answer:Since tan α=3x^{2}-1, We have tan α in [-1,+infty). When tan α in [0,+infty), α in [0, frac {pi}{2}); When tan α in [-1,0), α in [frac {3pi}{4}, pi); Therefore, α in [0, frac {pi}{2}) cup [frac {3pi}{4}, pi). So, the answer is: boxed{[0, frac {pi}{2}) cup [frac {3pi}{4}, pi)}. By the geometric meaning of derivatives, the slope of the tangent line is equal to the derivative at that point. Then, find the range of the slope according to the value range of the derivative. Finally, use the relationship between the slope and the angle of inclination (k=tan α) to find the range of α.

answer:To determine the mass percentage of hydrogen in water (H₂O), we need to consider the average atomic mass of hydrogen taking into account the natural abundance of its isotopes, and then calculate the mass percentage of hydrogen in a molecule of water. The average atomic mass of hydrogen (H) can be calculated as follows: Average atomic mass of H = (abundance of ¹H × atomic mass of ¹H) + (abundance of ²H × atomic mass of ²H) Given that the atomic mass of ¹H (protium) is approximately 1 amu and the atomic mass of ²H (deuterium) is approximately 2 amu, and their respective abundances are 99.98% and 0.02%, we can calculate the average atomic mass of hydrogen: Average atomic mass of H = (0.9998 × 1 amu) + (0.0002 × 2 amu) Average atomic mass of H = 0.9998 amu + 0.0004 amu Average atomic mass of H ≈ 1.0002 amu Now, let's calculate the mass percentage of hydrogen in water. A water molecule consists of two hydrogen atoms and one oxygen atom. The atomic mass of oxygen (O) is approximately 16 amu. The molar mass of water (H₂O) is the sum of the average atomic mass of hydrogen (multiplied by 2, since there are two hydrogen atoms) and the atomic mass of oxygen: Molar mass of H₂O = (2 × average atomic mass of H) + atomic mass of O Molar mass of H₂O = (2 × 1.0002 amu) + 16 amu Molar mass of H₂O = 2.0004 amu + 16 amu Molar mass of H₂O ≈ 18.0004 amu The mass percentage of hydrogen in water can be calculated by dividing the total mass of hydrogen in a water molecule by the molar mass of water, and then multiplying by 100 to get the percentage: Mass percentage of H in H₂O = [(2 × average atomic mass of H) / molar mass of H₂O] × 100 Mass percentage of H in H₂O = [(2 × 1.0002 amu) / 18.0004 amu] × 100 Mass percentage of H in H₂O = [2.0004 amu / 18.0004 amu] × 100 Mass percentage of H in H₂O ≈ 11.113% Therefore, the mass percentage of hydrogen in water, taking into account the presence and abundance of hydrogen isotopes, is approximately boxed{11.113%} .

answer:We start by simplifying the expression: 1. **Simplify each term:** - left(a+b+c+dright)^{-1} = frac{1}{a+b+c+d}. - left(a^{-1}+b^{-1}+c^{-1}+d^{-1}right) = frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{1}{d} = frac{ab+bc+cd+da+ac+bd}{abcd}. - left(ab+bc+cd+da+ac+bdright)^{-1} simplifies to frac{1}{ab+bc+cd+da+ac+bd}. - left[(ab)^{-1}+(bc)^{-1}+(cd)^{-1}+(da)^{-1}+(ac)^{-1}+(bd)^{-1}right] can be rewritten as frac{a+b+c+d}{abcd}. 2. **Combine the terms:** [ left(frac{1}{a+b+c+d}right)left(frac{ab+bc+cd+da+ac+bd}{abcd}right)left(frac{1}{ab+bc+cd+da+ac+bd}right)left(frac{a+b+c+d}{abcd}right) ] - Simplifying: [ left(frac{1}{a+b+c+d}right)left(frac{a+b+c+d}{abcd}right) = frac{1}{abcd} ] [ left(frac{ab+bc+cd+da+ac+bd}{abcd}right)left(frac{1}{ab+bc+cd+da+ac+bd}right) = frac{1}{abcd} ] - Final Simplification: [ frac{1}{abcd} cdot frac{1}{abcd} = frac{1}{(abcd)^2} ] Thus, the simplified expression is frac{1}{a^2b^2c^2d^2}, given as: [ a^{-2b^{-2}c^{-2}d^{-2}} ] The final answer is boxed{textbf{(B)} a^{-2}b^{-2}c^{-2}d^{-2}}

answer:1. We start by noting that the set (mathcal{P}) consists of planes in three-dimensional space with positive (x), (y), and (z) intercepts summing to one. The equation of such a plane can be written as: [ frac{x}{a} + frac{y}{b} + frac{z}{c} = 1 ] where (a + b + c = 1). 2. Given a point ((x, y, z)) with (min {x, y, z} > 0), it lies on exactly one plane in (mathcal{P}). We need to find the maximum possible integer value of (left(frac{1}{4} x^2 + 2y^2 + 16z^2right)^{-1}). 3. To maximize (left(frac{1}{4} x^2 + 2y^2 + 16z^2right)^{-1}), we need to minimize (frac{1}{4} x^2 + 2y^2 + 16z^2). Let: [ f(x, y, z) = frac{1}{4} x^2 + 2y^2 + 16z^2 ] 4. Using the constraint (frac{x}{a} + frac{y}{b} + frac{z}{c} = 1), we apply the method of Lagrange multipliers. Define the Lagrangian: [ mathcal{L}(x, y, z, lambda) = frac{1}{4} x^2 + 2y^2 + 16z^2 + lambda left( frac{x}{a} + frac{y}{b} + frac{z}{c} - 1 right) ] 5. Taking partial derivatives and setting them to zero, we get: [ frac{partial mathcal{L}}{partial x} = frac{1}{2} x + lambda frac{1}{a} = 0 implies lambda = -frac{1}{2} ax ] [ frac{partial mathcal{L}}{partial y} = 4y + lambda frac{1}{b} = 0 implies lambda = -4by ] [ frac{partial mathcal{L}}{partial z} = 32z + lambda frac{1}{c} = 0 implies lambda = -32cz ] 6. Equating the expressions for (lambda), we get: [ -frac{1}{2} ax = -4by implies ax = 8by ] [ -4by = -32cz implies by = 8cz ] 7. Solving these, we find: [ x = frac{8by}{a}, quad y = frac{8cz}{b} ] 8. Substituting (a = 2b) and (b = 2c) into (a + b + c = 1), we get: [ a = frac{4}{7}, quad b = frac{2}{7}, quad c = frac{1}{7} ] 9. Using these values, we find: [ x = frac{16}{49}, quad y = frac{4}{49}, quad z = frac{1}{49} ] 10. Substituting (x), (y), and (z) into (f(x, y, z)), we get: [ fleft(frac{16}{49}, frac{4}{49}, frac{1}{49}right) = frac{1}{4} left(frac{16}{49}right)^2 + 2 left(frac{4}{49}right)^2 + 16 left(frac{1}{49}right)^2 ] [ = frac{1}{4} cdot frac{256}{2401} + 2 cdot frac{16}{2401} + 16 cdot frac{1}{2401} ] [ = frac{256}{9604} + frac{32}{2401} + frac{16}{2401} ] [ = frac{256}{9604} + frac{48}{2401} ] [ = frac{256}{9604} + frac{192}{9604} ] [ = frac{448}{9604} = frac{1}{21.5} ] 11. Therefore, the maximum possible integer value of (left(frac{1}{4} x^2 + 2y^2 + 16z^2right)^{-1}) is: [ leftlfloor frac{1}{frac{1}{21.5}} rightrfloor = leftlfloor 21.5 rightrfloor = 21 ] The final answer is (boxed{21}).