I'm new to this forum and just discovered these solutions. Thanks for them. Here are some comments on other approaches, some of them significantly faster than yours or others mentioned in this thread.
For #24, no need to eliminate D before A and C. Every homogeneous system is consistent (eliminate A), and the set of solutions is a vector space, hence closed under addition (eliminate C).
For #31, you could just write a quick expression for det(A-lambda I) -- no need to simplify it fully -- and then test the given values to see whether they're roots. This might be slightly longer than your solution, but it's still quite fast.
For #44, no need to solve the ODE -- that's a time sink. Simply analyze the slope field when x<0. It helps first to look for the equilibria, where y'=0. This gives x-xy=0, so x=0 or y=1. Since y'=x(1-y) and y is initially less than 1, y' will be negative for x<0. So the solution will flow up to the equilibrium y=1 as x tends to negative infinity.
For #53, you don't need the Leibniz rule. We are told g(x) = int_0^x f(y)(y-x)dy. This is just
int_0^x yf(y)dy - x int_0^x f(y)dy
Now differentiate, using the fundamental theorem of calculus and the product rule. (Actually, now that I look at the question again, I see it does not state that f is continuous, so strictly speaking, the fundamental theorem doesn't apply, nor does Leibniz's rule. I'm not sure yet whether the test-writers simply made a mistake, or whether the problem is vastly more interesting than it seems.)
For #56, statement III asks whether there's a constant C so that |sin(x)-x|<=C|x^3| for all x. Your solution is to think geometrically. p-adic's, earlier in this thread, uses a bound for x>1 and then modifies it for x<1. But the easiest and most straightforward approach is simply to note that |sin(x)-x| is the absolute value of the remainder of the first-order Taylor approximation of sine. Since the Taylor series for sine is alternating, the absolute value of the remainder is bounded by the absolute value of the next term, |x^3/6|. This immediately gives a C, namely 1/6, that works for all values of x. The argument requires no case analysis.
For #57, statement II, you appeal to sin(1/x) as a counterexample. But the oscillatory behavior is overkill here. The function 1/x works just as well; it is continuous on (0,1) but the f(x_n) diverge to infinity.
Last edited by fluentmundo
on Wed Nov 11, 2015 5:43 am, edited 1 time in total.