answer:To solve this problem, we need to follow a series of steps based on the given information and the properties of hyperbolas. 1. **Identify the Equation of the Asymptote**: The given hyperbola C has asymptotes that can be derived from its standard form. For the hyperbola frac{x^2}{a^2}-frac{y^2}{b^2}=1, the equations of the asymptotes are y = pm frac{b}{a}x. Since a perpendicular line is drawn from F_1 to an asymptote and it's given that D is the foot of the perpendicular, we consider the asymptote y = -frac{b}{a}x for our calculations. 2. **Determine the Foci**: The foci of the hyperbola are located at F_1(-c, 0) and F_2(c, 0), where c = sqrt{a^2 + b^2}, based on the standard properties of hyperbolas. 3. **Calculate |DF_1| and |OD|**: Using the distance formula from a point to a line, we find that the distance from F_1 to the asymptote, which is |DF_1|, equals b. The distance from the origin O to D, denoted as |OD|, is found by considering the triangle ODF_1 and using the Pythagorean theorem, leading to |OD| = sqrt{c^2 - b^2} = a. 4. **Relation Between |DF_2| and |OD|**: It's given that |DF_2| = 2sqrt{2}|OD|. Substituting |OD| = a into this equation, we get |DF_2| = 2sqrt{2}a. 5. **Use Cosine Rule to Find a Relation**: By applying the cosine rule in triangles ODF_1 and ODF_2, and using the fact that cos angle F_1OD = -cos angle DOF_2, we establish a relationship: [ frac{a^2 + c^2 - b^2}{2ac} + frac{a^2 + c^2 - 8a^2}{2ac} = 0 ] Simplifying this equation, we find that c^2 = 5a^2. 6. **Calculate the Eccentricity**: The eccentricity e of a hyperbola is given by e = frac{c}{a}. Substituting c^2 = 5a^2 into this formula, we get e = sqrt{5}. Therefore, the eccentricity of the hyperbola C is boxed{sqrt{5}}, which corresponds to option C.

answer:# Solution: Part (1): Finding the Equation of the Tangent Line Given the function f(x) = ax - ln x where a in mathbb{R} and x > 0. - **Step 1:** Calculate the derivative of f(x). [ f'(x) = a - frac{1}{x} ] - **Step 2:** Let the point of tangency be (x_{0}, y_{0}). The slope of the tangent line passing through the origin is given by the derivative at x_0, so: [ frac{y_{0} - 0}{x_{0} - 0} = f'(x_{0}) = a - frac{1}{x_{0}} ] - **Step 3:** Since y_{0} = ax_{0} - ln x_{0}, we substitute and get: [ frac{ax_{0} - ln x_{0}}{x_{0}} = a - frac{1}{x_{0}} ] Simplifying this gives us: [ ax_{0} - ln x_{0} = ax_{0} - 1 ] Leading to: [ ln x_{0} = 1 ] Therefore, x_{0} = e and y_{0} = ae - 1. - **Step 4:** The equation of the tangent line is: [ y - y_{0} = f'(x_{0})(x - x_{0}) ] Substituting x_{0} = e, y_{0} = ae - 1, and f'(x_{0}) = a - frac{1}{e}, we get: [ y = left(a - frac{1}{e}right)x ] Simplifying, the equation of the tangent line is: [ y = frac{ae - 1}{e}x ] Hence, the equation of the tangent line passing through the origin is boxed{y = frac{ae - 1}{e}x}. Part (2): Proving the Equation f(x) + ax^{2} = 0 Has Only One Real Root When a < 0 - **Step 1:** Consider h(x) = ax - ln x + ax^{2}. We need to show that h(x) = 0 has only one real root when a < 0. - **Step 2:** Calculate the derivative of h(x): [ h'(x) = a - frac{1}{x} + 2ax = frac{2ax^{2} + ax - 1}{x} ] - **Step 3:** Let u(x) = 2ax^{2} + ax - 1. The derivative of u(x) is: [ u'(x) = 4ax + a = a(4x + 1) ] Since a < 0 and x > 0, u'(x) < 0 on (0, +infty), meaning u(x) is strictly decreasing. - **Step 4:** Since u(x) is strictly decreasing, we have u(x) < u(0) = -1 < 0 for all x > 0. This implies h'(x) < 0 on (0, +infty), meaning h(x) is strictly decreasing. - **Step 5:** As x rightarrow 0, h(x) rightarrow +infty, and as x rightarrow +infty, h(x) rightarrow -infty. Therefore, h(x) crosses the x-axis exactly once. Hence, when a < 0, the equation f(x) + ax^{2} = 0 has only one real root, which completes the proof. boxed{text{Proved}}

answer:1. **Let the number of pages in one English book be (x).** 2. **Let the number of pages in one literature book be (y).** 3. **According to the problem, we have the following system of equations:** [ x = y + 12 quad text{(1)} ] [ 3x + 4y = 1275 quad text{(2)} ] 4. **Substitute equation (1) into equation (2) to eliminate (x):** [ 3(y + 12) + 4y = 1275 ] 5. **Expand and simplify to solve for (y):** [ 3y + 36 + 4y = 1275 ] [ 7y + 36 = 1275 ] [ 7y = 1275 - 36 ] [ 7y = 1239 ] [ y = frac{1239}{7} ] [ y = 177 ] 6. **Substitute (y = 177) back into equation (1) to find (x):** [ x = 177 + 12 ] [ x = 189 ] 7. **Thus, one English book has (189) pages.** [ boxed{189} ]

answer:F(x)=t^{2}-10t-4=(t-5)^{2}-29, Since the range of the function t=f(x) is (0,8], 0 leqslant (t-5)^{2} < 25, -29 leqslant (t-5)^{2}-29 < -4, Hence, the answer is boxed{D}. From the range of the function t=f(x), we know that f(x) in (0,8]. The range of the function is found using the method of completion of the square. This problem tests the methods for finding the range of a function. In high school, methods for finding the range of a function include: 1, observation method, 2, completion of the square, 3, inverse function method, 4, discriminant method, 5, substitution method, 6, graphical method, 7, inequality method, 8, separation of variables method, 9, monotonicity method, 10, using derivatives to find the range of a function, 11, extreme value method, 12, construction method, 13, proportional method. The choice of method depends on the specific problem.