CSET Subtest I Practice Test

MCQ 1. A logistic growth model of the form y = A/(B + C·e ^–Kx) can be used to describe the proportion of households, y, that owns a VCR at year, x, where A, B, C and K are constants. Express x as a function of y.

MCQ 2. For what value(s) of K will the parabola y² – 6y + x + K = 0 have exactly 1 y–intercept? Find the intercept.

MCQ 3. For very large values of x, the graph of f(x) = x² / (x + 1) behaves as that of a line y = mx + b. Find m and b.

MCQ 4. The maximum safe load for a horizontal beam varies jointly as the width of the beam and the square of the thickness of the beam, and inversely with its length. How will the thickness of the beam have changed if the maximum safe load were halved while the width of the beam was doubled, and its length increased by a factor of 3?

MCQ 5. A swimming pool, rectangular in shape, is surrounded by a walkway of uniform width, say x. If the outer dimensions of the walkway are 16m by 10m and the area of the pool alone is 112m², find x.

MCQ 6. Find the Greatest Common Factor and Least Common Multiple of 5160, 5640, 4920, 4680 and 3720. [Express the LCM as a product of primes.]

MCQ 7. What is the point of intersection of f(x) = ½ x + 3/2 and its inverse?

MCQ 8. What shape does the graph |y| + |x| < 3 resemble?

MCQ 9. What does the expression: (x³/y³ – 1)(x/y – 1)^–1 /(x²/y² – 1)

simplify to?

MCQ 10. ΔCAB is right–angled at A with AC = 8 and AB = 12. Points F, E and D lie on sides AB, BC and CA respectively such that AFED is a rectangle. If AF ≈ x, find an expression for the Area of rectangle AFED in terms of x.

MCQ 11. If A is a 2X2 matrix, A = [a b]

[c d]

simplify: A1/|A| [d –b] where |A| denotes determinant(A).

[–c a]

MCQ 12. A company borrows $500,000 to expand its product line. Some of the money is borrowed at 7%, some at 8% and some at 10%. If the annual interest is $63,000 and the amount borrowed at 7% is 4 times that borrowed at 10%, express the system of equations to be solved in Matrix Form.

MCQ 13. The number of bacteria, in thousands, in a Petri dish is given by the function:

N(T) = 15T² – 100T + 1000, 5 < T < 20

where T denotes the temperature in Celsius. The temperature of the Petri dish is a function of time, t, in hours:

T(t) = 3t + 7, 0 < t < 5

(i) Find the number of bacteria in the Petri dish after 3 hours have elapsed.

(ii) If 1,500,000 bacteria are found in the Petri dish, how many hours must have elapsed?

MCQ 14. A parabola its x–intercepts at –1 and 3. If the minimum value of the function is –2 find its equation.

MCQ 15. Determine the constant term of the polynomial with integer coefficients of lowest degree with roots ±1 and (–3 + i√7)/2, and whose leading coefficient is 3.

MCQ 16. Find the magnitude and direction of the vector v = –√i + j.

MCQ 17. A computer manufacturer determines that it can sell 4000 machines at $750, and that for each $25 that the price is increased 20 fewer computers are sold. How much the manufacturer should charge to maximize his revenue?

MCQ 18. Find the number of integers between 50 and 500 [inclusive] divisible by 5 or 7.

MCQ 19. If $4500 compounded at 6.5% monthly for t years grows to $6800, write an expression for t using logs.

MCQ 20. A square sheet of paper of side, A, is snipped at a distance of C units from the corners to produce a regular octagon of side, B units. Find B and C in terms of A.


MCQ 21. There are 2 vectors a and b such that a = <–√2, –√2>

2 2

and b = –6a. Find the magnitude and direction of b.

MCQ 22. If point (3, 2) lies on the graph of the inverse of f(x) = 2x³ + x + A, find A.

MCQ 23. If 7 + g^–1(x – 1) = 9 and g(2) = 6, then find x.

MCQ 24. If f(x) = (x – 1)/(x – 2) and g(x) = (x – 3)/(x – 4), then find

the domain of f[g(x)].

MCQ 25. If two forces with magnitudes 4 lb and 3 lb act upon the same point with the angle between them being 120^o, then find the magnitude of the resultant force.

MCQ 26. The first few steps for finding the GCD between 726 and 275 are shown below. Complete the problem to find the GCD.

MCQ 27. If f(x) and g(x) are defined by
x –2 4 6 8
f(x) –3 0 3 4

x –3 –2 0 8
g(x) 2 4 6 –4

then find f(g^–1(4)) and g(f^–1(4)).

MCQ 28. What is the highest number of the points of intersection between If f(x) = x⁷ – 5x⁶ + 3x⁵ + 4x⁴ – 2x³ + 8x² + x – 6 and g(x) = x⁴ + 2x³ – 7x² +5x – 11?

MCQ 29. Find the equation of the line with the same x– and y–intercepts as x² + y² + 4x – 4y + 4 = 0.

MCQ 30. If i (=√–1) is a root of g(x) = 5x⁴ – 3x³ + ax² + bx – 7 then find the value of a and b.

FR #1: If the sum of the digits of a number is divisible by 3, then prove that the number itself is divisible by 3.

FR #2: A mattress manufacturer makes 2 kinds of mattresses, A and B. The production process involves two basic types of labor: filling and finishing. Mattress A requires 2 hours of filling and 1 hour of finishing, whereas B requires 3 hours of filling and ½ hour of finishing. The profit is $35 per mattress A and $20 per mattress B. The manufacturer’s employees can supply a maximum of 108 hours of filling work and 20 hours of finishing work per day. How many of mattress A and B should be made each day to maximize the profit?

FR #3: A window is in the shape of a semicircle surmounting a rectangle. If the perimeter of the window is P, find the dimensions of the rectangle and semicircle for which maximum light would be allowed.

FR #4: For the function below, state domain and range. Choosing any 8 points for x, draw a graph of an appropriate scale, using the information provided.

y = 1/3 log (x – 2) – 1;

Given: log 2 ≈ 0.3; log 3 ≈ 0.48.

CSET Subtest I Practice Test

Solutions

1. Solution. Cross–multiplying: By + Cy·e ^–Kx= A

=> By + Cy·e ^–Kx = A

=> Cy·e ^–Kx = A – By

=> e ^–Kx = (A – By)/Cy

Taking logarithms: –Kx = ln (A – By)/Cy

Or x = –1/K· ln [(A – By)/Cy]

2. Solution. At the y–intercept(s), x = 0 so that: y² – 6y + K = 0.

For the parabola to have exactly 1 y–intercept: b² – 4ac = 0 => 36 – 4K = 0.

Therefore, K = 9.

At K = 9, the y–intercept is: y² – 6y + 9 = 0

=> (y – 3)² = 0

=> y = 3

3. Solution. Since the degree of the numerator in f(x) = x² / (x + 1) is 1 more than that of the denominator, f(x) has no horizontal asymptotes. However, it has a slant / oblique asymptote.

To find this, simply divide x² by (x + 1) to get: f(x) = (x – 1) + 1/x².

Now, as x → ∞, f(x) → x – 1.

Therefore, the line is y = 1x – 1 with m = 1, b = –1.

4. Solution. Let S denote the maximum safe load for the beam, w, the width of the beam, T, its thickness and l, its length.

Then, S = kwT²/l where k is the constant of proportionality (*)

Initially, let w = S = l = 1unit.

Then, T = 1/√k [Substituting in (*) and solving]

Next, as indicated, let S = ½, w = 2 and l = 3.

Then, T = √(3/4k) [Substituting in (*) and solving]

= ½ √(3/k)

= ~ 0.85/√k [√3 ~ 1.7]

Therefore, the thickness would have dropped by 15%.

5. Solution. Outer dimensions of layout: 16 X 10.

Since the walkway has a uniform width of x, the dimensions of the pool: (16 – 2x) X (10 – 2x)

Given: (16 – 2x)(10 – 2x) = 112

Solving, x = 1m.

6. Solution.

10 |5160, 5640, 4920, 4680, 3720

4 | 516, 564, 492, 468, 372

3 | 129, 141, 123, 117, 93

43, 47, 41, 39, 31

Therefore, the GCF is: 10·4·3 = 120.

The LCM would be: 43·47·41·31·3·13·2·2·3·5.

7. Solution. f^–1(x) = (3x + 4)/(x – 1).

Solve f(x) and f^–1(x) simultaneously: set them equal, cross–multiply and simplify to yield a quadratic equation! The solution is: (2 + √2, 2 + √2) and (2 – √2, 2 – √2).

8. Solution. Graphing |y| + |x| < 3 is equivalent to: |y| < –|x| + 3 which can be broken into:

y < –|x| + 3 ≈ y < x + 3 and y < –x + 3

and

–y < –|x| + 3 ≈ y > x – 3 and y > –x – 3.

Graphing the 4 lines yields a shaded rhombus enclosed by (0, 3), (3, 0), (0, –3) and (–3, 0).

9. Solution.

(x³/y³ – 1)(x/y – 1)^–1 / (x²/y² – 1)

= [(x³ – y³)/y³)][(x – y)/y)]^–1 / [(x² – y² / y²)]

= [(x³ – y³)/y³)][y / (x – y)] / [(x² – y² / y²)]

= [(x – y)(x² + xy + y²)/y³)][y / (x – y)] / [(x + y)(x – y) / y²)]

= [(x – y)(x² + xy + y²)/y³)][y / (x – y)] .[y² / (x + y)(x – y)]

Cross–canceling yields: (x² + xy + y²) / x² – y²

10. Solution. Sketch and label figure to satisfy conditions. If AD ≈ y, then DE ≈ x [since AF = DE].

Using similar triangles, CD/CA = DE/AB so that (8 – y)/8 = x/12.

Simplifying, y = (24 – 2x)/3 (*)

Area of the rectangle, A = xy = (24x – 2x²)/3 [Using (*)]

11. Solution.

|A| = ad – bc

and 1/ |A| [d –b] = [d/ (ad – bc) –b/ (ad – bc)]

[–c a] [–c/ (ad – bc) a / (ad – bc)]

Finally, A [d –b]

|A| [–c a]

= [a b] [d/ (ad – bc) –b/ (ad – bc)]

[c d] [–c/ (ad – bc) a / (ad – bc)]

= [(ad – bc)/ (ad – bc) 0 ]

[0 (ad – bc)/ (ad – bc)]

= [1 0] = I

[0 1]

12. Solution.

If x, y and z are the amounts borrowed at 7%, 8% and 10%, respectively, then:

x + y + z = 500,000

0.07x + 0.08y + 0.1z = 63,000

x + 0y – 4z = 0

constitute the system of equations that illustrate the problem.

In matrix form: AX = B where

A = [1 1 1] X = [x] and B = [500,000 ]

[0.07 0.08 0.1] [y] [63,000 ]

[1 0 –4] [z] [0 ]

13. Solution. (i) After 3 hours, the temperature in the Petri dish is:

T(3) = 33 + 7 = 16 Celsius.

The number of bacteria (in thousands), then, is: N(16) = 1516² – 10016 + 1000 = 3240.

(ii) If 1500 (in thousands) bacteria are found in the Petri dish: N(T) = 1500 = 15T² – 100T + 1000.

Solving the quadratic function for T:

15T² – 100T – 500 = 0 or 3T² – 20T – 100 = 0 => T = 10.

Therefore, for the number of hours that must have elapsed: T(t) = 10 = 3t + 7 => t = 1

14. Solution. Let the parabola be represented as: y = a(x – x₁)(x – x₂), where x₁ and x_{2 ­}are its x–intercepts.

Then, using the given information: y = a(x + 1)(x – 3) (*)

Since the mid–point of the x–intercepts of a parabola lies on its Axis of Symmetry, and the minimum is given to be –2, the parabola’s vertex is V( ½(–1 + 3),–2)) => V(1,–2) (**)

Substituting (**) into (*) and solving for a: a = ½

The equation of the parabola is: y = ½ (x + 1)(x – 3)

15. Solution. If 1 and –3 + i√7/2 are the roots of a polynomial, f(x), then, by the Complex Conjugates Theorem, so is –3 – i√7/2.

Then, f(x) ought to be of degree 4 so that it may be of the lowest degree.

f(x) = a(x – 1)(x + 1)[x – (–3 + i√7/2)][x – (–3 – i√7/2)], where a is the leading coefficient to be determined.

=> f(x) = a(x² – 1)[(x + 3) – i√7/2][(x + 3) + i√7/2]

=> f(x) = a(x² – 1)[(x + 3)² – (i√7)/2)²]

=> f(x) = a(x² – 1)[(x² + 6x + 9) – (i²7)/4)]

=> f(x) = a(x² – 1)[(x² + 6x + 9) + 7/4] [since i² = –1]

=> f(x) = a(x² – 1)[(x² + 6x + 43/4]

Since the polynomial has to have integer coefficients, it may be re–written as:

f(x) = a/4(x² – 1)[(4x² + 24x + 43]

Here, the leading coefficient is a/4x⁴ and the constant term is –43a/4.

Since, the leading coefficient is given to be 3, a = 12.

This implies that the constant would be –43 (12)/4 = –129.

16. Solution. The vector u = –√3i + 1j, lies in quadrant II, and with the magnitude of a vector, denoted as |u|, is its length:

|u| = √((–√3)² + 1²) = 2. The direction of a vector u = ai + bj ~ <a, b> is the angle θ it makes with the positive x–axis. In general,

a = |u| cos θ and b = |u| sin θ so that tan θ = b/a => θ = tan^–1(b/a)

Here, θ = tan^–1(–1/√) ~ 150° ~ 5π/6 radians.

[Based on the point (a, b) for u = ai + bj ~ <a, b>, identify the correct quadrant for the vector and "adjust" the angle θ, since θ = tan^–1(b/a) gives the reference angle ie. the smallest angle the vector makes with the x–axis. The angle needed is the "total" angle that the vector makes with the positive x–axis!]

17. Solution. If n represents the number of $25 price increases, the Revenue function, R(n) = (4000 – 20n)(750 + 25n)

= 500(6000 + 170n – n²)

= –500n² + 85000n + 3000000

which is maximized for n = –b/2a

= –85000/(2 ∙ –500) = 85

So, the manufacturer should set the price at ($750 + $25∙85) = $2125 at which price $6612500 shall be made by selling 2300 computers.

18. Solution. The numbers divisible by 5 or 7 must include those divisible by 5 and by 7, but must exclude those divisible by 35 since these numbers would’ve been counted twice!

Now, if a is the 1^st term of an Arithmetic Series divisible by c, and b is the last term of the series divisible by c, then the number of terms between a and b (inclusive) divisible by c is: n = (b −a)/c + 1.

The 1^st number divisible by 7 between 50 and 500 is 56 [obtained by 50/7 ≈ 7, and then, adding 7 to 7∙7 = 49 since the integers lie between 50 and 500].

The last number divisible by 7 between 50 and 500 is 497 [obtained by 500/7 ≈ 71 and then, multiplying 71 by 7].

n₇ = (497 – 56)/7 + 1 = 64;

Similarly, n₅ = (500 – 50)/5 + 1 = 91 and n₃₅ = (490 – 70)/35 + 1 = 13.

So, the number of numbers between 50 and 500 [inclusive] divisible by 5 or 7 is: n = 91 + 64 – 13 = 142.

19. Solution. Since A = P(1 + r/n)^nt where

A = Amount,

P = Principal,

t = time (in # of years) and

n = number of times the Principal is compounded per year, here:

6800 = 4500·(1+ 0.065/12)^12t

so that (1 + 0.065/12)^12t = 68/45

Since (1 + 0.065/12) ≈ 1.005, and 68/45 ≈ 1.5:

=> 1.005^12t = 1.5

Writing in log form:

log _1.005 1.5 = 12t

=> t = 1/12·(log _1.005 1.5)

Changing the base to 10:

t = 1/12·[(log 1.5) / (log 1.005)

20. Solution. Clearly, examining the side of the square, side length of the octagon,
B = A – 2C      –––––––––––––––––––––––––––––––– (1)

Also, since off each corner of the square, an isosceles right triangle [≈ 45°–45°–90° triangle] is essentially being lopped off so that using Pythagoras' Theorem:
C² + C² = B²
or 2C² = B² –––––––––––––––––––––––––––––––– (2)
Substituting (1) in (2): 2C ² = (A – 2C)²
=> 2C² = A² – 4AC + 4C²
Rearranging and keeping in mind that A is a constant that would be given and C is the unknown: 2C² – 4AC + A² = 0
Solve the quadratic in C to get:
C = A ± A√(2)/2

which may be substituted in (1), if necessary to find B, the side of the octagon.

21. Solution. The magnitude of a vector v = ai + bj is |v| = √(a² + b²) and direction is given by tanθ = b/a. Since a = –√2/2i – √2/2j and b = –6a, therefore, b = 3√2i + 3√2j.

|b| = √[(3√2)² + (3√2)²] = 6 and tan θ = 1 => θ = 45° ~ π/4.

22. Solution. If (3, 2) lies on f^–1(x), then (2, 3) lies of f(x). Substitute to solve for A.

=> f(2) = 3

=> f(2) = 2(2)³ + 2 + A

= 18 + A = 3

So, A = –15.

23. Solution. Since 7 + g^–1(x – 1) = 9,

=> g^–1(x – 1) = 2

=> g(2) = x – 1

Since g(2) = 6, x = 7 [from above].

24. Solution. f(g(x)) = [(x – 3)/(x – 4) – 1] / [(x – 3)/(x – 4) – 2]

= 1/(–x + 5) [upon simplification]

For domain of f(g(x)), we find the domain of g(x): x ≠ 4, and that of f(g(x)): x ≠ 5. Therefore, the “merged” domain is D: [–∞, 4), (4, 5), (5, ∞).

25. Solution. For 2 vectors a and b with magnitudes, |a| and |b| acting on on a point, the resultant vector |R| given by the law of cosines:

R² = a² + b² – 2ab cos(180 – θ). Substitute for a = 3, b = 4 and cos(180 – 60) = cos 120 = –0.5, we get R = √37.

26. Solution. Since 726 = 275(2)  + 176 ...(1)
and 275 = 176(1) + 99...(2)
and 176 = 99(1) + 77...(3)

and 99 = 77(1) + 22...(4)
and 77 = 22(3) + 11...(5)
and 22 = 11(2)...(6)
therefore 11 is the required GCD.

27. Solution. For f(g^–1(4)), let g^–1(4) = x => g(x) = 4.

From the g(x) table: x = –2. Then, f(g^–1(4)) = f(–2) = –3.

Similarly, for g(f^–1(4)) let f^–1(4) = y so that f(y) = 4 => y = 8.

Then g(f^–1(4)) = g(8) = –4.

28. Solution. To find the points of intersection of f(x) and g(x), setting f(x) = g(x):

x⁷  5x⁶ + 3x⁵ + 4x⁴  2x³ + 8x² + x  6 = x⁴ + 2x³  7x² +5x  11
=> x⁷  5x⁶ + 3x⁵ + 3x⁴  4x³ + 15x²  4x + 5 = 0.

This function could have a maximum of 7 roots [implying 7 points of intersection].

29. Solution. Since, for the x–intercept, y = 0:

x² + 4x + 4 = 0

=> (x + 2)² = 0

=> x = –2

Likewise, for the y–intercepts, x = 0:

y² – 4y + 4 = 0

=> (y – 2)² = 0

=> y = –2

The equation of the line through (–2, 0) and (0, –2) is: y = x + 2.

30. Solution. If i is root of g(x), then –i is also a root.

Then g(i) = 0 and g(–i) = 0.

Substitute and solve the 2 equations simultaneously, keeping in mind, that:

i² = –1,

i³ = –i,

and i⁴ = 1.

Answer: a = –2, b = –3

FR 1. Solution. Let the number, N, be represented as

N ~ a_n­­a_n–1a_n–2…a₃a₂a₁

where the a_i represents the actual digits of N so that a_n­­a_n–1a_n–2…a₃a₂a₁ is actually a string of digits constituting N [like N = 7498, say].

Then the value of N is:

N = 10^n–1a_n + 10^n–2 a_n–1 + 10^n–3a_n–2+ . . . 10²­a₃ + 10a₂ + a₁ (1)

[just as 7498 = 7X1000 + 4X100 + 9X10 + 8]

Since the sum of the digits of N, a_n + _­­a_n–1 + a_n–2 + . . . a₃ + a₂ + a₁ is divisible by 3, that makes the sum a multiple of 3. That is,

a_n + _­­a_n–1 + a_n–2 + . . . a₃ + a₂ + a₁ = 3k, where k is an arbitrary number. (2)

=> a₁ = 3k − a_{n ­­}− a_n–1 − a_n–2 . . . − a₃ − a₂ (3)

Substituting (3) in (1) and simplifying:

N = (10^n–1 − 1)a_n + (10^n–2 − 1)a_n–1 +. . . 99^­a₃ + 9a₂ − 3k (4)

Now, 10^n–1 − 1 has (n − 1)9s, 10^n–2 − 1 has (n–2) 9s and so on, so that each term in (4) is divisible by 3.

Therefore, N itself must be divisible by 3!

FR 2. Solution. Let x be # of mattress A, and y be # of mattress B produced.

Objective Function, Profit, P = 35x + 20y Step 1

Filling constraint: 2x + 3y < 108

Finishing constraint: x + 0.5y < 20 Step 2

Also, x > 0 and y > 0

Plot the constraint functions from Step 2.

Obtain the corner points of the convex polygon to be (0, 0), (0, 36), (20, 0) and (3, 34) by solving the relevant pairs of Constraints simultaneously.

TIP! To plot the Constraint functions, choose the x–scale and y–scale wisely according to the following Rule of Thumb:

X–Scale  (XMax – XMin)/10 and

Y–Scale  (YMax – YMin)/10 where

XMax corresponds to the highest value of all X–Intercepts and

XMin corresponds to the lowest value of all X–Intercepts.

Likewise, for Ymax and Ymain.

Here, for example, XMax = 54, XMin = 0 = X–Scale  5 (or 6). Likewise, YMax = 40, Ymin = 0 and Yscale  4 (or 5)!

Evaluating the Profit at each of these points:

Vertices P = 35x + 20y

(0,0) 0

(0, 36) 720

(20, 0) 700

(3, 34) 785 => Maximum

Mattress A: 3 and Mattress B: 34 are the optimum values that need to be produced.

FR 3. Solution. Let r be the radius of the semicircle, and x be the height of the rectangle.

Then, Perimeter of the semicircle, P₁ = πr (1)

Perimeter of rectangular section, P₂ = 2x + 2r [we do not count the top of the rectangle!] (2)

The total perimeter, P = P₁ + P₂ = πr + 2x + 2r so that x = ½ (P – 2r – πr) (3)

The total Area of the window, A =A _semicircle +A_rectangle (4)

Now, A_semicircle = πr²/2 and A_rectangle = 2rx (5)

Substituting (5) in (4), A = πr²/2 +2rx (6)

Using (3) in (6): A = πr²/2 +2r.½(P – 2r – πr) [substituting x from (3)]

= –r² [½ (4+ π)] + Pr [upon combining like terms and taking LCMs]

which is a quadratic in r and is maximum when r = –b/2a = P/(4 + π) (7)

Substituting (7) into (3) and simplifying carefully after taking LCMs:

x = P/(4 + π) (8)

Substituting (7) and (8) in (6) and simplifying carefully: A = P²

2(4 + π)

FR 4. Solution. Given: log 2 ≈ 0.3 and log 3 ≈ 0.48.

Construct an x–y table, choosing the x–values cleverly using log properties!

For x = 3, y = 1/3·(log 1) −1 = 0 – 1 = −1

x = 4, y = 1/3·(log 2) − 1= 1/3· (0.3) −1 ≈ –0.9

x = 5, y = 1/3·(log 3) −1= 1/3· (0.48) −1 ≈ –0.84

x = 6, y = 1/3·(log 4) −1 = 1/3·(2 log 2) −1 ≈ –0.8

x = 10, y =1/3·(log 8) −1 = 1/3·(3 log 2) −1 ≈ –0.7

Other points one may choose are:

x = 8 [Use log 6 = log 2 + log 3]

x = 11 [Use log 9 = 2log 3]

x = 12 [Use log 10 = 1]

x =14 [Use log 12 = log (4·3) = log (2² ·3) = 2log 2 + log 3]

x = 18 [Use log 16 = log (2⁴) = 4log 2]

SKETCH!

The domain of the general log function, log f(x), is: all x such that f(x) > 0. Here, D: x > 2. The range is, R: all real numbers.