SAT II -Math Level 2 Test #02 Solution -...

SAT II SAT II -- Math Level 2Math Level 2SAT II SAT II -- Math Level 2Math Level 2SAT II SAT II -- Math Level 2Math Level 2Test #02 SolutionTest #02 SolutionTest #02 SolutionTest #02 SolutionTest #02 SolutionTest #02 Solution

SAT II SAT II -- Math Level 2 Test No. 2Math Level 2 Test No. 2SAT II SAT II -- Math Level 2 Test No. 2Math Level 2 Test No. 2

1. The positive zero of y = x2 + 2x – 3/5 is, to the nearest tenth, equal to1. The positive zero of y = x2 + 2x – 3/5 is, to the nearest tenth, equal to

(A) 0.8 (B) 0.7 + 1.1i (C) 0.7 (D) 0.3 (E) 2.2(A) 0.8 (B) 0.7 + 1.1i (C) 0.7 (D) 0.3 (E) 2.2

42 acbb 3Using Quadratic formula, with a = 1, b = 2, ,

2

42

a

acbbx

,

5

3cUsing Quadratic formula, with a = 1, b = 2, ,

2ax ,

5c

we get x = 0.3 for the positive root.we get x = 0.3 for the positive root.

Ans. (D)Ans. (D)


2. The solution set of x2 – 2x < 8 is given by the inequality2. The solution set of x2 – 2x < 8 is given by the inequality

(A) x < 4 (B) x > -2 (C) -2 < x < 4 (D) -2 ≤ x ≤ 4 (D) x > 4(A) x < 4 (B) x > -2 (C) -2 < x < 4 (D) -2 ≤ x ≤ 4 (D) x > 4

x2 - 2x - 8 < 0, or (x - 4)(x + 2) < 0,x - 2x - 8 < 0, or (x - 4)(x + 2) < 0,

Therefore -2 < x < 4Therefore -2 < x < 4

Ans. (C)Ans. (C)


3. If sin x = 0.6018, then sec x =3. If sin x = 0.6018, then sec x =

(A) 2.1290 (B) 1.2521 (C) 1.0818 (D) 0.9243 (E) 0.4890(A) 2.1290 (B) 1.2521 (C) 1.0818 (D) 0.9243 (E) 0.4890

sin x = 0.618, therefore, x = sin-1(0.6018) = 37°.sin x = 0.618, therefore, x = sin (0.6018) = 37°.

Now, sec x = 1/cos x = 1/ cos 37° = 1.25Now, sec x = 1/cos x = 1/ cos 37° = 1.25

Ans. (B)Ans. (B)


4. An equation of line l in Figure 6 is4. An equation of line l in Figure 6 is

(A) x = 2(A) x = 2(B) y = 2(B) y = 2(C) x = 0(C) x = 0(D) y = x + 2(D) y = x + 2(E) x + y = 2(E) x + y = 2

As shown in the figure, x = 2As shown in the figure, x = 2

Ans. (A)Ans. (A)


5. If 0 < x < 1, then5. If 0 < x < 1, then

(A) 0 < ln(1/x) < 1 (B) ln(1/x) > 1 (C) ln(1/x) < 0(A) 0 < ln(1/x) < 1 (B) ln(1/x) > 1 (C) ln(1/x) < 0(D) ln(1/x) > 0 (E) none of these is true(D) ln(1/x) > 0 (E) none of these is true

0 < x < 1, therefore, ln x < 0.

But, ln (1/x) = ln (x)-1 = (-1)·ln x.But, ln (1/x) = ln (x) = (-1)·ln x.

Since ln x < 0, we get ln (1/x) = (-)(-) > 0.Since ln x < 0, we get ln (1/x) = (-)(-) > 0.

Ans. (D)Ans. (D)


6. When (a + b)4 is expanded, what is the coefficient of6. When (a + b)4 is expanded, what is the coefficient ofthe third term?the third term?

(A) 10 (B) 11 (C) 6 (D) -11 (E) The sum cannot be determined.(A) 10 (B) 11 (C) 6 (D) -11 (E) The sum cannot be determined.

Refer to Binomial Expansion, .( )n

n n r ra b C a b Refer to Binomial Expansion, .

Using n = 4, r = 2, 0

( )n n r rn r

r

a b C a b

Using n = 4, r = 2,

we get the coefficient of the third term, C = 6.

0r

we get the coefficient of the third term, 4C 2 = 6.4 2

Ans. (C)Ans. (C)


7. What is the approximate x-intercept of ?3

)(

x

xf7. What is the approximate x-intercept of ?3

3)(

x

xxf

(A) -1.73 (B) 1.73 (C) -0.58 (D) 0.58 (E) 3

3x

(A) -1.73 (B) 1.73 (C) -0.58 (D) 0.58 (E) 3

To get the x – intercept,3x

we y = f (x) = 0, .03

3)(

x

xxfwe y = f (x) = 0,

Thus,

.03

)(

x

xf

.03 xThus, .03 x

.73.13 x

Ans. (A)

.73.13 x

Ans. (A)


8. If f (x) = and g(x) = 2x2 - 3, then f (g(2)) =4x8. If f (x) = and g(x) = 2x2 - 3, then f (g(2)) =4x

(A) 6.16 (B) 3.61 (C) 2.24 (D) 3.00 (E) 6.00(A) 6.16 (B) 3.61 (C) 2.24 (D) 3.00 (E) 6.00

This problem is about composite function.This problem is about composite function.

f (g(2)) means, first we get g(2) = 2(2)2 - 3 = 5, and then plug it into f (x) function.f (g(2)) means, first we get g(2) = 2(2)2 - 3 = 5, and then plug it into f (x) function.

3945)5( f

Ans. (D)

3945)5( f

Ans. (D)


9. Let the symbol △ be defined as A△B = cos A·sin B - sin A·cos B.△

9. Let the symbol △ be defined as A△B = cos A·sin B - sin A·cos B.△

△ △What is the approximate value of 76°△36°?

△ △What is the approximate value of 76°△36°?

△ △△

(A) .766 (B) .643 (C) .125 (D) 0 (E) -.643

△

(A) .766 (B) .643 (C) .125 (D) 0 (E) -.643

Just replacing A = 76°, and B = 36°,Just replacing A = 76°, and B = 36°,

we get cos (76°)·sin (36°) - sin (76°)·cos (36°) = (-)0.643.we get cos (76°)·sin (36°) - sin (76°)·cos (36°) = (-)0.643.

Ans. (E)Ans. (E)


10. The slope of a line perpendicular to the line10. The slope of a line perpendicular to the linewhose equation is x/4 + y/3 = 1 iswhose equation is x/4 + y/3 = 1 is

(A) 1/4 (B) -4/3 (C) -3/4 (D) 4/3 (E) -3(A) 1/4 (B) -4/3 (C) -3/4 (D) 4/3 (E) -3

x/4 + y/3 = 1 into a standard eq., we get y = (-3/4)x + 3.x/4 + y/3 = 1 into a standard eq., we get y = (-3/4)x + 3.

Now, to be perpendicular to this line, we need the slope m, which is a negative reciprocal of -3/4.Now, to be perpendicular to this line, we need the slope m, which is a negative reciprocal of -3/4.

41

3

4

3

1

m

3

4

3

m

4

Ans. (D)Ans. (D)


33 5 5x 11. Approximately, what is ?3

3 2

3 5 5lim2 3 4x

xx x

11. Approximately, what is ?3 2lim

2 3 4x x x

(A) 3.02 (B) 3.13 (C) 3.24 (D) 3.35 (E) 3.46(A) 3.02 (B) 3.13 (C) 3.24 (D) 3.35 (E) 3.46

To find the limit of this eq., notice that the degrees of numerator and denominator areTo find the limit of this eq., notice that the degrees of numerator and denominator are

both the same degrees of 3.both the same degrees of 3.53

Thus, the limit on this will be the ratio of the highest degree coefficient, which is .35.32

53Thus, the limit on this will be the ratio of the highest degree coefficient, which is

Ans. (D)

.35.32

Ans. (D)


12. The maximum value of 3·sin x·cos x + 3/2 is12. The maximum value of 3·sin x·cos x + 3/2 is

(A) 3/2 (B) 5/2 (C) 3 (D) 7/2 (E) 6(A) 3/2 (B) 5/2 (C) 3 (D) 7/2 (E) 6

Using graphing calculator, we get max = 3.Using graphing calculator, we get max = 3.

Or, since 3sin x·cos x + 3/2 = 3/2(2sin x·cos x) + 3/2, Or, since 3sin x·cos x + 3/2 = 3/2(2sin x·cos x) + 3/2,

and 2sin x·cos x = sin 2x, we get 3/2(sin 2x) + 3/2.and 2sin x·cos x = sin 2x, we get 3/2(sin 2x) + 3/2.

The amplitude of this function is 3/2, but the function has been shifted up by 3/2.The amplitude of this function is 3/2, but the function has been shifted up by 3/2.

Therefore, the maximum value becomes (3/2 + 3/2) = 3.Therefore, the maximum value becomes (3/2 + 3/2) = 3.

Ans. (C)Ans. (C)


13. As angle x increases from 0 to 2π radians, sin x increases in13. As angle x increases from 0 to 2π radians, sin x increases in

(A) no quadrants(A) no quadrants(B) the first and third quadrants only(B) the first and third quadrants only(C) the second and fourth quadrants only(C) the second and fourth quadrants only(D) all four quadrants(D) all four quadrants(E) the first and fourth quadrants only(E) the first and fourth quadrants only

3

2

3

2

2

2

As shown in this graph, y = sin x increases in (I) and (IV). As shown in this graph, y = sin x increases in (I) and (IV).

Ans. (E)Ans. (E)


14. The solution set of 2x – 5y < 3 lies in which quadrants?14. The solution set of 2x – 5y < 3 lies in which quadrants?

(A) I only (B) I and II (C) I, II, and III(A) I only (B) I and II (C) I, II, and III(D) II, III, and IV (E) I, II, III, and IV(D) II, III, and IV (E) I, II, III, and IV

2x - 5y < 3, or 2x - 3 < 5y → (2/5)x - 3/5 < y2x - 5y < 3, or 2x - 3 < 5y → (2/5)x - 3/5 < y

As shown here, the graph covers all four quadrants. As shown here, the graph covers all four quadrants.

Ans. (E)Ans. (E)


15. If f (x) = 2·ex - 4 and g(x) = ln x, then f (g(9)) = 15. If f (x) = 2·ex - 4 and g(x) = ln x, then f (g(9)) =

(A) 6.83 (B) 12 (C) 14 (D) 45.98 (E) 568.17(A) 6.83 (B) 12 (C) 14 (D) 45.98 (E) 568.17

As in question #8, we get g(9) = ln 9, and then f (g(9)) = f (ln 9) = 2·(9ln e) - 4 = 18 - 4 = 14.As in question #8, we get g(9) = ln 9, and then f (g(9)) = f (ln 9) = 2·(9 ) - 4 = 18 - 4 = 14.

(note: aln b = bln a)(note: a = b )

Ans. (C)Ans. (C)


1 16. Find the positive value of .

2

1sincos 1

16. Find the positive value of 3

3

.2

sincos

(A) 1/2 (B) 1 (C) (D) (E) 22

33(A) 1/2 (B) 1 (C) (D) (E) 22

1Here, .30

1sin 1 Here, .30

2sin

.3

30cos .2

330cos

2

Ans. (C)Ans. (C)


17. How many different hands, each consisting of four cards, can be 17. How many different hands, each consisting of four cards, can be drawn, from a deck of 52 different cards?drawn, from a deck of 52 different cards?

(A) 132,600 (B) 1.3 × 1067 (C) 270,725 (D) 1.4 × 1010 (E) 2652(A) 132,600 (B) 1.3 × 10 (C) 270,725 (D) 1.4 × 10 (E) 2652

To solve this problem,To solve this problem,

we need to understand the following ideas on probability and combination.we need to understand the following ideas on probability and combination.

1) The total possible number of combination on n objects is n!1) The total possible number of combination on n objects is n!(eg) (A, B, C) = 3!, because A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, C-B-A, total of 6 = 3!(eg) (A, B, C) = 3!, because A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, C-B-A, total of 6 = 3!(A, B, C, D) = 4!, (A, B, C, D, E) = 5! … etc.(A, B, C, D) = 4!, (A, B, C, D, E) = 5! … etc.

2) The total number of combination on n objects,2) The total number of combination on n objects,but with the repetition of k and r objects is n!/(k!r!)but with the repetition of k and r objects is n!/(k!r!)(eg) COFFEE → (6 letters)/(2F’s)(2E’s) = 6!/(2!2!)(eg) COFFEE → (6 letters)/(2F’s)(2E’s) = 6!/(2!2!)

Minimum → (7 letters)/(3m’s)(2i’s) = 7!/(3!2!)Minimum → (7 letters)/(3m’s)(2i’s) = 7!/(3!2!)

3) The total number of combination on n objects,3) The total number of combination on n objects,but with k group with n, and n2… member in them is k!(n1!)(n2!)…but with k group with n, and n2… member in them is k!(n1!)(n2!)…(eg) A, B, C, D, E into 2 grouping of (A, B), (C, D, E) will be(eg) A, B, C, D, E into 2 grouping of (A, B), (C, D, E) will be(2! grouping)(2! on A, B)(3! on C, D, E) = (2!)(2!)(3!)(2! grouping)(2! on A, B)(3! on C, D, E) = (2!)(2!)(3!)

( Next Slide ►► )( Next Slide ►► )


( ►► Continued from previouse slide)( ►► Continued from previouse slide)

4) Permutation is with a consideration of "Ordering", P = n!/(n – r)!4) Permutation is with a consideration of "Ordering", nPr = n!/(n – r)!(eg) Suggest I choose 3 students with the best score out of 10 students in my Math IIc class,(eg) Suggest I choose 3 students with the best score out of 10 students in my Math IIc class,and then buy them lunch, steak for the best, hamburger for the second best,and then buy them lunch, steak for the best, hamburger for the second best,and tacos for the third best. Then, this will a permutation, because "Order" mattersand tacos for the third best. Then, this will a permutation, because "Order" mattersby having different lunches with their rank, thus P = 10!/(10 – 3)! = 10!/7!by having different lunches with their rank, thus 10P3 = 10!/(10 – 3)! = 10!/7!

5) Combination is without a consideration of "ordering", C = n!/((n - r)!r!)5) Combination is without a consideration of "ordering", nCr = n!/((n - r)!r!)(eg) Suppose that I choose 3 students with the top best scores in my Math IIc class, and then(eg) Suppose that I choose 3 students with the top best scores in my Math IIc class, and thenbuy them lunch, but this time, all hamburgers. Then, in this case, "order" does not matter,buy them lunch, but this time, all hamburgers. Then, in this case, "order" does not matter,because they all eat hamburgers. Thus, it is Combination, C = 10!/((10 – 3)!3!) = 10!/(7!3!).because they all eat hamburgers. Thus, it is Combination, 10C3 = 10!/((10 – 3)!3!) = 10!/(7!3!).Now, notice that C and P are exactly the same, except that C is being divided by 3!,Now, notice that 10C3 and 10P3 are exactly the same, except that 10C3 is being divided by 3!,while P isn’t. That is because in C case, those 3 students selected will havewhile 10P3 isn’t. That is because in 10C3 case, those 3 students selected will havethe same lunch, hamburger – hamburger –hamburger, or H - H - H → repetition of 3,the same lunch, hamburger – hamburger –hamburger, or H - H - H → repetition of 3,therefore, being divided by 3!.therefore, being divided by 3!.

Based on the above ideas on Combination, we choose 4 cards out of 52 cards,Based on the above ideas on Combination, we choose 4 cards out of 52 cards,

but without any particular ordering.but without any particular ordering.

Therefore, it will be combination, C = 52!/(48!4!) = 270,725.Therefore, it will be combination, 52C4 = 52!/(48!4!) = 270,725.52 4

Ans. (C)Ans. (C)


18. The area of a triangle with sides 4, 5, and 8 is18. The area of a triangle with sides 4, 5, and 8 is

(A) 7.5 (B) 8.18 (C) 3.75 (D) 13.0 (E) 2.4(A) 7.5 (B) 8.18 (C) 3.75 (D) 13.0 (E) 2.4

Here, we are given the lengths of 3 side only,Here, we are given the lengths of 3 side only,

without any angle information. Therefore, wewithout any angle information. Therefore, we

use Heron’s formula to find the area, that isuse Heron’s formula to find the area, that is

5.8854

cbas 5.8

2

854

2

cbas

22

))()(( csbsasSArea ))()((

csbsasSArea

18.8)85.8)(55.8)(45.8(5.8

Ans. (B)Ans. (B)

Now, in addition to Heron’s formula, we also needNow, in addition to Heron’s formula, we also need

to review following formulas for Math IIc - exam:

1) The Law of Sine: (sin A)/a = (sin B)/b = (sin C)/c1) The Law of Sine: (sin A)/a = (sin B)/b = (sin C)/c2) The law of Cosine: c2 = a2 + b2 – 2ab·cos C2) The law of Cosine: c2 = a2 + b2 – 2ab·cos C3) Area with 2 side, and angle θ:3) Area with 2 side, and angle θ:

Area = (1/2)bc·sin θArea = (1/2)bc·sin θ


19. One side of a given triangle is 6 inches. Inside the triangle a line 19. One side of a given triangle is 6 inches. Inside the triangle a line segment is drawn parallel to this side, cutting off a trianglesegment is drawn parallel to this side, cutting off a trianglewhose area is one-thirds that of the given triangle.whose area is one-thirds that of the given triangle.Find the length of this segment in inches.Find the length of this segment in inches.

(A) 2 (B) (C) (D) (E) 932 33 36(A) 2 (B) (C) (D) (E) 932 33 36

△Since the ratio of the area △DBE is 1:1/3,1

Since the ratio of the area △DBE is 1:1/3,

we get the ratio of the line segment .1

:1: DEAC

△

we get the ratio of the line segment

Ans. (B)

.3

:1: DEAC

Ans. (B)

(Note)(Note)

Length : 1 → 2 → 3Length : 1 → 2 → 3

Area : 12 → 22 → 32Area : 1 → 2 → 3

Volume : 13 → 23 → 33Volume : 1 → 2 → 3


20. Given the statement, "Only if it snows will the class be cancelled",20. Given the statement, "Only if it snows will the class be cancelled",if it is known that it did not snow, what conclusion can be drawn?if it is known that it did not snow, what conclusion can be drawn?

(A) The class is cancelled.(A) The class is cancelled.(B) The class is not cancelled.(B) The class is not cancelled.(C) The class might be cancelled.(C) The class might be cancelled.(D) No conclusion can be drawn.(D) No conclusion can be drawn.(E) None of the above.(E) None of the above.

As mentioned in the previous test #1, it is a logic problem with negation.As mentioned in the previous test #1, it is a logic problem with negation.

Ans. (B)Ans. (B)


52 3 x21. What is the domain of the function ?

52)(

3

xxf21. What is the domain of the function ?

52)(

3

xxf

(A) x < -1.36 (B) -1.36 ≤ x ≤ 1.36 (C) x > 1.36

52 x

(A) x < -1.36 (B) -1.36 ≤ x ≤ 1.36 (C) x > 1.36(D) x ≥ 1.36 (E) x < -1.36 or x > 1.36(D) x ≥ 1.36 (E) x < -1.36 or x > 1.36

In most case,In most case,

a domain problem in Rational Function involves with the condition of its denominator.a domain problem in Rational Function involves with the condition of its denominator.

In this problem, the denominator and at the same time, 2x3 – 5 > 0.,052 3 xIn this problem, the denominator and at the same time, 2x – 5 > 0.,052 x

533

2

5or 052 xx

Ans. (C)

2or 052 xx

Ans. (C)


22. The hyperbola 9x2 – 4y2 = 16 intersects the x-axis at22. The hyperbola 9x2 – 4y2 = 16 intersects the x-axis atapproximately which of the following points?approximately which of the following points?

(A) (2, 0) (B) (0, 2) (C) (1.33, 0) (D) (0, 1.33) (E) (0, -1.33)(A) (2, 0) (B) (0, 2) (C) (1.33, 0) (D) (0, 1.33) (E) (0, -1.33)

This problem involves with the topics on Conic Sections.This problem involves with the topics on Conic Sections.

However, the question only asks about the x-intercept, when y = 0,However, the question only asks about the x-intercept, when y = 0,

we only need to plug in y = 0 into equation.we only need to plug in y = 0 into equation.

9x2 – 4(0)2 = 16

∴

9x2 – 4(0)2 = 16

∴∴ x2 = 16/9∴ x = 16/9

∴

33.1416

x

∴

∴

33.13

4

9

16 x

∴ (1.33, 0)

39

∴ (1.33, 0)

Ans. (C)

∴

Ans. (C)

∴

(Note)(Note)

For the future reference,For the future reference,

I want to go over topics on Conic Sections as follows:I want to go over topics on Conic Sections as follows:




1) Circle: Let P1 be the plane that cuts the cone in a1) Circle: Let P1 be the plane that cuts the cone in ahorizontal direction. Then it creates a circle,horizontal direction. Then it creates a circle,formed by its intersection.formed by its intersection.Given (x - h)2 + (y - k)2 = r2,Given (x - h) + (y - k) = r ,

• center = (h, k)• center = (h, k)• radius = r• radius = r

2) Ellipse: Let P2 be the plane that2) Ellipse: Let P2 be the plane thatcuts the cone in a tilted direction.cuts the cone in a tilted direction.Then it creates an elliptical figure,Then it creates an elliptical figure,formed by intersection.formed by intersection.Given (x – h)2/a2 + (y – k)2/b2 = 1,Given (x – h) /a + (y – k) /b = 1,

• center = (h, k)• center = (h, k)• length of Major axis = 2a, when a > b• length of Major axis = 2a, when a > b• length of Minor axis = 2b• length of Minor axis = 2b• vertex = (±a + h, o + k)• vertex = (±a + h, o + k)• Foci = (h ± c, k), where c2 = a2 – b2• Foci = (h ± c, k), where c = a – b• d1 + d2 = 2a x2/a2 + y2/b2 = 1, a > b• d1 + d2 = 2a x2/a2 + y2/b2 = 1, a > b

a < ba < b




3) Hyperbola: Let P3 be the plane that cuts the cone in a vertical direction.3) Hyperbola: Let P3 be the plane that cuts the cone in a vertical direction.Then it creates a hyperbolic figure, formed by its intersection. Then it creates a hyperbolic figure, formed by its intersection. Given (x – h)2/a2 – (y – k)2/b2 = 1, or (y – k)2/b2 – (x – h)2/a2 = 1,Given (x – h) /a – (y – k) /b = 1, or (y – k) /b – (x – h) /a = 1,

• center = (h, k)• center = (h, k)• vertex = (±a + h, o + k)• vertex = (±a + h, o + k)• Asymptotes: y - k = ±(b/a)(x – h),• Asymptotes: y - k = ±(b/a)(x – h),• where the slope b/a is rise/run.• where the slope b/a is rise/run.• Foci = (±c + h, o + k),• Foci = (±c + h, o + k),

where c2 = a2 + b2where c = a + b• d1 – d2 = 2a• d1 – d2 = 2a

x2/a2 – y2/b2 = 1x2/a2 – y2/b2 = 1

y2/b2 – x2/a2 = 1y /b – x /a = 1




4) Parabola: Let P4 be the plane that cuts the cone in a parallel to slant edge direction.4) Parabola: Let P4 be the plane that cuts the cone in a parallel to slant edge direction.Then it creates a parabolic figure, formed by its intersection.Then it creates a parabolic figure, formed by its intersection.Given 4P(y - k) = (x - h)2, or 4P(x - h) = (y - k)2,Given 4P(y - k) = (x - h) , or 4P(x - h) = (y - k) ,

• vertex = (h, k)• vertex = (h, k)• Foci = (o + h, p + k ) or (p + h, o + k)• Foci = (o + h, p + k ) or (p + h, o + k)• Directrix line: y = -p + k or x = -p + h• Directrix line: y = -p + k or x = -p + h


23. The length of the vector that could correctly be used to23. The length of the vector that could correctly be used torepresent in the complex plane the number isi27 represent in the complex plane the number isi27

(A) 11 (B) (C) (D) (E) 11 23 5 13(A) 11 (B) (C) (D) (E) 11 23 5 13

The length of a vector, or the magnitude of a vector,The length of a vector, or the magnitude of a vector,

z = a + bi = < a, b > is given by .22 babiaz z = a + bi = < a, b > is given by .22 babiaz

Therefore, .112727 22 izTherefore, .112727 22 iz

Ans. (B)Ans. (B)


24. In a cube, the ratio of the longest diagonal to a side is24. In a cube, the ratio of the longest diagonal to a side is

(A) (B) (C) (D) (E)3:2 2:6 2:6 2:3 1:3(A) (B) (C) (D) (E)3:2 2:6 2:6 2:3 1:3

3111 222 c 3111 c

211 22 b

Thus,

211 b

1:3: acThus,

Ans. (E)

1:3: ac

Ans. (E)


25. What is the measure of the smallest angle in a25. What is the measure of the smallest angle in aright triangle with sides of lengths 5, 12 and 13?right triangle with sides of lengths 5, 12 and 13?

(A) 21.04° (B) 22.62° (C) 24.62° (D) 42.71° (E) 67.34°(A) 21.04° (B) 22.62° (C) 24.62° (D) 42.71° (E) 67.34°

Referring to the topics of Trig. Functions previously,Referring to the topics of Trig. Functions previously,

we may use the Law of cosine, c2 = a2 + b2 – 2ab·cos C.we may use the Law of cosine, c2 = a2 + b2 – 2ab·cos C.

Here, the shortest side is 5.Here, the shortest side is 5.

cos1312213125 222 C cos1312213125 222 C

62.2251312

cos222

1C

62.22

13122

51312cos 1C

Ans. (B)

13122

Ans. (B)


26. What is the period of the curve whose equation is y = 4sin x·cos x?26. What is the period of the curve whose equation is y = 4sin x·cos x?

(A) 60° (B) 120° (C) 180° (D) 360° (E) 720°(A) 60° (B) 120° (C) 180° (D) 360° (E) 720°

y = 4sin x·cos x = 2(2sin x·cos x).

∴

y = 4sin x·cos x = 2(2sin x·cos x).

But 2sin x·cos x = sin 2x.

∴

But 2sin x·cos x = sin 2x.

∴ y = 2·sin 2x.

∴

∴ y = 2·sin 2x.

∴

∴

∴ p = 2 π/2 = π = 180°

∴

∴

Ans. (C)

∴

Ans. (C)


27. Let f (x) be a polynomial function.27. Let f (x) be a polynomial function.If and , then f (x) is divisible by0)2( f 0)2( fIf and , then f (x) is divisible by0)2( f 0)2( f

(A) x – 3 (B) x2 – 2 (C) x2 + 2 (D) x2 - 3x + 2 (E) x2 + 3x + 2(A) x – 3 (B) x – 2 (C) x + 2 (D) x - 3x + 2 (E) x + 3x + 2

Refer to the previous test #1, we have stated Factor Theorem,Refer to the previous test #1, we have stated Factor Theorem,

that is, if f (x) has f (a) = 0, then it must have a factor of (x - a).that is, if f (x) has f (a) = 0, then it must have a factor of (x - a).

Here, and

∴

0)2( f .0)2( fHere, and

∴ we have factors.

0)2( f .0)2( f

∴ we have factors.)2)(2( xx∴

Therefore, f (x) must be division by (x2 - 2).

∴

Therefore, f (x) must be division by (x - 2).

Ans. (B)Ans. (B)


28. How many real roots does the following equation have?28. How many real roots does the following equation have?5ex + e-x – 4 = 05e + e – 4 = 0

(A) 0 (B) 1 (C) 2 (D) 4 (E) an infinite number(A) 0 (B) 1 (C) 2 (D) 4 (E) an infinite number

Using a Graphing utility, let y1 = 5ex + e-x - 4, and then check how many x-intercept it has.Using a Graphing utility, let y1 = 5e + e - 4, and then check how many x-intercept it has.

There is no real root.There is no real root.

Ans. (A)Ans. (A)


29. If f (x) = ln x and g(x) = f (x) ·f -1(x), what does g(2) equal?29. If f (x) = ln x and g(x) = f (x) ·f -1(x), what does g(2) equal?

(A) 8.1 (B) 7.5 (C) 8.3 (D) 5.1 (E) 7.4(A) 8.1 (B) 7.5 (C) 8.3 (D) 5.1 (E) 7.4

f (x) = y = ln x.

∴

f (x) = y = ln x.

Now, to get the inverse f -1(x) by switching x and y's, x = ln y or y = ex = f -1(x).

∴

∴

Now, to get the inverse f (x) by switching x and y's, x = ln y or y = e = f (x).

∴ g(x) = f (x)·f -1(x) = (ln x)·(ex)

∴

∴ g(x) = f (x)·f -1(x) = (ln x)·(ex)

∴ g(2) = (ln 2)·(e2) = 5.1

∴

∴ g(2) = (ln 2)·(e2) = 5.1

Ans. (D)

∴

∴

Ans. (D)


30. If the 3rd term of a geometric progression is and3 a30. If the 3rd term of a geometric progression is and3 a

the 6th term is , what is the 12th term of the progression?3 2athe 6th term is , what is the 12 term of the progression?a

(A) a (B) 2a (C) a4/3 (D) a5/3 (E) a2(A) a (B) 2a (C) a (D) a (E) a

Recall the Geometric series with the following formulas:Recall the Geometric series with the following formulas:

an = a1r n-1, r = an/an-1, sn = a1((1 - rn)/(1 - r))

And also, the infinite series, s = a (1/(1 - r)), | r | < 1.

₆

an = a1r , r = an/an-1, sn = a1((1 - r )/(1 - r))

And also, the infinite series, s = a1(1/(1 - r)), | r | < 1.

₆

∴

1

Here, a3 = a1/3 and a₆ = a2/3.

∴

Here, a3 = a and a₆ = a .

∴ a6/a3 = a1r5/a1r

2 = r3, which is, a2/3/a1/3 = r3, or a1/3 = r3 → r = a1/9.

₆

∴ a6/a3 = a1r /a1r = r , which is, a /a = r , or a = r → r = a .

Therefore, a12 = a6·r6 = a2/3·(a1/9)6 = a2/3·a2/3 = a4/3.

∴

Therefore, a12 = a6·r6 = a2/3·(a1/9)6 = a2/3·a2/3 = a4/3.

Ans. (C)

∴

Ans. (C)

(Note) (Note)

For Arithmetic series,For Arithmetic series,

a = a + (n - 1)d, d = a - a , s = ((a + a )/2)·n,an = a1 + (n - 1)d, d = an - an-1, sn = ((a1 + an)/2)·n,where d is the "Common Difference”where d is the "Common Difference”


31. If and f (a, b, 0) = f (0, a, b), then a/b =zyxzyxf 653),,( 31. If and f (a, b, 0) = f (0, a, b), then a/b =zyxzyxf 653),,(

(A) 0.32 (B) 2.7 (C) 8.6 (D) 0.12 (E) 1.18(A) 0.32 (B) 2.7 (C) 8.6 (D) 0.12 (E) 1.18

zyxzyxf 653),,(

→ eq. (1)

zyxzyxf 653),,(

053)0,,( baba → eq. (1)

→ eq. (2)

∴

053)0,,( baba

baba 650),,0( → eq. (2)

∴

baba 650),,0(

∴ By equating (1) = (2), we get bababa )56()53(or ,6553 ∴

18.156

a

∴

18.135

56

b

a

Ans. (E)

35 b

Ans. (E)


132. What is the inverse of ?

1)( xf32. What is the inverse of ?

5)(

2

xxf

5 x

1y

15y

15y

15y 2 (A) (B) (C) (D) (E)

25

1

xy

2

15

xy

2

15

xy

2

15

xy 52 xy(A) (B) (C) (D) (E)

25 x2x 2x 2x

1.

1)(

2 xfy .

5)(

2

xxfy

11To get the inverse f -1(x) by switching x and y's, we get .

1or

12

2

xxTo get the inverse f (x) by switching x and y's, we get

Now, by reciprocal of both sides,

.5

or 5

22

yx

yx

Now, by reciprocal of both sides, 5 y

111 ).(1

5or ,1

551 1

22

22

2xf

xy

xyy

x

222 xxx

Ans. (C)Ans. (C)


33. Find the value of P if (x + 1) is a factor of f (x) = 3x4 + 2x3 – Px2 – 2x + 1.33. Find the value of P if (x + 1) is a factor of f (x) = 3x4 + 2x3 – Px2 – 2x + 1.

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

If (x + 1) is a factor of f (x), then f (-1) = 0.

∴

∴

If (x + 1) is a factor of f (x), then f (-1) = 0.

∴ f (-1) = 3 - 2 - p + 2 + 1 = 0.

∴

∴ f (-1) = 3 - 2 - p + 2 + 1 = 0.

∴ p = 4

∴

∴ p = 4

∴

∴

Ans. (D)


34. In how many different orders can 10 students34. In how many different orders can 10 studentsarrange themselves around a table?arrange themselves around a table?

(A) 9 (B) 81 (C) 181,440 (D) 362,880 (E) 387,420,489(A) 9 (B) 81 (C) 181,440 (D) 362,880 (E) 387,420,489

Here, the circular permutation is (n – 1)!

∴

Here, the circular permutation is (n – 1)!

∴ 9! = 362,880∴ 9! = 362,880

Ans. (D)

∴

Ans. (D)


35. If [x] is defined to represent the greatest integer less than or equal to 35. If [x] is defined to represent the greatest integer less than or equal to x, and f (x) = | x – [x] – 1/2 |, what is the period of f (x)?x, and f (x) = | x – [x] – 1/2 |, what is the period of f (x)?

(A) 1 (B) 1/2 (C) 2 (D) 4 (E) f is not a periodic function.(A) 1 (B) 1/2 (C) 2 (D) 4 (E) f is not a periodic function.

f (x) = [x] is what is called "Integer Function",f (x) = [x] is what is called "Integer Function",

and defined as the greatest integer, x ≤ [x] < x + 1.and defined as the greatest integer, x ≤ [x] < x + 1.

(ie) [3.5] = 3, [12.57] = 12, [-2.5] = -3, ... etc.(ie) [3.5] = 3, [12.57] = 12, [-2.5] = -3, ... etc.

Here, the quicker way of solving it is by using a graphing utility.Here, the quicker way of solving it is by using a graphing utility.

y = abs(x - int(x) - 0.5) shows that y repeats its cycle every one unit.y1 = abs(x - int(x) - 0.5) shows that y1 repeats its cycle every one unit.

Ans. (A)Ans. (A)


36. If the equation 3x3 + 9x2 + px – q = 0 has 3 equal roots, then36. If the equation 3x3 + 9x2 + px – q = 0 has 3 equal roots, then

(A) q = 0 (B) p = 0 (C) q = 1 (D) each root = 1 (E) each root = -1(A) q = 0 (B) p = 0 (C) q = 1 (D) each root = 1 (E) each root = -1

For a given cubic function,For a given cubic function,

f (x) = ax3 + bx2 + cx + d, the sum of the three roots,f (x) = ax3 + bx2 + cx + d, the sum of the three roots,x + x + x = (-)b/a and the product of the roots, .

dxxx x1 + x2 + x3 = (-)b/a and the product of the roots, .

a

dxxx 321

∴

a

Here, the sum of 3 equal roots is x1 + x2 + x3 = x1 + x1 + x1 = 3x1 = (-)9/3 = (-)3.

∴

Here, the sum of 3 equal roots is x1 + x2 + x3 = x1 + x1 + x1 = 3x1 = (-)9/3 = (-)3.

∴ x = (-)1∴ x1 = (-)1

Ans. (E)

∴

Ans. (E)


37. Each term of a sequence, after the first, is inversely proportional to 37. Each term of a sequence, after the first, is inversely proportional to the term preceding it. If the first two terms are 3 and 9,the term preceding it. If the first two terms are 3 and 9,what is the 100th term?what is the 100th term?

(A) 3 (B) 9 (C) 3100 (D) 399 (E) The 100th term cannot be determined.(A) 3 (B) 9 (C) 3100 (D) 399 (E) The 100th term cannot be determined.

a = 3, a = 9, a , a , ...a1 = 3, a2 = 9, a3, a4, ...

∴

Since the second term 9 is inversely proportion to the proceeding term 3,

∴

Since the second term 9 is inversely proportion to the proceeding term 3,

we get 9 = k·(1/3), ∴ k = 27.we get 9 = k·(1/3), ∴ k = 27.

Therefore the general term an = k·(1/an-1) = 27(1/an-1).

∴

Therefore the general term an = k·(1/an-1) = 27(1/an-1).

Using this sequence, a = 27·(1/a = 27·(1/9) = 3,

∴

Using this sequence, a3 = 27·(1/a2) = 27·(1/9) = 3,

a = 27·(1/a ) = 27·(1/3) = 9, ... etc., which creates the sequence of 3, 9, 3, 9, ...

∴

a4 = 27·(1/a3) = 27·(1/3) = 9, ... etc., which creates the sequence of 3, 9, 3, 9, ...

∴Every even term is 9. ∴ a100 = 9.Every even term is 9. ∴ a100 = 9.

Ans. (B)

∴

Ans. (B)


38. The point whose polar coordinates are (3, 45°) is38. The point whose polar coordinates are (3, 45°) isthe same as the point whose polar coordinates arethe same as the point whose polar coordinates are

(A) (-3, 45°) (B) (-3, 225°) (C) (3, -45°) (D) (-3, 135°) (E) (3, 225°)(A) (-3, 45°) (B) (-3, 225°) (C) (3, -45°) (D) (-3, 135°) (E) (3, 225°)

As shown in the figure,As shown in the figure,

the point Q is in the opposite side of the point P.the point Q is in the opposite side of the point P.

Now, for the point Q(3, 225°) to beNow, for the point Q(3, 225°) to be

the same point as P(3, 45°),the same point as P(3, 45°),

we need to changewe need to change

r = (+)3 of point Q to r = (-)3

∴

r = (+)3 of point Q to r = (-)3

to move the direction backward!!

∴

to move the direction backward!!

∴ (-3, 225°) = (3, 45°)∴ (-3, 225°) = (3, 45°)

In general, (r, θ) = (-r, θ + 180°).

∴

In general, (r, θ) = (-r, θ + 180°).

∴

Ans. (B)Ans. (B)


39. If log x2 ≥ log 8 + (1/2)log x, then39. If log x2 ≥ log 8 + (1/2)log x, then

(A) x ≥ 2 (B) x ≤ 2 (C) x ≤ 4 (D) x ≥ 4 (E) x ≥ 1(A) x ≥ 2 (B) x ≤ 2 (C) x ≤ 4 (D) x ≥ 4 (E) x ≥ 1

Referring to the properties of logarithmic function,Referring to the properties of logarithmic function,

log AB = log A + log B, log A/B = log A - log B, we getlog AB = log A + log B, log A/B = log A - log B, we get

32222

2 xx .080881

81

8or ,0log8loglog 3222 xxxxxx

x

x

x

xxx

Since we get

88 xx

,0x .083 xSince we get,0x .083 x

.4 83 xx

Ans. (D)

.4 8 xx

Ans. (D)

However, this algebraic manipulation takes too long.However, this algebraic manipulation takes too long.

Therefore, my recommendation is to use a graphing utility,Therefore, my recommendation is to use a graphing utility,

let y1 = log x2, and y2 = log 8 + (1/2)log x, and then,1 2

look for the solution for y1 ≥ y2 → x ≥ 4.look for the solution for y1 ≥ y2 → x ≥ 4.


40. A right circular cylinder is circumscribed about a sphere.40. A right circular cylinder is circumscribed about a sphere.If Vs represents the volume of the sphere and Vc representsIf Vs represents the volume of the sphere and Vc representsthe volume of the cylinder, thenthe volume of the cylinder, then

2 2 2 2 2(A) (B) (C) (D) (E)VcVs

2 VcVs

2 VcVs

2 VcVs

2 VcVs

2(A) (B) (C) (D) (E)VcVs

3 VcVs

3 VcVs

3 VcVs

3 VcVs

3

3 3 3 3 3

Refer to the figure shown,Refer to the figure shown,

Vc = πr2h = πr2(2r) = 2πr3.

∴

Vc = πr2h = πr2(2r) = 2πr3.

V = 4/3πr3

∴

Vs = 4/3πr3

∴ V = 2/3(V )∴ Vs = 2/3(Vc)∴

Ans. (A)


2 2 3x x 41. If , what must the value of k be in order2 2 3 when 3 ( ) 3,

x x xf x x 41. If , what must the value of k be in order when 3 ( ) 3,

, when 3

xf x xk x

for f (x) to be a continuous function?

, when 3k x

for f (x) to be a continuous function?for f (x) to be a continuous function?

(A) 2 (B) 4 (C) 0 (D) -2(A) 2 (B) 4 (C) 0 (D) -2(E) No value of k will make f (x) a continuous function.(E) No value of k will make f (x) a continuous function.

In order to be a continuous function, the discontinuity at x = 3 must be removed.In order to be a continuous function, the discontinuity at x = 3 must be removed.

That means, for the given function,That means, for the given function,

f (x) = (x2 - 2x - 3)/(x - 3) = (x - 3)(x + 1)/(x - 3) must be reduced tof (x) = (x2 - 2x - 3)/(x - 3) = (x - 3)(x + 1)/(x - 3) must be reduced to

f (x) = (x - 3)(x + 1)/(x - 3) = (x + 1), x ≠ 3.

Now, with this function f (x) = (x + 1), f (3) = 3 + 1 = 4,Now, with this function f (x) = (x + 1), f (3) = 3 + 1 = 4,

and k must be equal to f (3) for f (x) to be a continuous function.and k must be equal to f (3) for f (x) to be a continuous function.

Ans. (B)Ans. (B)


42. Write the complex number in polar form.i342. Write the complex number in polar form.i3

(A) (B)

sincos2

i

sincos2

i(A) (B)

6sin

6cos2

i

6sin

6cos2

i

(C) (D)

66 66

77

1111

(C) (D)

6

7sin

6

7cos2

i

6

11sin

6

11cos2

i

(E)

66

66

1111 (E)

6

11sin

6

11cos2

i

66

Let where,3 ibiaz .1)( ,3 ba1

Let where

Then, from the course of Polar coordinates,

,6

1tan ,24 122

barThen, from the course of Polar coordinates, ,63

tan ,24 bar

or in the 4th quadrant.11

or in the 4th quadrant.66

Therefore, using z = a + bi = r(cos θ + i sin θ) in polar form,

.11

sin11

cos23

iiz .6

11sin

6

11cos23

iiz

Ans. (E)

66 Ans. (E)


43. If , then θ =32sincos 22

43. If , then θ =32cossin

sincos

(A) 15° (B) 30° (C) 45° (D) 60° (E) 75°

cossin

(A) 15° (B) 30° (C) 45° (D) 60° (E) 75°

sincos 22 Using a graphing utility, let and then find the intersection.,322 y,

sincos 22

1

yUsing a graphing utility, let and then find the intersection.

The result is θ = 15°.

,322 y,cossin

1

y

The result is θ = 15°.

Ans. (A)

cossin

Ans. (A)


44. What is the degree measure of the angle x44. What is the degree measure of the angle xof a triangle that has sides of lengthof a triangle that has sides of lengthof 5 and 7 as shown in the figure (1)?of 5 and 7 as shown in the figure (1)?

(A) 44.43° (B) 54.31°(A) 44.43° (B) 54.31°(C) 71.37° (D) 25.69°(C) 71.37° (D) 25.69°(E) 34.31°(E) 34.31°

Using the Law of sine, we get 5/sin 30° = 7/sin x°Using the Law of sine, we get 5/sin 30° = 7/sin x°

730

10

7

5

30sin7sin x

7

105sin7sin x

43.447

sin 1x 43.4410

sinx

Ans. (A)Ans. (A)


45. In △ ABC, a = 2, b = 5, ∠C = 60°.△

45. In △ ABC, a = 2, b = 5, ∠C = 60°.△

△ ∠What is the area of △ ABC?△ ∠

What is the area of △ ABC?△ ∠

△

(A) 4.3

△

(A) 4.3(B) 3.6(B) 3.6(C) 3.2(C) 3.2(D) 2.9(D) 2.9(E) 2.3(E) 2.3

3511Area of .

2

3560sin52

2

1sin

2

1 cabABCArea of

Ans. (A)

.2

60sin522

sin2

cabABC

Ans. (A)


46. If f (x, y) = x3 – xy2 + 3y4,46. If f (x, y) = x3 – xy2 + 3y4,which of the following is (are) true?which of the following is (are) true?

I. f (x, y) = f (x, -y)I. f (x, y) = f (x, -y)II. f (x, y) = f (-x, y)II. f (x, y) = f (-x, y)III. f (x, y) = f (-x, -y)III. f (x, y) = f (-x, -y)

(A) I only (B) II only (C) III only (D) I and III only (E) I, II, and III(A) I only (B) II only (C) III only (D) I and III only (E) I, II, and III

Since f (x, y) = x3 - xy2 + 3y4 has y–terms with even powers,Since f (x, y) = x3 - xy2 + 3y4 has y–terms with even powers,

f (x, -y) should have the same result as f (x, y),f (x, -y) should have the same result as f (x, y),

while f (-x, y) should have effect on its x–terms with odd powers.while f (-x, y) should have effect on its x–terms with odd powers.

Therefore, the only right choice is (I).Therefore, the only right choice is (I).

Ans. (A)Ans. (A)


47. Figure 5 shows a cube with edge of47. Figure 5 shows a cube with edge of Blength 4 centimeters. If points A and C

Blength 4 centimeters. If points A and Care midpoints of the edges of the cube,are midpoints of the edges of the cube,what is the area of region ABCD?what is the area of region ABCD?

A

(A) 15.71 cm2

A

(A) 15.71 cm2

(B) 17.25 cm2 C(B) 17.25 cm2

(C) 19.60 cm2

C

(C) 19.60 cm2

(D) 20.00 cm2(D) 20.00 cm2

(E) 22.63 cm2(E) 22.63 cm2

DD

.522042 22 AD

∴ area of □

.522042 AD

222052 cmABCD ∴ area of □

Ans. (D)

2052 cmABCD ∴

Ans. (D)


48. If (x, y) represents a point on the graph of which of the ,12 xy48. If (x, y) represents a point on the graph of which of the ,12 xy

following could be a portion of the graph of the set of points (x, y2)?following could be a portion of the graph of the set of points (x, y )?

(A) (B) (C) (D) (E) (A) (B) (C) (D) (E) 2y 2y 2y 2y

2yy y y 2y y

1 1 1 1 1

x1

x1

x1

x1

x11 1

For a given function, we transform our coordinates from (x, y) to (x, y2).,12 xyFor a given function, we transform our coordinates from (x, y) to (x, y2).,12 xy

2 Therefore, letting Y = y2, we get ,121222 xxyYTherefore, letting Y = y , we get

or Y = 2x + 1 in (x, Y) coordinate system.

or Y = 2x + 1 in (x, Y) coordinate system.

Ans. (A)Ans. (A)


49. If the parameter is eliminated from the equations49. If the parameter is eliminated from the equationsx = 4t2 + 1 and y = 2t, then the relation between x and y isx = 4t + 1 and y = 2t, then the relation between x and y is

(A) y = x – 1 (B) y = 1 – x (C) y2 = x – 1 (D) y2 = (x – 1)2 (E) y2 = 4x - 4(A) y = x – 1 (B) y = 1 – x (C) y = x – 1 (D) y = (x – 1) (E) y = 4x - 4

In this problem of parametric equation, we need to eliminate "t" from x and y equations,In this problem of parametric equation, we need to eliminate "t" from x and y equations,

and set up the two equation into one equation with x and y variables only.and set up the two equation into one equation with x and y variables only.

∴

Since y = 2t, and x = 4t2 + 1, we replace 4t2 with y2.

∴

∴

∴ x = 4t2 + 1 = (2t)2 + 1 = y2 + 1

∴

∴ x = 4t + 1 = (2t) + 1 = y + 1

∴ y2 = x – 1

∴

∴ y2 = x – 1

Ans. (C)

∴

∴

Ans. (C)


50. If 8 people shook hands with each other,50. If 8 people shook hands with each other,how many handshakes were exchanged?how many handshakes were exchanged?

(A) 8 (B) 16 (C) 21 (D) 28 (E) 56(A) 8 (B) 16 (C) 21 (D) 28 (E) 56

Shaking hands implies between 2 people.Shaking hands implies between 2 people.

Therefore, C = 28.Therefore, 8C2 = 28.

Ans. (D)

SAT II -Math Level 2 Test #02 Solution -...

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