

(1/36) * (1/6)^{n - 2} = (1/6^2) * (1/6)^{n - 2} = (1/6)^{n}

Yes, that's correct. Therefore:

[n(n - 1)/2] * (1/6)^n = 1/216

Multiply both sides by 6^n:

n(n - 1)/2 = 6^n / 216

Which is:

n(n - 1)/2 = 6^{n - 3}

Yes, because 216 is 6^3. So 6^n / 6^3 = 6^{n - 3}

So, the equation is n(n - 1)/2 = 6^{n - 3}

Now, let me try plugging in values for n. Since n has to be at least 2, because we are talking about exactly two dice showing numbers other than 1. So n must be ≥ 2.

Let me try n = 3:

Left-hand side: 3*2/2 = 3

Right-hand side: 6^{3 - 3} = 6^0 = 1

So 3 ≠ 1. Not equal.

n = 4:

Left-hand side: 4*3/2 = 6

Right-hand side: 6^{4 - 3} = 6^1 = 6

Oh! 6 = 6. So n = 4 is a solution. Let me check if there are other solutions.

n = 5:

Left-hand side: 5*4/2 = 10

Right-hand side: 6^{5 - 3} = 6^2 = 36

10 ≠ 36. Not equal.

n = 2:

Left-hand side: 2*1/2 = 1

Right-hand side: 6^{2 - 3} = 6^{-1} = 1/6 ≈ 0.166...

Not equal.

n = 1:

But n must be at least 2, since we need exactly two dice showing numbers other than 1. So n=1 is invalid.

n=0? Also invalid.

What about n=5? As above, LHS=10, RHS=36. Not equal.

n=6:

LHS: 6*5/2 = 15

RHS: 6^{6 - 3} = 6^3 = 216

15 ≠ 216.

n=7:

LHS: 7*6/2 = 21

RHS: 6^4 = 1296

Still not equal.

So, seems like n=4 is the only solution here. Let me check n=3 again. Wait, n=3 gives LHS=3, RHS=1. Not equal. So n=4 is the only integer solution here.

But let me check if there's a possibility of another solution for higher n? For example, maybe n= something else?

But as n increases, the RHS is 6^{n - 3}, which grows exponentially, while the LHS is quadratic in n. So for n ≥ 5, RHS is growing much faster than LHS. Therefore, after n=4, RHS is way larger than LHS. For n=4, they are equal. For n=3, LHS is larger than RHS. For n=2, LHS is 1, RHS is 1/6. So, seems like n=4 is the only solution.

But let me check n=4 in the original probability equation to make sure.

Original equation: probability is C(4, 2)*(5/6)^2*(1/6)^{4 - 2} = ?

Compute that:

C(4, 2) is 6.

(5/6)^2 is 25/36.

(1/6)^{2} is 1/36.

Multiply them all together: 6 * (25/36) * (1/36) = 6 * 25 / (36 * 36) = 150 / 1296.

Simplify 150/1296: divide numerator and denominator by 6: 25 / 216. Which is exactly the given probability. So yes, n=4 is correct.

Therefore, the answer is 4. Wait, but let me check if there's another possible n? For example, maybe negative numbers? But n has to be positive integer. So no. So n=4 is the only solution.

But let me think again. Maybe I made a mistake in simplifying the equation? Let me check again.

Original equation:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216

For n=4:

C(4,2)=6, (5/6)^2=25/36, (1/6)^{2}=1/36. Multiply them: 6*(25/36)*(1/36)= 6*25/(36*36)= 150/1296=25/216. Correct.

So that works. So n=4 is correct. Therefore, the answer is 4.

But let me check n=5 again. If n=5, then the probability would be C(5,2)*(5/6)^2*(1/6)^{3} = 10*(25/36)*(1/216) = 10*25/(36*216). Let me compute that:

Denominator: 36*216 = 7776. Numerator: 250. So 250/7776. Simplify: divide numerator and denominator by 2: 125/3888 ≈ 0.03215... whereas 25/216 ≈ 0.1157... So not equal. So n=5 is not.

Similarly, n=3: C(3,2)=3, (5/6)^2*(1/6)^1= 25/36 * 1/6 = 25/216. Multiply by 3: 75/216 = 25/72 ≈ 0.347... which is bigger than 25/216 ≈ 0.1157. So not equal.

So n=4 is the only solution. Therefore, the answer is 4.

But just to make sure, let me check if there's another way to approach this problem, maybe using combinations?

Alternatively, think about the total number of possible outcomes when rolling n dice: 6^n.

The number of favorable outcomes is the number of ways to choose 2 dice to show numbers other than 1, and the remaining (n - 2) dice to show 1. So:

Number of favorable outcomes = C(n, 2) * 5^2 * 1^{n - 2}

Because for the two chosen dice, each has 5 possibilities (numbers 2-6), and the rest have 1 possibility (number 1).

Therefore, the probability is [C(n, 2) * 5^2 * 1^{n - 2}] / 6^n = [C(n, 2) * 25] / 6^n

Set equal to 25/216:

[C(n, 2) * 25] / 6^n = 25/216

Divide both sides by 25:

C(n, 2)/6^n = 1/216

Multiply both sides by 6^n:

C(n, 2) = 6^n / 216 = 6^{n - 3}

Which is the same equation as before: C(n, 2) = 6^{n - 3}

Which again leads to n(n - 1)/2 = 6^{n - 3}

So same equation. Therefore, same result. So n=4 is the solution.

Therefore, I think the answer is 4. Let me check once more if there's any other possible n. For example, n=0? Not possible. n=1? No. n=2? Let's check:

For n=2, the probability would be C(2,2)*(5/6)^2*(1/6)^0 = 1*(25/36)*1 = 25/36 ≈ 0.694, which is not 25/216. So no. So n=4 is the only solution.

Therefore, the answer is 4. So I think that's it.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that the probability of **exactly two out of $ n $ fair 6-sided dice** showing a number **other than 1** is $ \frac{25}{216} $. We are to find the value of $ n $.

---

### Step 1: Understanding the Probability Model

Each die has 6 faces, and the probability of rolling a number **other than 1** is:

$$
p = \frac{5}{6}
$$

The probability of rolling a **1** is:

$$
1 - p = \frac{1}{6}
$$

We are looking for the probability that **exactly two** of the $ n $ dice show a number other than 1. This is a classic **binomial probability** scenario, where:

- $ k = 2 $ (number of successes),
- $ p = \frac{5}{6} $ (probability of success),
- $ n $ is the total number of trials (dice rolled),
- $ 1 - p = \frac{1}{6} $ (probability of failure).

The binomial probability formula is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

Substituting the known values:

$$
P(2) = \binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2}
$$

We are told that this equals $ \frac{25}{216} $, so:

$$
\binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

---

### Step 2: Simplify the Equation

Recall that:

$$
\binom{n}{2} = \frac{n(n - 1)}{2}
$$

So the equation becomes:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{25}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by 25:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{216}
$$

Now simplify the constants:

$$
\left( \frac{1}{36} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6^2} \right) \cdot \left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6} \right)^n
$$

So the equation becomes:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{6} \right)^n = \frac{1}{216}
$$

Multiply both sides by $ 6^n $:

$$
\frac{n(n - 1)}{2} = \frac{6^n}{216}
$$

Note that $ 216 = 6^3 $, so:

$$
\frac{n(n - 1)}{2} = 6^{n - 3}
$$

---

### Step 3: Solve the Equation

We now solve:

$$
\frac{n(n - 1)}{2} = 6^{n - 3}
$$

Try small integer values of $ n $:

- $ n = 2 $: LHS = $ \frac{2 \cdot 1}{2} = 1 $, RHS = $ 6^{-1} = \frac{1}{6} $ → Not equal
- $ n = 3 $: LHS = $ \frac{3 \cdot 2}{2} = 3 $, RHS = $ 6^0 = 1 $ → Not equal
- $ n = 4 $: LHS = $ \frac{4 \cdot 3}{2} = 6 $, RHS = $ 6^1 = 6 $ → **Equal**
- $ n = 5 $: LHS = $ \frac{5 \cdot 4}{2} = 10 $, RHS = $ 6^2 = 36 $ → Not equal

As $ n $ increases, the RHS grows exponentially while the LHS grows quadratically, so no further solutions exist.

---

### Step 4: Verify the Solution

Let’s verify $ n = 4 $ in the original probability expression:

$$
P(2) = \binom{4}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^2 = 6 \cdot \frac{25}{36} \cdot \frac{1}{36} = \frac{150}{1296} = \frac{25}{216}
$$

This matches the given probability.

---

### Final Answer

$$
\boxed{4}
$$

Solution: For $n$ dice, there are $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to choose two of them. For each of these ways, there is a $\left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n-2}$ chance that all but the chosen two dice will roll a 1. Therefore, we need to find the value of $n$ for which $\frac{25n(n-1)}{2 \cdot 6^n}=\frac{25}{216}$, or $108n(n-1)=6^n$. Plugging in values for $n$, we can see that $n=\boxed{4}$ works and no value of $n$ less than 4 works. Now we just need to prove that no values of $n$ greater than 4 work. Note that if $n \geq 5$, then $n < 3^{n-3}$ and $n-1 < 2\cdot 2^{n-3}$. We can multiply these inequalities to get that when $n \geq 5$, we have $n(n-1) < 2\cdot 6^{n-3}$, or $108n(n-1)<6^n$.


Explanation:
Okay, so Karla drove her vehicle 12,000 miles last year. The question is asking how many gallons of gasoline she would have saved if she had driven a fuel-efficient hybrid car instead of an SUV. The hybrid averages 48 miles per gallon, and the SUV averages 15 miles per gallon. Alright, let me think about how to approach this.

First, I need to figure out how much gasoline she used with the SUV and then how much she would have used with the hybrid. Then, subtract the hybrid's gasoline usage from the SUV's to find the savings. That makes sense. Let me write that down.

So, the formula for gasoline used is total miles divided by miles per gallon. That gives gallons. So, for the SUV, gallons used would be 12,000 miles divided by 15 miles per gallon. Let me calculate that. 12,000 divided by 15. Hmm, 15 times 800 is 12,000, right? Because 15 times 100 is 1,500, so 15 times 800 is 12,000. So, that would be 800 gallons for the SUV.

Now, for the hybrid car, it's 48 miles per gallon. So, the gasoline used would be 12,000 divided by 48. Let me do that division. Hmm, 48 times 250 is 12,000? Let me check. 48 times 200 is 9,600. Then 48 times 50 is 2,400. So 9,600 + 2,400 is 12,000. Yes, so 250 gallons for the hybrid.

Therefore, the difference between the SUV and the hybrid is 800 gallons minus 250 gallons. That would be 550 gallons. So, she would have saved 550 gallons. Let me just verify my calculations again to make sure I didn't make a mistake.

First, SUV: 12,000 / 15. Let me do 12,000 divided by 15. Dividing both numerator and denominator by 5 gives 2,400 / 3, which is 800. Yep, that's correct. For the hybrid: 12,000 / 48. Let me divide numerator and denominator by 12 first. 12,000 divided by 12 is 1,000. 48 divided by 12 is 4. So now it's 1,000 / 4, which is 250. That checks out. So 800 - 250 is indeed 550.

Wait, but let me think again. Is there another way to approach this problem? Maybe by calculating the difference in fuel efficiency and then applying it somehow? But I don't think that's straightforward because the fuel efficiency is different. The total gallons saved depends on the total miles driven and the difference in fuel efficiency. But maybe I can think in terms of gallons per mile?

For the SUV, it's 1/15 gallons per mile, and for the hybrid, it's 1/48 gallons per mile. So the difference in gallons per mile is 1/15 - 1/48. Then multiply that by total miles to get total gallons saved. Let me try that method to cross-verify.

Calculating 1/15 - 1/48. To subtract these fractions, find a common denominator. The least common denominator of 15 and 48. Let's see, prime factors of 15 are 3 and 5. Prime factors of 48 are 2^4 and 3. So LCM would be 2^4 * 3 * 5 = 16 * 3 * 5 = 240. So converting both fractions:

1/15 is equal to 16/240 (since 15*16=240). 1/48 is equal to 5/240 (since 48*5=240). Therefore, 16/240 - 5/240 = 11/240 gallons per mile. Then multiply by total miles, 12,000 miles. So 12,000 * 11 / 240. Let me compute that.

First, simplify 12,000 / 240. 240 times 50 is 12,000. So 12,000 / 240 = 50. Then multiply by 11: 50 * 11 = 550. So same result, 550 gallons. Okay, that confirms the answer. So regardless of the method, I get 550 gallons saved.

Therefore, I think that's the correct answer. Let me just check once more for any possible miscalculations.

Alternatively, maybe I can think about how many gallons each car uses. For the SUV, 12,000 divided by 15. Let me do 12,000 divided by 15 step by step. 15 goes into 120 (the first three digits) 8 times (15*8=120). Then bring down the next 0, making it 0. So 15 goes into 0 zero times. Then bring down the next 0, making it 0 again. So total is 800. Yep. For the hybrid, 12,000 divided by 48. Let me do 48 into 12,000. 48 into 120 is 2 times (48*2=96), remainder 24. Bring down the 0, making it 240. 48 into 240 is 5 times (48*5=240). So that's 25. Then bring down the next 0, which is 0. So total is 250. Yep. So 800 - 250 = 550.

Therefore, I think I can be confident that the answer is 550 gallons.

Another way to think about it: if the SUV gets 15 mpg, then for every gallon, it can go 15 miles. So, 12,000 miles would require 12,000 / 15 = 800 gallons. The hybrid gets 48 mpg, so for every gallon, it can go 48 miles. Therefore, 12,000 miles would require 12,000 / 48 = 250 gallons. The difference is 800 - 250 = 550 gallons saved.

All methods lead to the same answer. Therefore, I think the answer is 550 gallons.

**Final Answer**
Karla would have saved \boxed{550} gallons of gasoline last year.

To determine how many gallons of gasoline Karla would have saved last year by driving a fuel-efficient hybrid car instead of an SUV, we need to compare the fuel consumption of both vehicles over the same distance of 12,000 miles.

---

### Step 1: Calculate Gasoline Used by the SUV

The SUV averages **15 miles per gallon**. To find out how many gallons it used for 12,000 miles:

$$
\text{Gallons used by SUV} = \frac{12,000 \text{ miles}}{15 \text{ miles per gallon}} = 800 \text{ gallons}
$$

---

### Step 2: Calculate Gasoline Used by the Hybrid

The hybrid averages **48 miles per gallon**. To find out how many gallons it would use for the same distance:

$$
\text{Gallons used by hybrid} = \frac{12,000 \text{ miles}}{48 \text{ miles per gallon}} = 250 \text{ gallons}
$$

---

### Step 3: Calculate the Difference (Gasoline Saved)

To find the amount of gasoline saved:

$$
\text{Gasoline saved} = 800 \text{ gallons} - 250 \text{ gallons} = 550 \text{ gallons}
$$

---

### Final Answer

$$
\boxed{550}
$$

[Assistant]
Explanation:
Okay, so Karla drove her vehicle 12,000 miles last year. The question is asking how many gallons of gasoline she would have saved if she had driven a fuel-efficient hybrid car instead of an SUV. The hybrid averages 48 miles per gallon, and the SUV averages 15 miles per gallon. Alright, let me think about how to approach this.

First, I need to figure out how much gasoline she used with the SUV and then how much she would have used with the hybrid. Then, subtract the hybrid's gasoline usage from the SUV's to find the savings. That makes sense. Let me write that down.

So, the formula for gasoline used is total miles divided by miles per gallon. That gives gallons. So, for the SUV, gallons used would be 12,000 miles divided by 15 miles per gallon. Let me calculate that. 12,000 divided by 15. Hmm, 15 times 800 is 12,000, right? Because 15 times 100 is 1,500, so 15 times 800 is 12,000. So, that would be 800 gallons for the SUV.

Now, for the hybrid car, it's 48 miles per gallon. So, the gasoline used would be 12,000 divided by 48. Let me do that division. Hmm, 48 times 250 is 12,000? Let me check. 48 times 200 is 9,600. Then 48 times 50 is 2,400. So 9,600 + 2,400 is 12,000. Yes, so 250 gallons for the hybrid.

Therefore, the difference between the SUV and the hybrid is 800 gallons minus 250 gallons. That would be 550 gallons. So, she would have saved 550 gallons. Let me just verify my calculations again to make sure I didn't make a mistake.

First, SUV: 12,000 / 15. Let me do 12,000 divided by 15. Dividing both numerator and denominator by 5 gives 2,400 / 3, which is 800. Yep, that's correct. For the hybrid: 12,000 / 48. Let me divide numerator and denominator by 12 first. 12,000 divided by 12 is 1,000. 48 divided by 12 is 4. So now it's 1,000 / 4, which is 250. That checks out. So 800 - 250 is indeed 550.

Wait, but let me think again. Is there another way to approach this problem? Maybe by calculating the difference in fuel efficiency and then applying it somehow? But I don't think that's straightforward because the fuel efficiency is different. The total gallons saved depends on the total miles driven and the difference in fuel efficiency. But maybe I can think in terms of gallons per mile?

For the SUV, it's 1/15 gallons per mile, and for the hybrid, it's 1/48 gallons per mile. So the difference in gallons per mile is 1/15 - 1/48. Then multiply that by total miles to get total gallons saved. Let me try that method to cross-verify.

Calculating 1/15 - 1/48. To subtract these fractions, find a common denominator. The least common denominator of 15 and 48. Let's see, prime factors of 15 are 3 and 5. Prime factors of 48 are 2^4 and 3. So LCM would be 2^4 * 3 * 5 = 16 * 3 * 5 = 240. So converting both fractions:

1/15 is equal to 16/240 (since 15*16=240). 1/48 is equal to 5/240 (since 48*5=240). Therefore, 16/240 - 5/240 = 11/240 gallons per mile. Then multiply by total miles, 12,000 miles. So 12,000 * 11 / 240. Let me compute that.

First, simplify 12,000 / 240. 240 times 50 is 12,000. So 12,000 / 240 = 50. Then multiply by 11: 50 * 11 = 550. So same result, 550 gallons. Okay, that confirms the answer. So regardless of the method, I get 550 gallons saved.

Therefore, I think that's the correct answer. Let me just check once more for any possible miscalculations.

Alternatively, maybe I can think about how many gallons each car uses. For the SUV, 12,000 divided by 15. Let me do 12,000 divided by 15 step by step. 15 goes into 120 (the first three digits) 8 times (15*8=120). Then bring down the next 0, making it 0. So 15 goes into 0 zero times. Then bring down the next 0, making it 0 again. So total is 800. Yep. For the hybrid, 12,000 divided by 48. Let me do 48 into 12,000. 48 into 120 is 2 times (48*2=96), remainder 24. Bring down the 0, making it 240. 48 into 240 is 5 times (48*5=240). So that's 25. Then bring down the next 0, which is 0. So total is 250. Yep. So 800 - 250 = 550.

Therefore, I think I can be confident that the answer is 550 gallons.

Another way to think about it: if the SUV gets 15 mpg, then for every gallon, it can go 15 miles. So, 12,000 miles would require 12,000 / 15 = 800 gallons. The hybrid gets 48 mpg, so for every gallon, it can go 48 miles. Therefore, 12,000 miles would require 12,000 / 48 = 250 gallons. The difference is 800 - 250 = 550 gallons saved.

All methods lead to the same answer. Therefore, I think the answer is 550 gallons.

**Final Answer**
Karla would have saved \boxed{550} gallons of gasoline last year.

To determine how many gallons of gasoline Karla would have saved last year by driving a fuel-efficient hybrid car instead of an SUV, we need to compare the fuel consumption of both vehicles over the same distance of 12,000 miles.

---

### Step 1: Calculate Gasoline Used by the SUV

The SUV averages **15 miles per gallon**. To find out how many gallons it used for 12,000 miles:

$$
\text{Gallons used by SUV} = \frac{12,000 \text{ miles}}{15 \text{ miles per gallon}} = 800 \text{ gallons}
$$

---

### Step 2: Calculate Gasoline Used by the Hybrid

The hybrid averages **48 miles per gallon**. To find out how many gallons it would use for the same distance:

$$
\text{Gallons used by hybrid} = \frac{12,000 \text{ miles}}{48 \text{ miles per gallon}} = 250 \text{ gallons}
$$

---

### Step 3: Calculate the Difference (Gasoline Saved)

To find the amount of gasoline saved:

$$
\text{Gasoline saved} = 800 \text{ gallons} - 250 \text{ gallons} = 550 \text{ gallons}
$$

---

### Final Answer

$$
\boxed{550}
$$

Solution: Karla used $12000/15 = 800$ gallons to drive 12000 miles. Had she driven the hybrid, she would have used $12000/48 = 250$ gallons. Therefore, she would have saved $800 - 250 = \boxed{550}$ gallons.