Randy P. GubatonMNHS

Each question is copied verbatim from the test paper to preserve the originality of the item. For the purpose of study, some questions are rephrased with the author’s comment in Note or COMMENT.

Trial Version

The accuracy of the answer in selected solutions (with Note)is verified using The Geometer’s Sketchpad v. 3.10N and 4.06

1

MTAP 2012 – Math 4Detailed Solution

Randy P. GubatonMaloco NHS

Solve each item on scratch paper and write the answer on the blank. Give equation of

lines asax+by+c=0 .

1. definition of function

Which of the following is/are not functions?a. y = 5x – 3 b. y = x2 – 3x + 4c. x = y2 – 3y + 2

Solution:

a] is a linear a function. It takes the

form of the line ax+by+c=0 .

b] is a quadratic function. It takes the

general form ax 2+bx+c=0 .

c] is a quadratic relation. It is not a function. Its graph is a parabola that opens to the right. It fails as a function by the vertical test.

d] is a radical function. Its graph is in QI.

Answer: a, b, and d.

if n is an even integer, g( x )≥0 .

2 x−3≥02 x≥3x≥3

2

The domain isx≥ 32 .

2. domain, linear function

What is the domain of y = 4 – 4x?

Solution:

The domain is all permissible values of x.

We can assign any real number as x and get a unique number y as a result.

The domain is R (all real numbers).

3. range of linear equation

What is the range of y = 8x – 3?

Solution:

The range is the set of all resulting values of y when x is used in a function.

In the equation y=8 x−3 , when x is assigned any real number, y gives a unique real number.

The range is R (all real numbers).

4. composition of function

If f(x) = 4x – 3 and g(x) = √4 x+1 , what is (fog)(6)?

2

Solution:

Evaluate ( f ∘g ) (6 ) given f ( x )=4 x−3

and g ( x )=√4 x+1 .

( f ∘g ) (6 )=f (g (6 ) )

Since

g (6 )=√4 (6 )+1=√24+1=√25=5

therefore

f (g (6 ) )=f (5 )=4 (5 )−3=20−3=17

5. equation of line, parallel

Write an equation of the line through (3,–5) and the parallel to the line 3x – 5y – 12 = 0.

Solution:

Since the required line is parallel to the given line, their slopes are equal. Finding the slope of the given line by rewriting it to the slope-intercept form of the line y=mx+b .

3 x−5 y−12=03 x−12=5 y35

x−125= y

Since the resulting equation is of the form y=mx+b where m is the slope, we can

easily see thatm=35 .

With point (3, –5) and slopem=35 , using

the point-slope form of the line,

y− y1=m (x−x1)y−−5=3

5(x−3 )

5 ( y+5 )=(3 x−95 ) (5 )

5 y+25=3 x−93 x−5 y−25−9=03 x−5 y−34=0

6. form y = mx + b

Write 7x – 3y = 21 in the form y = mx + b.

Solution:

7 x−3 y=217 x−21=3 yy=7

3x−21

3

y=73

x−7

7. distance between two points

Find the length of the line segment joining P( -7,8) & Q(5,-1).

Solution:

The distance d between two points is

d=√( x2−x1)2+( y2− y1 )

2

¿√ (−7−5 )2+(8−−1 )2

¿√ (−12 )2+(9 )2=√81+144¿√225=15 units

3

8. inequality, absolute

Solve for x: |2x + 3| ≤ 9.

Solution:

|E|<b is equivalent to −b<E<b .

Rewrite |2 x+3|≤9 in this form and solve for x.

|2 x+3|≤9−9≤2 x+3≤9

−9−3≤2 x+3−3≤9−3−12≤2 x≤6−6≤x≤3

9. y-intercept, parabola

If y = x2 – 6x + c and the graph of the function passes through A(2,3), what is the value of c? Solution:

The parabola contains the point A(2, 3). Substitute these values to the given equation.

y=x2−6 x+c3=22−6 (2 )+c3=4−12+c3=−8+c3+8=c11=c

10. standard form, quadratic

Write y = 2x2 + 4x – 3 is the form y = a(x – h)2 + k.

Solution:

Rewrite into the standard form

y=a ( x−h )2+k by completing the

square.

y=2 x2+4 x−3¿ (2 x2+4 x )−3

¿2 (x2+2 x )−3

¿2( x2+2 x+(22 )2 )−3−2(22 )

2

¿2 (x2+2 x+1 )−3−2

y=2 ( x+1 )2−5

11. vertex of parabola

What is the vertex of the graph of y = 2x2 + 4x – 3?

Solution:

Find the axis of symmetry first.

x=− b2a=− 4

2 (2 )=− 4

4=−1

Substitute the result to the original equation.

y=2x2+4 x−3¿2 (−1 )2+4 (−1 )−3¿2−4−3¿−5

The vertex (h , k )=(−1 ,−5 ) .

12. roots, quadratic

Solve for the roots of 2x2 – 9x – 5 = 0.

4

Solution:

By factoring,

2 x2−9 x−5=(2 x+1 ) ( x−5 )2 x+1=02 x=−1x=−1

2

x−5=0x=5

The roots are −12 , 5.

13. quadratic equation, given roots

Find a quadratic equation with integral coefficient whose roots are 4 and –5/6.

Solution:

x=4x−4=0

x=−56

6 x=−56 x+5=0

Set the values as factors in an equation and expand.

( x−4 ) (6 x+5 )=06 x2−24 x+5 x−20=06 x2−19x−20=0

14. y-intercept of line

What is the y-intercept of 2x – 3y = 6?

Solution:

b is the y-intercept of the line y=mx+b .

Rewrite the given equation in the above form.

2 x−3 y=62 x−6=3 y23

x−63= y

23

x−2= y

The y-intercept b is –2.

15. odd/even function

Evaluate f(x)= 2x4 + 2x2 + 5 at x = 2. From the result, can you say what kind of function it is?

Solution:

Evaluate f (2 )andf (−2 ) .

If f (2 ) ,

f ( x )=2x4+2 x2+5f (2 )=2 (2 )4+2 (2 )2+5¿2 (16 )+2 (4 )+5¿32+8+5¿45

If f (−2 ) ,

f ( x )=2x4+2 x2+5f (−2 )=2 (−2 )4+2 (−2 )2+5¿2 (16 )+2 (4 )+5¿32+8+5¿45

Sincef ( x )=f (−x ) , the function is even.

5

16. synthetic division

If the polynomial P(x) = x3 – 3x2 – x + 3 is written in the form P(x) = Q(x)D(x)+R(x), if D(x) = x+1, what is Q(x)?

Solution:

D ( x )=x+1 takes the general form

D ( x )=x−r . It follows that

−r=1 ⇒ r=−1

Use this as divisor to P ( x ) by synthetic

division.

[−1 ] 1 −3 −1 3−1 4 −3

1 −4 3 0

Q ( x )=x2−4 x+3 .

17. remainder theorem

What is the remainder when 5x4 + 3x3 – 4x2 + 5x – 3 is divided by x + 1?

Solution:

[−1 ] 5 3 −4 5 −3−5 2 2 −7

5 −2 −2 7 −10

The remainder is –10.

18. equation, polynomial function

Find a polynomial function with integral coefficients and of lowest degree whose roots are -2, 2, 1.

Solution:

The roots are

x1=−2 x2=√2

x3=−√2

x4=1

Setting this up in an equation,

f ( x )= (x+2 ) ( x−√2 ) ( x+√2 ) ( x−1 )f ( x )=x 4+x3−4 x2−2 x+4

Note: Answer Key says

f ( x )= (x+2 ) (x2−2 ) ( x−1 ) .

19. perfect square polynomial

What polynomial of lowest degree must be multiplied to 2x3 – 5x2 – 4x + 12 to make it perfect square?

Solution:

A polynomial is a perfect square polynomials if its extreme terms, when arranged, are perfect squares.

The leftmost term is 2 x3. The lowest

perfect term to it is4 x4. Finding the

quotient of the two values,

4 x4

2 x3=2 x

The rightmost term is 12. Its nearest perfect square multiple is 36. Finding the quotient of the two values,

3612=3

6

The required factor is the sum of the two values above.

The factor is2 x+3 .

20. simplifying exponential expr.

Simplify and express as a product of powers (integral or fractional) of primes: (24)1/4(32)(15)2/3

Solution:

(24 )1/4 (√32 ) (15 )2/3=(8⋅3 )14 (8⋅4 )

12 (3⋅5 )

12

¿(23(14 )⋅314 )(23(12 )⋅2

2(12 ))(3(23 )⋅5(

23 ))

¿234⋅2

32⋅2

22⋅3

14⋅3

23⋅5

23

¿234⋅2

64⋅2

44⋅3

312⋅3

812⋅5

23

¿2134 ⋅3

1112⋅5

23

21. quadratic equation

For what value of k will x–5 be a factor of kx3 – 17x2 – 4kx + 5?

Solution:

x−5 ⇒ x=5

By factor theorem,

kx 3−17 x2−4kx+5=0k (5 )3−17 (5 )2−4k (5 )+5=0125 k−425−20 k+5=0105 k−420=0105 k=420k=4

22. roots, polynomial equation

Find the roots of x3 – 2x2 – x + 2 = 0.

Solution:

By synthetic division,

[1 ] 1 −2 −1 21 −1 −2

[−1 ] 1 −1 −2 0−1 2

[2 ] 1 −22

1 0

The roots are –1, 1 and 2.

23. simplifying exponential expr.

Simplify (32)2/5.

Solution:

(32 )2/5=3225=(2 )

5(25 )=22=4

24. domain, exponential function

What is the domain of the function f(x) = 2x+2?

Solution:

7

Test several values of x.

Ifx=−3 ,

2x+2=2−3+2=2−1=12

Ifx=−2 ,

2x+2=2−2+2=20=1

Ifx=0 ,

2x+2=20+2=22=4

Ifx=1 ,

2x+2=21+2=23=8

6

4

2

-5 5

The graph shows the values of x are continuous from negative infinity to

positive infinity(−∞ , ∞ ) .

The domain is R, or the set of all real numbers.

25. range, exponential function

What is the range of the function f(x) = 2x–3?

Solution:

Test several values of x.

Ifx=−1 ,

2x−3=2−1−3=2−4= 1

24= 1

16

Ifx=0 ,

2x−3=20−3=2−3= 1

23=1

8

Ifx=3 ,

23−3=20=1

Ifx=4 ,

24−3=21=2

5

5

The graph shows the values of y are continuous from above the x-axis to positive infinity.

The range is y>0 .

26. inverse of linear function

What is the inverse of f(x) = 5 – 3x?

Solution:

Let y=f −1 (x ).8

Interchange y to x and solve for y.

f ( x )=5−3 x⇓y=5−3 x⇓x=5−3 y

x=5−3 y3 y=−x+5

y=− x+53

y=−13

x+ 53

f−1( x )=−13

x+ 53 .

27. domain, logarithm

What is the domain of y = logbx?

Solution:

In exponential form,

b y=x

We cannot find a number x where x=0 . It is always more than 0.

The domain is x>0 .

28. exponential form

Write in the exponential form: log(0.001) = -3

Solution:

In common logarithm, the base is 10, not usually written.

log (0 . 001 )=−3log10 0 . 001=−3

10−3=0 . 001

29. exponential decay

The price of a car depreciates at the rate of 8% per year. Give the formula that will give the value of a car that cost P600,000 at the end of three years.

Solution:

P ( t )=A (1−r )t

P (3 )=600 ,000 (1−0.08 )3

P (3 )=600 ,000 (0 .92 )3

30. exponential equation

Solve for x: 92x + 3 = 27x + 4.

Solution:

92 x+3=27x+4

32 (2 x+3 )=33 (x+4 )

2 (2x+3 )=3 ( x+4 )4 x+6=3 x+12x=6

31. exponential

The half-life of a radioactive substance is 75 years. How much of a 100-g of the substance remains after 300 years?

Solution:

y=100 ( 12 )t

75

After 300 years,

9

y=100 (12 )t75=100(12 )

30075 =100(12 )

4

¿100(116 )=6 .25

6.25 grams remains after 300 years.

32. logarithmic expression

Simplify and express in terms of the logarithms of the individual variables: logb[a(3mn) / p2]

Solution:

logb [a ( 3√mn) / p2]

¿ logb[a (mn )13

p2 ]¿ logb

[a (mn )13 ]− logb P2

¿ logba+1

3logb m+ logbn−2 logb P

Note: Answer Key says logb a+( 1

3 ) [ logb m+logb n ]

33. logarithmic expression

Express as single logarithm: 2log(x – 3) + log(x+1) – log(x + 3)

Solution:2 log ( x−3 )+ log ( x+1 )−log ( x+3 )=log (x−3 )2+ log ( x+1 )−log (x+3 )

¿ log( x−3 ) ( x+ 1 )x+3

34. logarithmic equation

Solve for x: log3(x – 2) + log3(x + 4) = 3.

Solution:

Apply the logarithm of a product.

log3 ( x−2 )+log3 (x+4 )=3

log3 ( x−2 ) (x+4 )=3

Change to exponential form and solve for x.

log3 ( x−2 ) (x+4 )=3

33=( x−2 ) ( x+4 )27=x2+4 x−2 x−827=x2+2 x−80=x2+2 x−350=( x−5 ) ( x+7 )

x−5=0x=5

x+7=0x=−7

Check each value of x.

If x=5 , the result is true.

log3 (5−2 )+ log3 (5+4 )=3log33+log3 9=3

log331+ log3 32=31+2=33=3

If, the result is false.

log3 (−7−2 )+ log3 (−7+4 )=3

log3 (−9 )+ log3 (−3 )=3

10

Since the logarithm of a negative number does not exist, x=−7 is not a solution.

x=5 .

Note: Answer Key says x=5 , x=−7 .

35. logarithm of a product

If log2 = 0.3010 log3 = 0.4771 log7 = 0.8451, find log 42.

Solution:

Factor 42 into primes.

42=2×3×7

Apply the logarithm of a product.

42=2×3×7log 42=log (2×3×7 )log 42=log 2+log 3+ log7

Substitute the given values to the equation.

log 42=log 2+log 3+ log7=log 2+ log 3+log 7=0.3010+0 .4771+0 .8451log 42=1. 6232

36. coordinates of pt. on unit circle

If (x, 7/4) is on the unit circle, find the value of x in Q1.

Solution:

Find x by Pythagorean Theorem.

x2+ y2=r2

x2+(√74 )

2

=12

x2+716=1

x2=1616−7

16

x2=916

x=±√916=±3

4

Since x is in the first quadrant, it is positive.

x=34 .

37. magnitude of arc

How many degrees is the angle formed by a ray that makes 3 1/5 complete rotations counterclockwise?

Solution:

In a circle, 1 rotation is 360°. 15 of a

rotation is

15×360 °=72 °

315 rotation is

3 (360 ° )+72°=1 ,080 °+72°=1 , 152 °

38. clock problem

The minute hand of a clock is 12 cm long. Find the length of the arc traced by the minute hand as it moved from its position at 3:00 to 3:40.

7/4

x

1

12

11

Solution:

One rotation of the minute hand of the clock is 360°.

In 5 minutes, the minute hand traces

360 °÷12=30°

From 3:00 to 3:40, the hand has traced

30 °×8=240 °

Convert this to radians.

240 °× π180 °

=10⋅3⋅2⋅4⋅π10⋅3⋅2⋅3

=43

π

Find the length of the arc.

s=θr=4

3π (12 )=16 π=50 .22 cm

39. function of an angle

If is a real number representing an arc of length 28 /6, in which quadrant is the terminal point of ?

Solution:

Divide the unit circle into 12 congruent arcs. Starting from the initial point A going counterclockwise around the circle,

A B= 16

π , A C= 26

π , etc.

We continue going around until we stop at

the terminal point E with arclength 286

π .

E is at quadrant II.

40. exact value of angles

What is the exact value of (sin /3)(cos /6)?

Solution:

(sinπ3 )(cos

π6 )=√3

2⋅√3

2=3

4

41. angle of elevation

What is the height of a tree if 150 meters from its base, its top is sighted at an angle of 300?

Solution:

tanθ=opp .adj.

tan30 °=h150

150 tan 30 °=h150 (0 .5774 )=h86 .61 m=h

tan30=sin 30 °cos°

=12

√3

2

¿1

√3⋅√3

√3

¿√33¿0 .5774

The tree is 86.51 meters high.

Note: Answer Key says 28.87 m.

42. angle of depression

An observer is on top of a lighthouse, 60

meters above sea level. The observer sighted a ship at an angle of depression of 300. How far is the ship from the lighthouse?

60

x

30

30

lighthouse

ship12

Solution:

tanθ=opp .adj .

tan30 °=60x

x=60tan 30°

x=60(0 .57735 )

103 . 92 m=x

tan30=sin 30 °cos°

=12

√3

2

¿1

√3⋅√3

√3

¿√33¿0 . 5774

The ship is 103.92 meters from the

lighthouse.

43. 30°-60°-90° triangle

Triangle MNO has a right angle at N. If MN = 15 cm and ∠O = 300, find the length of NO.

Solution:

In a 30°-60°-90°

triangle, the ratio of the sides is

x : √3 :2 x

Find ON by proportion.

ON

√3=

151

ON=15√3 m

44. cosine law

Two sides of ABC are AB = 15 cm and BC= 12 cm. If ∠B = 600, how long is AC. Give the answer as a simplified radical.

Solution:

Solve for b using the cosine law.

Note: Answer Key

says 3√31 cm .

b2=a2+c2−2ac cos βb2=122+152−2 (12 ) (15 ) (cos60 ° )b2=144+225−360(12 )b2=369−180=189b=√189=√9⋅21=3√21 cm

45. geometric mean

In ABC, AB = 13 cm, AC = 15 cm and BC = 14 cm. Find the length of the altitude from A to BC.

Solution:

In right triangle ADB,

h

15030

151

2 3 30

M N

O

c = 15 a = 12

b

60

A

B

C

15

13h

14 - x

x

A C

B

D

13

h2+x2=132

h2+x2=169h2=−x2+169

In right triangle ADC,

h2+(14−x )2=152

h2+196−28 x+x2=225h2+x2−28 x=29h2=−x2+28 x+29

Equate and .

−x2+28 x+29=−x2+16928 x+29=16928 x=169−2928 x=140x=140

28

x=5

Substitute the result to .

h2+x2=132

h2+(5 )2=132

h2+25=169h2=169−25h2=144h=√144=12 cm

46. trigonometric equation

On 0° ≤ ≤ 180°, find all the values of for which 4 cos2 = 3.

Solution:

4 cos2θ=3cos2 θ=3

4

√cos2 θ=±√3

4

cosθ=±√3

√4

cosθ=±√32

The values of for which

cosθ=±√32 are

{30°, 150°, 210°, 330°}.

But given the restriction 0 °≤θ≤180 ° , the permissible values are

{30°, 150°}.

47. projectile

An object is projected upward from the ground. After t seconds, its distance in feet above the ground is s = 144t – 16t2. After how many seconds will the object be 128 feet above ground in coming down?

Solution:

After t seconds, its distance from the ground is 128 feet, or

s=144 t−16 t2

128=144 t−16 t2

14

Solve for t.

128=144 t−16 t2

8=9 t−t2

t2−9 t+8=0( t−8 ) ( t−1 )=0t−8=0t=8

t−1=0t=1

The object is 128 feet above ground going up after 1 second, and reaches the same height after 8 seconds after launch going down.

The object will be 128 feet above ground after 8 seconds after launch going down.

48. volume of cube

The diagonal NQ of cube MNOPQRST

measures √12cm. What is the volume of the cube?

Solution:

Let x be the length of the cube.

Find MO by Pythagorean Theorem.

x2+x2=MQ2

2 x2=MQ2

√2x2=√MQ2

x √2=MQ

Solve for x in NMO by Pythagorean Theorem.

x2+(2√ x )2=√122

x2+4 x=12x2+4 x−12=0( x+6 ) ( x−2 )=0

x12

2 xM

N

Q

x+6=0x=−6

x−2=0x=2

We reject x=−6 since it is negative. So, x=2 .

Find the volume usingx=2 .

V=a3

V=23=8 cm3

49. arithmetic series

The nth term of a series is 9n + 7. What is the result of subtracting the kth term from the (k + 1)th term?

Solution:

Multiply by 9.

Add 7 to the result.

Subtract the nth term (in terms of k) from the result.

xx

x

NO

M P

Q

RS

T

x 2

x

x

T

M

Q

P

15

50. sum of roots

Find k so that the sum of roots of 3x2 – (3k + 2)x + 18 = 0 is 6.

Solution:

The given equation is in the general form

ax 2+bx+c=0 . Rewrite it in the form

x2+ ba

x+ ca=0 .

3 x2− (3 k+2 ) x+18=0⇓

x2−3k+23

x+183=0

We see that

ba=−

3 k+23

−ba=3 k+2

3

The roots of quadratic function are

and

x2=−b−√b2−4 ac

2 a

The sum of the roots of a quadratic function is

x1+x2=−b+√b2−4 ac2 a

+−b−√b2−4 ac2 a

x1+x2=−2b2 a

=−ba

Since the sum of the roots is 6, then, equating and ,

3 k+23

=6

3 k+2=183 k=16k=16

3

16

a

acbbx

2

42

1