answer:Given a > 1 and x in [1,2], - y=log _{a}x is an increasing function, - y=frac{4}{x} is a decreasing function. Thus, the function f(x)=log _{a}x- frac{4}{x} is an increasing function. Hence, the maximum value of f(x) is attained when x=2, denoted as log _{a}2 - 2 = 0. Solving for a, we obtain: boxed{a=sqrt{2}}. The choice is boxed{B}. By using the monotonicity of the function f(x)=log _{a}x- frac{4}{x} (a > 1), we can find the maximum value on the interval [1,2]. Then, we can set up and solve an equation to obtain the value of a. This problem tests your understanding of maximum values of a function, monotonicity of functions, and solving equations. It is of moderate difficulty.

answer:First, find a common denominator for frac{7}{12} and frac{3}{8}. The least common multiple of 12 and 8 is 24. Rewrite the fractions: - frac{7}{12} = frac{7 cdot 2}{12 cdot 2} = frac{14}{24} - frac{3}{8} = frac{3 cdot 3}{8 cdot 3} = frac{9}{24} Add the fractions: [ frac{14}{24} + frac{9}{24} = frac{14 + 9}{24} = frac{23}{24} ] Now, multiply this result by frac{2}{3}: [ frac{23}{24} cdot frac{2}{3} = frac{23 cdot 2}{24 cdot 3} = frac{46}{72} ] Simplify frac{46}{72} by dividing numerator and denominator by their greatest common divisor, which is 2: [ frac{46 div 2}{72 div 2} = frac{23}{36} ] Thus, the final answer is boxed{frac{23}{36}}.

answer:First, we need to find text{gcd}(300, 315). Since 315 is 300 + 15, we perform a Euclidean algorithm: [ text{gcd}(300, 315) = text{gcd}(300, 315-300) = text{gcd}(300, 15). ] We further reduce this as: [ text{gcd}(300, 15) = text{gcd}(300 - 20 times 15, 15) = text{gcd}(0, 15) = 15. ] Therefore, the gcd of the numbers 2^{300}-1 and 2^{315}-1 will be 2^{gcd(300, 315)} - 1 = 2^{15} - 1. Calculating 2^{15} - 1: [ 2^{15} = 32768 implies 2^{15} - 1 = 32768 - 1 = 32767. ] Thus, the greatest common divisor of 2^{300}-1 and 2^{315}-1 is boxed{32767}.

answer:We evaluate each case by expressing p and q as odd integers, say p = 2a+1 and q = 2b+1: 1. **Option (A) p^2 + 5q:** - Substituting p = 2a+1 and q = 2b+1, we get: [ p^2 + 5q = (2a+1)^2 + 5(2b+1) = 4a^2 + 4a + 1 + 10b + 5 = 4a^2 + 4a + 10b + 6 ] - This is an even number since 4a^2 + 4a + 10b is even, plus 6. 2. **Option (B) 2p^2 - 4q:** - Substituting, we get: [ 2p^2 - 4q = 2(2a+1)^2 - 4(2b+1) = 8a^2 + 8a + 2 - 8b - 4 = 8a^2 + 8a - 8b - 2 ] - This is also even as all terms, except -2, are clearly even and subtracting 2 keeps it even. 3. **Option (C) 5p^2 + 2q^2:** - Substituting, we find: [ 5p^2 + 2q^2 = 5(4a^2 + 4a + 1) + 2(4b^2 + 4b + 1) = 20a^2 + 20a + 5 + 8b^2 + 8b + 2 = 20a^2 + 20a + 8b^2 + 8b + 7 ] - Odd, as 20a^2, 20a, 8b^2, 8b are all even, and 5 + 2 is 7, which is odd. 4. **Option (D) (pq + 2)^2:** - Substituting, we get: [ pq + 2 = (2a+1)(2b+1) + 2 = 4ab + 2a + 2b + 1 + 2 = 4ab + 2a + 2b + 3 ] - This expression squared remains odd since pq + 2 simplifies to an even number plus 3. 5. **Option (E) p^2 + pq + q^2:** - Substituting, we get: [ p^2 + pq + q^2 = (2a+1)^2 + (2a+1)(2b+1) + (2b+1)^2 = 4a^2 + 4a + 1 + 4ab + 2a + 2b + 1 + 4b^2 + 4b + 1 ] - The entire expression simplifies to an odd number, as all other terms are even and 1+1+1 is added. The only expressions that must be an odd integer for odd p and q are textbf{(C) 5p^2 + 2q^2 text{ and } textbf{(E)} p^2 + pq + q^2}. The final answer is C) boxed{textbf{(C)} 5p^2 + 2q^2} E) boxed{textbf{(E)} p^2 + pq + q^2}