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Precalculus Individual January 2011 Palm Harbor University High School Pre-­‐calculus Individual 2011 Answers and Solutions 1. B The 3rd decimal place is 1 (consult any calculator or book). 2. E The conic is degenerate (upon completing the square, the right hand side equals 0). We therefore have 2 intersecting lines. 3. D This is also sin^2(k). If we use the identity sin^2(90-­‐k)=cos^2(k), we can manipulate the arguments to find convenient expressions of the form sin^2(a)+cos^2(a)=1. This we can do 89/2 ways. Since the sin (90) = 1, we have 91/2. 4. A Letting the expression be x and then taking the tan of both sides, we see that these angles have arctan values of ½ and 1/3. Using the tan sum formula, we see that the equation simplifies to tan(x)=1, so x is pi/4. 5. C (2) is neither because this operation will make a function with even and odd exponents. (3) is neither because and odd plus an even, once again, will produce a function with odd and even exponents. (5) is even because –p(x)/-­‐q(x)= p(x)/q(x). (7) This will be even because no matter what g(x), f(-­‐x)=f(x). Also, g(-­‐
x)=g(x), so two even functions added is still even. (8) is even because –p(x) times –q(x)=p(x)q(x). (5+7+8)-­‐(0)=20 ⎡1 1⎤
6. E Only one number must be zero, in any matrix where n > 2. The matrix ⎢
⎥ ⎣1 1⎦
has determinant of zero, so if we expand by cofactors to find the determinant of a larger matrix, we will have all of the terms being multiplied by a zero. 7. D When finding the determinant, the signs of all of the products will be the €
opposite, thus the determinant is the additive inverse. 8. A A two by two matrix with all i’s has determinant zero, so this will cause all of the determinants to find the cofactor matrix to be zero. In this cofactor matrix, all elements will be zero, so the determinant is zero. 9. B The requested ratio will always be equal to the diameter of its circumsbribed circle. Explanation: If we place a new point, Q, onto the circle between P and H, such that angle QHU is right and QU is a diameter of the circle, the measure of angle HQU will be equal to the measure of HPU because both angles subtend to the same arc. By definition, sin(HQU)=HU/QU. Since HQU and HPU have equal measures, sin(HQU)=sin(HPU). Thus, sin(HPU)=HU/QU. Or rather QU=HU/sin(HPU). Because QU is a diameter, its length is 2r, or sqrt(3). (If the triangle is obtuse, even though HQU is not equal to HPU, their sin values will be. If the triangle is right, then the hypotenuse must be a diameter.) 10. A By the law of cosines, all of sides will have length sqrt(2-­‐sqrt(3)). To find the perimeter multiply this by 12. 11. B To use the double angle formula, divide 3 by 2 to get 1.5sin(2x). The amplitude of this is 1.5. 12. A After completing the square, we see that a=3 and b=2 (in this case, a was “under” the y, and b was “under” the x). The formula for the length of the latus rectum is 2b^2/a. Substituting: 2*4/3=8/3 13. A Let z=a+bi. Then a+bi+sqrt(a^2+b^2)=sqrt(10)+2i. Equating real parts and imaginary parts, b=2 so a+sqrt(a^2+4)=sqrt(10). Subtract a from both sides and square. Then a^2+4=10-­‐2sqrt(10)a+a^2 => -­‐6=-­‐2sqrt(10)a. Then a=3/sqrt(10). Then 4+9/10=49/10=4.9 14. C A complex number solution follows. There are other geometric approaches. Place a corner of the rectangle on the origin of the complex plane. Let the other vertices be at the points 1, i, and 1+i. Then to maximize the area of the triangle, Precalculus Individual January 2011 Palm Harbor University High School €
let one vertex be on the origin and let the other vertices be on the opposite sides at the points 1+yi and x+i. The second point can be obtained by a 60 degree rotation about the origin, or (1+yi)(cos60+isin60)=x+i. Then 1/2+isqrt(3)/2-­‐
ysqrt(3)/2+yi/2=x+i. Then equate the real parts and the imaginary parts: 1/2-­‐
ysqrt(3)/2=x and sqrt(3)/2+y/2=1. Solve the second equation for y. y=2-­‐
sqrt(3). Then |y|=sqrt(4-­‐4sqrt(3)+3+1)=sqrt(8-­‐4sqrt(3)). The area of an equilateral triangle is s^2*sqrt(3)/4. The side length is the last expression, so (8-­‐
4sqrt(3))*sqrt(3)/4= 2sqrt(3)-­‐3. So a=2, b=3, and c=3. 2+3+3=8. 15. B To simplify this (since we are taking the absolute values later, we need not worry about signs), we can let a=sin of some angle, b=cos of the same angle (say u). Do the same thing with c and d (for v). Then ad-­‐bc is sin(u-­‐v)=5/9. So bd+ac=cos(u-­‐v). The cos is sqrt(1-­‐(12/13)^2), or 5/13. 5+13=18. 16. E If the line was made between the positive x axis and a point to create the same angle, the sin would be positive. 17. E Upon putting the angle in standard position, the coordinates will flip, so the cotangent will be -­‐4/3. 18. A We will find the sin of two times this angle (using 2sin(x)cos(x)) and then reciprocate it. We have sin(x)=3/5, and cos(x)=-­‐4/5. Multiplying these two by 2 gives -­‐24/25. The reciprocal is -­‐25/24 19. D Let this expression equal x, and square it. We get 7+sqrt13+2sqrt(49-­‐13)+7-­‐
sqrt(13)=x^2. Or 14+2sqrt(36). This is x^2=26. X must be positive because this is a sum with positive numbers. Thus this expression is \sqrt{26}. 20. B This is the minimum number of times the graph must cross the line y=x. A third degree polynomial goes to positive and negative infinity, so it must cross the line at least once. 21. B By b^2-­‐4ac, the conic is a parabola (it is zero). This means that its eccentricity is 1. 22. E Add sqrt(2)/2+1/2=(sqrt(2)+1)/2. 23. D Substitute u = sqrt(x^2-­‐7x-­‐9). Then the original equation becomes u^2-­‐u-­‐6=0. This factors as (u-­‐3)(u+2)=0. U=3 or -­‐2, but it can’t be negative since it is a square root. So sqrt{x^2-­‐7x-­‐9}=3, so x^2-­‐7x-­‐9=9  x^2-­‐7x-­‐18=0  (x-­‐
9)(x+2)=0, so x is -­‐2 or 9; both of which are solutions so the product is -­‐18 D. 24. C Define a ≡ b (mod n) to mean that b is the remainder when a is divided by n. Then, by Fermat’s little theorem: 132 36 ≡ 1(mod 37) . Then the original problem becomes 132 2×36+3 ≡ (132 36 ) 2 × 132 3 ≡ 12 × 132 3 (mod 37) . But 132 = 3 × 37 + 21, so 12 × (3 × 37 + 21) 3 ≡ 213 ≡ 9261 ≡ 250 × 37 +11 ≡ 11(mod 37) . 25. B I. is a solution to x^2-­‐2=0. Pi is transcendental, so squaring a transcendental €
number will yield another transcendental number. € Same is true for III. IV is not a €
number, but an idea, thus it can’t be a transcental number. V is obviously not transcendental. So 2 are. 26. B That means 1-­‐i is also a root. So, 1+i+1-­‐i=2, but the sum of the roots needs to equal 0 by Vieta, so the third root is -­‐2. C=-­‐product of the roots. (1+i)(1-­‐i)(-­‐
2)=2*-­‐2=-­‐4. And –(-­‐4)=4. 27. D The left side is equal to zero, which can first be achieved by the left side when |cos(2x)|=1, or when x=90. The remainder when 90/7 is 6. 28. D This is the sum of the numbers from 1 to 1000, or 1000(1001)/2=500500 29. E The function will never attain a maximum because tan goes to infinity. 30. A If we compute 8110 in base 4, we get 11014, so our polynomial is x^3+x^2+1 because (4)^3+(4)^2+1. So, if we plug in 5 we get 151. Precalculus Individual January 2011 Palm Harbor University High School Answers 1. B 2. E 3. D 4. A 5. C 6. E 7. D 8. A 9. B 10. A 11. B 12. A 13. A 14. C 15. B 16. E 17. E 18. A 19. D 20. B 21. B 22. E 23. D 24. C 25. B 26. B 27. D 28. D 29. E 30. A