answer:Start by analyzing the given equation: [ x^4 + y^2 = 4y + 4. ] Step 1: Rearrange the equation Isolate the terms involving ( y ): [ y^2 - 4y - x^4 + 4 = 0. ] Step 2: Complete the square Complete the square for ( y ): [ (y-2)^2 - 4 + x^4 = 4 ] [ (y-2)^2 = 4 + 4 - x^4 ] [ (y-2)^2 = 8 - x^4. ] Step 3: Analyze the powers of ( x ) Since ( x^4 ) is also a power, but lower than the original, the possible values of ( x^4 ) for integer ( x ) are limited to the powers of integers. However, ( (y-2)^2 geq 0 ) implies: [ 8 - x^4 geq 0 ] [ x^4 leq 8. ] The integers ( x ) that satisfy this are ( x = 0, pm 1, pm 2 ) since: - ( 0^4 = 0 ), - ( 1^4 = 1 ), - ( (-1)^4 = 1 ), - ( 2^4 = 16 ) (exceeds 8, so only ( x = 0, pm 1 )). Step 4: Substitute possible values of ( x ) and solve for ( y ) - **For ( x = 0 )**: [ (y-2)^2 = 8 ] [ y-2 = pm sqrt{8} ] [ y = 2 pm 2sqrt{2} ] (Not integers, discard) - **For ( x = pm 1 )**: [ (y-2)^2 = 7 ] [ y-2 = pm sqrt{7} ] [ y = 2 pm sqrt{7} ] (Not integers, discard) Conclusion The possible integer solutions do not exist as the square roots do not result in integers. Thus, the number of ordered pairs ((x, y)) that satisfy the equation is (0). The final answer is boxed{textbf{(A)} 0}

answer:Since log_{a}2 < log_{b}2 < 0 = log_{b}1, it follows that log_{2}b < log_{2}a < 0. Therefore, according to the monotonicity of the logarithmic function, we have 0 < b < a < 1. Hence, the relationship among 0, 1, a, b is 0 < b < a < 1. Thus, the answer is boxed{0 < b < a < 1}.

answer:1) The inequality 2f(x) < 4-|x-1| is equivalent to 2|x+2|+|x-1| < 4. We have three cases to consider: - When xleqslant -2, we have -2(x+2)-x+1 < 4, which simplifies to -frac{7}{3} < x. - When -2 < x < 1, we have 2(x+2)-x+1 < 4, which simplifies to -2 < x-1. - When xgeqslant 1, we have 2(x+2)+x-1 < 4, which has no solution. Combining the solutions from the first two cases, we get the solution set -frac{7}{3} < x < -1. 2) Since |x-a|-f(x)=|x-a|-|x+2|leqslant |x-a-x-2|=|a+2|, the maximum value of |x-a|-f(x) is |a+2|. Also, since m+n=1(m > 0,n > 0), we have (frac{1}{m}+frac{1}{n})(m+n)=frac{n}{m}+frac{m}{n}+2geqslant 2+2=4, so the minimum value of frac{1}{m}+frac{1}{n} is 4. For the inequality |x-a|-f(x)leqslant frac{1}{m}+frac{1}{n} to always hold, we must have |a+2|leqslant 4. Solving this inequality gives us -6leqslant aleqslant 2. Therefore, the range of values for the real number a is boxed{[-6,2]}.

answer:Analysis: This problem examines the graph and properties of the function y=Asin (omega x+varphi). Within the interval left( dfrac{pi}{3}+varphi, dfrac{2pi}{3}+varphi right), there is exactly one kpi, kin mathbb{Z}. By considering the given options, we can find the appropriate solution. Step-by-step solution: 1. We know that the center of symmetry of the function y=sin left( 2x+varphi right), left(0 < varphi < dfrac{pi }{2} right) lies within the interval left( dfrac{pi }{6},dfrac{pi }{3} right). 2. This means that the range of 2x + varphi should be left( dfrac{pi}{3}+varphi, dfrac{2pi}{3}+varphi right) and there should be exactly one kpi within this range, where k is an integer. 3. By evaluating the given options, when varphi = dfrac{5pi}{12}, pi lies within the specified range, making it the correct choice. The final answer is: boxed{D: dfrac{5pi }{12}}