Some might recognize this parametric plot, however I will derive an expression for y as a function of x. Each point in the xy plane on the graph swept by the parametric plot is given by (cos t, sin t).
As we all know, [; \sin t = \pm \sqrt[]{1-\cos^2 t} ;]. From this, we see that [; y = \pm \sqrt[]{1-x^2} ;]. Immediately, we recognize this as the equation of a circle with unit radius. As we know, the arc length of a circle from 0 to θ is rθ (with θ being in radians). This leads us immediately to (B).
As with all problems considering tangent lines, we must first take a derivative.
[; y(x) = x + e^x ;] so [; \frac{dy}{dx} = 1 + e^x ;].
[; \frac{dy}{dx}(0) = 1 ;]. The equation of a line is y(x) = mx + b. We know our slope to be 1, so we get: y(x) = x + b. We still have to determine b. The way to determine b comes from the fact that the tangent line must be in contact with our graph at the specified point (0, 1). So, 1 = 1*0 + b. b = 1, giving y = x + 1. (B)
[; y(x) = x + e^x ;] so [; \frac{dy}{dx} = 1 + e^x ;].
[; \frac{dy}{dx}(0) = 1 ;]. The equation of a line is y(x) = mx + b. We know our slope to be 1, so we get: y(x) = x + b. We still have to determine b. The way to determine b comes from the fact that the tangent line must be in contact with our graph at the specified point (0, 1). So, 1 = 1*0 + b. b = 1, giving y = x + 1. (B)
For this problem, I did it by example because the result easily holds in general. The intersection of V and W cannot have more than two dimensions since each only has two dimensions. So the only possible dimensions are 0, 1 and/or 2. If V = W, then the dimension of the intersection of V and W is is obviously 2, so now we have to determine if 0 or 1 are possible. 0 is easily possible if V = span{[; e_1, e_2 ;]} and W = span{[; e_3, e_4 ;]}. In this case, V and W are completely disjoint, so their intersections would be the empty set, which has dimension 0. The only possibly answer is then (D).
To begin this problem, I started with the fact that our function is strictly increasing, so it can have at most one root. This immediately eliminates (C), (D) and (E), leaving only two options for us to check. Therefore all we have to do is check the values of our function at x = 0 and x = 1.
f(0) = -1 and f(1) = e^1 - 1 ≈ 1.718. Because our function is less than 0 at x = 0 and greater than 0 at x = 1, the zero lies somewhere in this interval. Therefore our answer is (B).
There are two techniques to determine b, both of which you are undoubtedly familiar with. At the point x = 2, f(x) is 0. Also at this point, f'(x) is 0. So you can choose to take either path; personally, I would go with the critical value approach. Either will lead you to the conclusion that b = -12. Therefore f(x) = 3x^2 - 12x + 12. Inputting x = 5 gives a value of 27. (B)
On first inspection, one would assume that the circle that contacts the bottom of the parabola and intersects the sides would be the circle with the most intersections. However it is conceivable that one could have a circle that intersects four times, so we will have to check each.
x^2 = -y^2 + r^2 (General form of each solution.)
y + 4 = -y^2 + r^2. y^2 + y + (4-r^2) = 0.
[; y = \frac{-1 \pm \sqrt[]{1^2 - 4(1)(4-r^2)}}{2} ;]
[; y = \frac{-1 \pm \sqrt[]{-15 + 4r^2}}{2} ;]
This solution will only have real solutions if 4r^2 - 15 > 0. So, out of our bunch, r must be 2, 3, 4 or 5.
If r = 2, [; y = 0 ;].
If r = 3, [; y = \frac{-1 \pm \sqrt[]{21}}{2} ;].
If r = 4, [; y = \frac{-1 \pm 7}{2} ;], so [; y = 3 ;] or [; y = -4 ;].
If r = 5, [; y = \frac{-1 \pm \sqrt[]{85}}{2} ;]. The square root of 85 is approximately 9. y = -5 is not on the curve for y = x^2 - 4, so this solution can be ruled out immediately (and so can our exact solution by proxy).
So we only have to check what multiplicity the roots r = 3 and r = 4 have. The root y = -4 for r = 4 is a double root (the parabola crosses the y axis at y = -4) so it has 3 roots. r = 3 has four roots and can easily be checked. Our answer is then (C).
[; \int_{-3}^3 |x+1| dx ;] can be represented as the sum of the area of two triangles. One with base 2 and height 2 and another with base 4 and height 4. The area of these two is 10. (C).
For this problem, two legs of the triangle have length r. We wish to maximize area, so we will first seek an expression for the area in terms of θ.
[; A = \frac{1}{2} bh = \frac{1}{2} (2 r \cos\theta) r \sin\theta = \frac{1}{2} r^2 \sin(2\theta) ;]
The radius has unit length and sin(θ) ≤ 1, so the maximum value of A will occur when sin(θ) = 1, giving A = 1/2. (A).
This problem can be approached heuristically. The easy part of this problem is to start with K. The integrand for K is not less than 1, so the integral is a sum of numbers greater than (or equal to) one, so it must evaluate to being greater than 1. So we have already eliminated (B) and (D).
Now, [; 0 \le \sqrt[]{1-x^4} \le 1 ;] over the interval we are interested in. The area of the square [0,1]x[0,1] is 1. And since the area under our function is completely contained within this square, it must be less than 1. This immediate removes (E). The same argument holds for [; \sqrt[]{1-x^8} ;].
If 0 < x ≤ 1 and 0 < m < n with m, n integers, then x^m ≤ x^n. If we let m = 8 and n = 4, then x^8 ≤ x^4. If we subtract 1 from each side we get x^8 - 1 ≤ x^4 - 1. Multiplying everything by -1, we see that 1 - x^8 ≥ 1 - x^4. From this it follows that sqrt(1-x^8) ≥ sqrt(1-x^4) over the interval, so the area under sqrt(1-x^8) is larger than that of sqrt(1-x^4) so our answer is (A).
This problem is a simple area-under-the-curve problem and we want to determine the point at which the total area between the graph and the x-axis is a maximum. If we just inspect the graph, we can see that at x = 2, the area is a maximum, giving us that (B) is our answer. Of course you could show this is true by finding the total area at a few points, but time is of the essence!
At first, we might be tempted just to round down or up and approximate and hope we're near one of the answer choices. However, in this case the rounded result is fairly close to the median of two answers, so that is a bust. Instead, if we look closely at the numbers we can come up with a good approximation.
[; \sqrt[]{\frac{3}{2}} 266^{3/2} = \sqrt[]{\frac{3}{2}(266)(266)^2} = \sqrt[]{399 (266)^2} ;]
Since 399 is very close to 400, so we will approximate it with 400. So we end up with 20*266 = 5320, giving us (E).
[; A = \left ( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right ) ;]
From the assumption that the rows and columns sum to k, we get the following set of equations:
a + b = k a + c = k
b + d = k c + d = k
We see that a = d and b = c, giving that
[; A = \left ( \begin{array}{cc} a & b \\ b & a \\ \end{array} \right ) ;]
We wish to find the eigenvectors of A, so we wish to solve [; Av = \lambda v ;].
[; \left ( \begin{array}{cc} a-\lambda & b \\ b & a - \lambda \\ \end{array} \right) ;] [; \left(\begin{array}{c} v_1 \\ v_2 \\ \end{array} \right) ;] [; = \left(\begin{array}{c} 0 \\ 0 \\ \end{array} \right) ;]
If this matrix has an inverse, then the only possible solution is the zero vector, so it must not have an inverse, which says that (a- λ)^2 = b^2. Therefore λ = a ± b. If λ = a + b, we have the following system:
[; \left( \begin{array}{cc} a & b \\ b & a \\ \end{array} \right) ;] [; \left( \begin{array}{c} v_1 \\ v_2 \\ \end{array} \right) ;] [; = (a+b)\left(\begin{array}{c} v_1 \\ v_2 \\ \end{array} \right) ;]
So, [; a v_1 + b v_2 = a v_1 + b v_1 ;]. This implies that [; v_1 = v_2 ;], which leads us to III being an eigenvector and leaving only (C) and (E).
If λ = a - b, we will get that [; v_1 = - v_2 ;] which does not correspond to any of the other solutions, so our answer must be (C).
This is a simple optimization problem. Let A be the area and P be the perimeter. P = x = 2h + b, where h is the height and b is the base, and A = bh.
b = x - 2h, so A(h) = (x-2h)h = -2h^2 + xh.
If we want the maximum value of A, we will have to find when its derivative is 0. A'(h) = -4h + x = 0 implies that h = x/4. Therefore, A = x^2/8, (B).
Whenever I come across a problem like this, I typically do an induction proof (or just prove to myself that the pattern works). It is highly nonsensical to try to compute 7^25 in a reasonable amount of time, but there are techniques to use. Notice the following pattern in the ones digit of powers of 7:
7^0 = 1
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
... Etc.
Why do we choose to look at it this way? Well, if we only want to know the ones digit in 7^25, we only need to look at the ones digit of the powers of 7 because when multiplying two integers, only the multiplication of the ones digits of each number contribute to the ones digit of the resulting number. As we can see, the ones digits follow a nice pattern: 1, 7, 9, 3, 1, 7, 9, 3, ... . We see that the ones digit can be identified by the quantity n mod 4. In our case, 25 mod 4 = 1, so we only need to look at 7^1, which gives us a ones digit of 7. (D).
This question is a simple know-your-definitions-and-theorems question. (E) is false - all others are true. Can we construct an example for which (E) is false? A simple example would be f(x) = x if x < 0 and x^2 if x ≥ 0. From the left of 0, the derivative is 1, however from the right it is 0. So, the derivative is not defined at x = 0.
No comments:
Post a Comment