MTAP Reviewer 4th Year

MTAP Reviewer – 4th Year 1. What is the domain of y = Solution: 3x – 2 > 0 2 x> 3

3x  2 ? [ x > 2/3 ]

3x 2 ? [ y > 0 ] 2 3. Which of the following is a function? [ c (because the exponent of y is odd.) ] a. x2 + y2 + 2x – 3y – 8 = 0 b. y2 = 8x – 3 c. x2 = 8y + 3

2. What is the range of y 

4. If f(x) =

4x  5 and g(x) = x2 – 1, find the domain of

f (x) . [ x > 5/4, x ≠ 1 ] g(x)

Solution: Domain of f(x) = x > 5/4 Domain of g(x) = x  R, x ≠ ± 1 Domain of

f (x) : x > 5/4, x ≠ 1 g(x)

5. With the same functions in #4, find the domain of (f o g)(x). [ x < - 3/2 or x > 3/2 ] Solution: (f o g)(x) =

4(x2  1)  5  4x2  9

D(f o g)(x) : 4x2 – 9 > 0  x < - 3/2 or x > 3/2 x 6. What is the equation of the line given by the table? [ 2x – y – 5 = 0 ]

2

4 6

8

y - 1 3 7 11

Solution: Get 2 points and use the formula y = m(x – x1) + y1. Using (4, 3) and (6, 7).

73 2 64 Solving for the equation of the line, Solve for the slope m. m =

y = 2(x – 4) + 3  y = 2x – 8 + 3  2x – y – 5 = 0 7. What is the equation of the line with an x – intercept of 3 and a slope of – 2/3? [ 2x + 3y – 6 = 0 ] Solution: 2 y =  x  3  0  3y = –2x + 6  2x + 3y – 6 = 0 3

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8. The x – and y – intercepts of a line are – 2 and 6 respectively. Find the equation of the line. [ 3x – 2y + 6 = 0 ] Solution: x y Use the formula   1  0 where a is the x – intercept and y – intercept. a b

x y   1  0  3x – 2y + 6 = 0 2 6 9. Find the intersection of 2x + 5y = 0 and 5x – 2y + 29 = 0. [ ( – 5, 2) Solution: 2x + 5y = 0  10x + 25y = 0 5x – 2y = – 29  10x – 4y = – 58 By elimination, y = 2 and x = – 5 10. How are graphs of 5x – 2y + 6 = 0 and 10x – 4y + 15 = 0 related? [ parallel (because the slopes are equal) ] 11. Six less than three times a number is the same as 26 less than four times the number. What is the number? [ 20 ] Solution: 3n – 6 = 4n – 26  n = 20 12. The vertices of a triangle are at A(0, 0), B(4, 3) and C(– 2, 5). Find its perimeter. [ 5  29  2 10 ] Solution: Use the distance formula. dAB =

(4  0)2  (3  0)2  5

dAC =

(2  0)2  (5  0)2  29

dBC =

(2  4)2  (5  3)2  40  2 10

Perimeter = 5  29  2 10 13. Find the vertex of the parabola y = 2x2 + 3x – 2. [ ( - 3/4, - 25/9 ) ] Solution:   b 4ac  b2   V(h,k)   , 4a   2a

h=

3 3  2(2) 4

k=

4(2)(2)  32 25  4(2) 8

14. Solve for the inequality 2x2 + 7x – 15 > 0. [ x < - 5 or x > 3/2 ] Solution: 2x2 + 7x – 15 > 0 (2x – 3)(x + 5) > 0 x < - 5 or x > 3/2

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15. Find the quadratic equation whose roots are 3/2 and – 2/3. [ 6x2 – 5x – 6 = 0 ] Solution: x = 3/2 x = – 2/3  (2x – 3)(3x + 2) = 0  6x2 – 5x – 6 = 0

16. Find the quadratic equation whose roots are

1 3 . 2

[ 2x2 + 2x – 1 = 0 ]

Solution: x=

1 3  2x + 1 =  3  (2x + 1)2 = (  3 )2  4x2 + 4x + 1 = 3 2

4x2 + 4x – 2 = 0  2x2 + 2x – 1 = 0 17. For what value/s of k will 2x2 – kx + 5 = 0 have 2 distinct roots? [ k  2 10 or k  2 10 ] Solution: b2 – 4ac > 0 (-k)2 – 4(2)(5) > 0 k2 – 40 > 0 k  2 10 or k  2 10 18. A ball is thrown vertically upward at an initial speed v0 of 200 ft/s. If after t seconds, the height of the ball is d = v0t – 16t2, find the maximum height the ball is going to attain? [ 625 ft ] Solution: v0 = 200  d = 200t – 16t2 Solve for k =

4(16)(0)  200 2 4ac  b2  625 ft  k= 4(16) 4a

19. Find the remainder when x4 – 5x3 – 6x2 + 2x – 3 is divided by x – 1. [ - 11 ] Solution: Use synthetic division 1

1

-5 1

-6 -4

2 - 10

-3 -8

1

-4

- 10

-8

- 11

20. If 4x3 – 16x2 + 29x – 25 is expressed as (2x – 3)Q(x) + R, what is Q(x)? [ Q(x) = 2x2 + 5x + 22 ] Solution: Long Division: (4x3 – 16x2 + 29x – 25) ÷ (2x – 3)  (2x2 + 5x + 22) + 41 Therefore, Q(x) = 2x2 + 5x + 22.

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