(x 3)(x + 5) = (x 3)(x 1) = x + 5

Transcription

1 RMT 3 Calculus Test olutions February, 3. Answer: olution: Note that Answer: 3 3) + 5) = 3) ) = = =. olution: We have that f) = b and f ) = ) + b = b + 8. etting these equal to each other, we see that b = Answer:, ) and, ) olution: ince this is in an indeterminate form, we can use L Hospital s Rule to obtain sin ) ae a b =. However, the numerator goes to zero, so the denominator must also go to zero to give us another indeterminate form. This implies that a = b. Using L Hospital s Rule again, we have that cos ) a e a =. The numerator goes to, so the denominator must go to. Therefore, a = b = ±, giving us a, b) =, ) and, ).. Answer: e + Let w = so that w = and d = w dw. Then the integral becomes To find this integral, use integration by parts: u = w du = dw; dv = e w dw v = e w we w dw = uv v du = we w e w dw = w )e w. we w dw. Evaluating w )e w at our its of integration yields e Answer: 5 olution : For any function f with f) =, we know that f) = f) f) = f ). sin5), tan), and log + ) are all at =, and their derivatives at are 5,, and, respectively. o, divide numerator and denominator by 5 and re-arrange to get sin 5) tan 3 ) log + )) 5 = sin5) ) log+) ) 3 tan) ) 5 = = 5.

2 RMT 3 Calculus Test olutions February, 3 olution : Recall from Taylor series that if f) =, then f) f ) when is small. This allows us to write sin 5) tan 3 ) 5) ) 3 log + )) 5 = ) 5 = Answer: 3 Bring the sum into the integral, so we have π 3 sin k d. k= The integrand is a geometric series, so the answer is π 3 π sin d = 3 π ) sec d = tan tan) = Answer: 3 e olution: ince f n) )/n + + C) is independent of n, we can say that it is equal to g). Multiplying by n + + C), we have that f n) ) = n + + C)g). Taking a derivative with respect to, we obtain f n+) ) = n + + C)g ) + g). However, this is equal to n C)g) by the problem statement. Canceling terms, we obtain that g) = g ). The only class of functions that is its own derivative is ae, so we have that g) = ae for some constant a). Now, f ) = + C + )ae, so f ) = gives us that a = /C + ). We also have that f) d = + C C + e d = C + e ). Integration by parts gives us which simplifies to e )C + C + = C + e ), C = 3 e, from which it follows that the answer is 3 e. 8. Answer: +log) olution : First, epress and y as functions parametrized by m. We have the system y = m + y = m.

3 RMT 3 Calculus Test olutions February, 3 olving for y, we get y = m m. Hence, maimizing y is equivalent to maimizing. By +m +m differentiating with respect to m, we see that the maimum occurs when m =, at the point, ). Now we just need to compute the integral. However, this parametric form is not convenient. Instead, by drawing the line y =, we notice that the integral splits up into a right isosceles triangle and a region between the line y = and the y-ais. This suggests that we should convert to polar coordinates. In fact, f) is equivalent to the graph rθ) =, since a line tanθ) at angle θ to the -ais has slope tanθ). The area we wish to compute is π/ rθ) dθ = π/ cotθ) dθ = [logsinθ))]π/ π/ = log/ )) = log). ) We add this area to the area of the triangle, which is =, so our final answer is + log). olution : We begin as before to find a, but present a different method of computing the integral. olving for in terms of y, we get that + y = /y = y + y 3 = = = ± y. y We only care about the region where ± y = y, since, y. Hence, we take = y y. Notice that we can compute the desired quantity as ) y dy, y since within the square bounded by the coordinate aes and, y, the area between the curve and the -ais plus the area between the curve and the y-ais sum to the area of the whole square.

4 RMT 3 Calculus Test olutions February, 3 Now, using the substitution u = y, we get y dy = y = = = u)u u du u)u u) + u) du u + u du + u du The answer is 9. Answer: /3 minus this quantity, so report = [u log + u)] = log). + log) olution : Observe that by pulling a factor of cos out of the denominator, we can write the given integral as d π/ + tan ) cos = sec d + tan ). We now substitute u = tan + : du = sec tan d = sec u ) d.. Thus, our integral is equal to which simplifies to u u du = [ u + 3 u 3 ] u 3 u du, = 3. olution : Let I be the value of the given integral. Note that I = ) ) sin) + cos) d, sinθ) ) which is the polar area bounded by the curve rθ) = + cosθ) and the and y aes for θ [, π/]. Converting to Cartesian coordinates, we get ) = r sinθ) + cosθ) ) = r sinθ) + r cosθ) = + y = = y = ) = +.

5 RMT 3 Calculus Test olutions February, 3 Therefore, I = = + d [ + 3 3/ = + 3 = 6 ] = I = 3. Answer: π π π olution : Observe that cos ) looks like a bunch of spikes, centered at, π, π,..., each with mass I n = cos ) d. We can integrate by parts to see that I k = = k ) cos ) k d = [ ] π/ cos ) k sin + k ) cos ) k cos ) d = k )I k I k ). cos ) k sin d Therefore, I k = k k I n k = I n = k= ) k k I = π n k= k k. As n, the spikes get sharper and sharper; this means that the denominator of the integrand gets concentrated at =, π, π,.... Therefore, we epect that as n, n ) k cos ) n ) k I n k d k kπ = π π = π π π. k= k= k= olution : We present a more rigorous approach here. First, rewrite the problem into the following form: For a positive integer n, let a n = d. Also let c = n n k= k ), which is a positive finite constant. Evaluate c a n. Let B = {, π, π,... }. The idea is that the numerator of the integrand approaches a function with cπ area concentrated infinitely closely to each point in B. Therefore the it should be d = cπ π = cπ π, B where the last equality follows by the formula for summing geometric series. We will soon get to a more precise way of thinking about the area being concentrated infinitely closely to points in B, but first let s see why the numerator should have cπ area around each

6 RMT 3 Calculus Test olutions February, 3 point in B. ince the area around each point in B is the same cos is periodic), we need only consider the area around. We can apply integration by parts to find a formula for the area around in each term of the sequence. The recurrence is d = ) n ) d. Repeatedly applying this formula, we get d = π n n k ). Taking the it as n, the area around goes to cπ. o the answer makes sense. Now we will prove it more rigorously. For / B, the integrand goes to as n because cos <. o for any open set sufficiently disjoint from B, we might guess that d =. If we require that every point in is at least ɛ > away from any point in B, then this is indeed true. There are two ways to see this. The fanciest way to see it is to use the dominated convergence theorem which says that if a sequence of functions f n converges pointwise to a funciton f and if there is some function φ with f n ) < ϕ) for all and ϕ <, then f n = f. To apply this theorem, we let f n be the integrand of the n-th term of the sequence. To construct ϕ, notice that since every point in is at least ɛ away from every point in B, there is some δ < so that cos < δ for all. o is bounded by nδ for all. ince nδ has a finite it as n, is bounded by some finite number B for all. o we can let ϕ) = B/. Then f n ) < ϕ) for all and ϕ <, just as we need to apply the theorem. o we apply the theorem to get d = d = d =. But we of course don t epect you to know the dominated convergence theorem, so we can also prove this using a bare hands method that is actually easier. Bare hands is usually much harder than the dominated convergence theorem proof. That is why people use the dominated convergence theorem. But we have arranged for this problem to work with bare hands.) As we argued above, there is some δ < so that cos < δ for all. Then < nδ for all. o we have a bound d nδ d. ince the integral converges, this goes to as n so we again have the desired result. The upshot of all this is that we can now define B ɛ to be the points in [, ) that are within ɛ of B and have d = B ɛ d. k=

7 RMT 3 Calculus Test olutions February, 3 To calculate the integral on the right, notice that it is just the sum over all integers k of kπ+ɛ d = ɛ kπ ɛ d. kπ ɛ By the formula for summing geometric series, the sum of this over all integers k is ɛ d = π ncos ) π ɛ d. ) o we have reduced the problem to calculating the following it: ɛ ɛ d. To do this, bound the it above and below by taking the highest and lowest possible values of out of the integral: ɛ ɛ ɛ ncos ) ɛ d ɛ ɛ d ɛ ncos ) d. By a very similar argument as above, nothing outside ɛ, ɛ) contributes to the integrals in our bounds and therefore ɛ π/ ncos ) d = ncos ) d. ɛ We have already calculated the right hand side: it is cπ. o we can plug this back into our bounds to get ɛ ɛ cπ ɛ d ɛ cπ Plugging this bound into ) gives ɛ cπ π ncos ) π d ɛ cπ π π. ince ɛ was arbitrary, taking ɛ forces as desired. d = cπ π π, Finally, you might be interested in knowing why c is a positive finite constant. This is not necessary to solve the problem, but it is necessary to be sure that the problem makes sense. And it is interesting.) To see this, let b n = n n k= k ). Then log b n = n log n + log k ) = n log n + k + O k )). k= Here O ) denotes some function of k whose absolute value is always less than C for some k k C big enough. The fact that log k ) = k + O ) follows from the taylor epansion of log k k= ɛ

8 RMT 3 Calculus Test olutions February, 3 around. It is well known that n k= k = log n + γ + O n ) where γ is some constant. Plugging this in gives log b n = log n log n γ + O n n ) + O k ) = γ + O n n ) + O k ). ince k= converges to some constant, n k k= O ) converges to some finite constant α as k n. Therefore log b n γ+α as n. This is some finite number, so c = ep γ+α) is a positive finite number as desired. k= k=