Subtest I Final Review Study Guide

The above Octagon–based problem has been spotted by certain chaps and I hasten to equip you with its resolution.

Problem. There's a regular octagon with side length B to be constructed from a square of side A by cutting off C units from both ends of each of its 4 sides. The figure above illustrates the problem very ably, only on the test one may expect C ≈ x, say, and A to be provided as a number. The object is to find C, and then, B.

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 + 4C²Rearranging and keeping in mind that A is a constant that would be given and C is the unknown:

2C² – 4AC + A² = 0

You may 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.

Practice: Try A = 4, say, and start from the beginning to find B and C independently!

Update 2: Rational Roots Theorem with Fractional Roots

In the past, it has always been my impression – since such had been my experience! – that if a Q were asked to Find all Rational Roots of f(x) , that while performing Synthetic Division for the possible Rational Roots "generated" by the Rational Roots Theorem, one may "safely" assume that the first root would be a reasonably small whole number [±1, ±2, ±3, ±4, etc.] even though there might be smaller fractional roots for f(x).

In other words, one might be "spared" of the inconvenience of trying fractional roots of f(x) on the Test.

Well, I am mistaken.

Problem. Candidates have just provided feedback that in recent tests, in the Free Response section, one of the Qs has indeed had Fractional [p/ q] and Irrational [a ±√b] Roots only.

Solution. This is merely to call your attention to the fact that you might receive such a problem, too – not that it is assured!

To be forewarned is to be fore–armed, as they say.

So essentially, work as usual, only keeping in mind that if you exhaust all integers as possible Rational Roots – and ruled out the possibility of "errors" in your work! – then do proceed to perform Synthetic Division with the fractional roots, starting with the smaller ones, naturally.

I've been informed that the Synthetic Division with the possible fractional roots yield "nice" integers in the Rows...

I do have Qs in my Subtest I Q Bank that have only Fractional and Irrational Roots – for instance, SET 37 #8 – but for immediate practice, try one or more (!) of the following Qs very similar to those you might encounter:

Practice 1: Find all Rational Roots of f(x) = 3x³ – x² –15x + 5.
Practice 2 : Find all Rational Roots of f(x) = 4x⁴ + 7x² – 2
Practice 3: Find all Rational Roots of f(x) = 3x³ + 4x² – 7x + 2
Practice 4 : Find all Rational Roots of f(x) = 4 x⁵ – 8x⁴ – x + 2

Update 3: Absolute Value Function Graphs

Absolute–Value Function graphs of the kind y = a|x – h| + k are found in textbooks [a V–shaped graph with the Vertex at (h, k), and the slope and orientation (up / down?) provided by a]. For “work”, make a table of values, starting at the Vertex, and choosing a few ones to the left and right of x = h. Complicated–seeming (?!) ones with the absolute value being attached to both y and x have been seen recently. The “logic” remains the same, though.

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

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

Update 4: Elementary algebraic simplification Qs have been seen, of the type:

What does the expression (x³/y³ – 1)(x/y – 1)^–1/(x²/y² – 1) simplify to?

If you believe you have discovered a mistake, FIRST, consult this site! If it fails to address your doubt, THEN, email me!

I shall respond with alacrity [oh, it just means "quickly"!]

Note: In a massive document – of 200+ pages – it is inevitable that typos and other errors creep in. I appreciate the exasperation felt when answers / solutions don't match, and I apologize profusely for the same.

Still, I rely on the "cooperation" [read: prompt feedback!] and "compassion" [read: forgiveness!] on my patrons to extirpate these as swiftly as I can!

Tip! Do focus on the "larger" concept addressed by the problem, and do not agonize excessively with the answer.

Stress process over product: in life – as well as on the CSET and in a Math classroom, in general – the journey is as, if not more, important than the destination!

Remember: it is indeed embarrassing to me when these mistakes are pointed out – I've proofread the content a couple of times but clearly, I am not infallible!

Yet, a sense of perspective is important: I've organized these documents single–handedly whereas textbook authors have assistance in typing and proof–reading. And still, as you shall acknowledge, even formal texts [especially, first editions!] are replete with errors [they are in fact often recognized in the acknowledgments as long lists of names!].

Be that as it may, I actively solicit your input so that future editions may be spared!

Here are useful CSET links to good resources. Please examine each link carefully for nuggets of wisdom!

SECOND, study the NOTES in front of the Subtest I Q Bank thoroughly!

FINALLY, use the following study–guide Tips and Suggestions for your preparation!

Note: This alone does NOT purport to absolutely sufficient for Subtest I preparation! Naturally, I'd hope that you'd have got extensive practice through working through most of the Qs in my Q Bank! But as a "general review " the set of Qs outlined below is simply peerless!

1. Examine this link once again to go typical Free Response Q topics, this for the Top 8 Subtest I Topics you ought to have mastered.

A. Find:
a) the equation of the parabola passing through (0,5), (2,5) and (-2,21).
b) the equation of the line with x-intercept 5/8 making a triangular area of 25/16 in the 4th Quadrant.
c) at what point(s) does the line intersect the parabola?

Solution. Let the equation be y = ax² + bx + c. Substitute the given points and solve for a, b and c.

For 5 b), sketch a figure as described and denote the y–intercept as (0, k) say. Find k by using the information about the area of the triangle, A = ½ bh ≈ ½ (x–intercept)∙(y–intercept); then, find the equation of the line.

B. If 1, 1/2 and -3 are x-intercepts of a polynomial function f(x) = ax³ + bx² + cx + 3, find f(x) and graph it.
Solution. Since 1, 1/2 and –3 are the x–intercepts, (1, 0), (1/2, 0) and (–3, 0) lie on the polynomial f(x). Plug in to solve the 3 equations in 3 unknowns a, b and c simultaneously. Alternately, and more quickly, y = p (x – 1)(x – ½)(x+3) is the general form of the cubic polynomial with the given roots / x–intercepts. Since the constant term is 3, p∙(–1)∙(– ½)∙(3) = 3 => p = 6. So the polynomial is readily: y = 6(x – 1)(x – ½)(x +3) which can be easily graphed.

C. If f(x) = x² - 7x + k, find ALL value(s) of k such that f(x) has
a) No x-intercepts
b) Exactly 1 x-intercept
c) 2 x-intercepts

Solution. For #3, f(x) = x² – 7x + k to have NO, ONE and TWO x–intercepts respectively, the discriminant, D = b² – 4ac for x² – 7x + k should be <, = and > ZERO, respectively. Set the discriminant to each and find the value(s) of k!

D. On the same graph, plot the following functions:
a) f(x) = x² - 2x - 8
b) g(x) = r(f(x))
c) h(x) = 1/f(x)
Solution. a) and c) are straightforward [graphing a parabola and a simple rational expression, respectively]. For b), g(x) = √x² – 2x – 8 = √(x – 4) (x + 2). Use this to determine the domain of g(x) ie. the values for which g(x) is defined (which would be x < –4 or x > 2). Then, simply construct an x–y table of, say 6 values, for x values satisfying the Domain. Make approximations to simplify for y – since a calculator is not permitted! – and sketch.

E. Graph the rational function: f(x) = (x² + 2x - 15)/x(x² - 7x + 12) stating its
a) Domain
b) x- and y-intercepts
c) Asymptotes
Solution. This is a straightforward graphing of a Rational Function!

3. Go over the Algebra Competencies Checklist – sent as part of the Subtest I Question Bank (at the very front). Make repeated "sweeps" to identify your weak areas. Get a book and remedy them, if possible!

From the Q Bank go over the following Free Response Qs another time (remember: some of these Q can appear as MCQs!). Also, you do NOT have to actually do every Q below: simply make sure you know how to do each! (You might simply start off a Q and abandon it once you're certain you can finish it!!)

a) Proof of Quadratic Formula: you must know how to derive the roots of the Quadratic Equation, ax² + bx + c = 0 (by Completing the Square!), and use the result to prove that the sum of the roots is –b/a and the product of the roots is c/a. The latter 2 results have been supplied in my Solutions document [SET 1 #1 a, b]. A variation of this Q is SET 8 #1.

Problem. If x₁ and x₂ are the roots of the quadratic equation ax² + bx + c = 0, then it can be shown that the Sum of the Roots, x₁+ x₂ = –b/a and the Products of the Roots, x_1•x₂= c/a. Solve this system of equations to find x₁and x₂. Then find the difference of the roots, x₁ – x₂. [x₁, x₂ = {–b ± (√b² – 4ac)}/2a; x₁ – x₂ = (√b² – 4ac)/2a]

b) Elementary skills pertaining to quadratic functions are routinely tested. The relationship between the Discriminant, D = b² – 4ac of a quadratic equation [> 0, = 0 and < 0] and the roots / x–intercepts of a quadratic function (ie. Parabola) [2, 1 and no real roots, respectively] is very important.

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

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

c) Writing the equation of a parabola in Standard [y = ax² + bx + c], Vertex [y = a(x – h)² + k] and Intercept [y = a(x – x₁)(x – x₂ )] forms is a basic skill, including graphing parabolas and labeling the Axis of Symmetry, Vertex, the intercepts and stating the Domain and Range. Set 5 #8, SET 8 2 &3, SET 10 #11 & 12, SET 11 #20, SET 12 #14–16, SET 13 #7, SET 15 #11–18, SET 16 #1–5 test these skills.

Remember that it is the vertex of a parabola [in particular, the y–coordinate] that determines the Range of a Quadratic Function!

d) Quadratic Modeling questions – that is, word problems that involve quadratic functions – SET 8 #8 & 9, SET 9 #6, SET 10 #8–10, SET 11 #15, SET 13 #7, SET 16 #6, SET 19, #1, SET 22 #2–5, SET 27 #3, SET 35 #3, SET 36 #1–4, SET 38 #6 & 7 appear as Free Response Qs (less so as MCQs). All of them follow the same format: finding the dimensions of a figure for a given objective (maximize / minimize a quadratic function, usually, Area) or subject to certain conditions (for a given Area, for instance).

Problem. 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.

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

· Use the given info to write an equation (invent your own variables if necessary, only keep their creation to a minimum: do NOT create a, y and z when x and y shall suffice!)

· Use the variables to write an expression that the Q desires (Area or Perimeter usually)

· Since the expression shall invariably involve 2 variables, use the equation (in the previous step) to make a substitution (ie. solving for 1 variable in step 2 and substituting this in step 3 so that the expression contains 1 variable ONLY!)

· If done right (?!), the expression (now in 1 variable alone!) should be a quadratic (ie. a parabola) in that variable.

· To maximize or minimize the expression, the optima occurs at the 'vertex': –b/2a.

· Simplify carefully, and substitute back into the equation to determine the dimensions of the figure.

!! Especially try Qs involving rectangle inscribed in a semicircle / circle; finding dimensions of a window to allow maximum light; and swimming–pool walkways.!!

e) Miscellaneous word problems calling for quadratic functions are SET 17 #2 &7, SET 27 #1 & 2, SET 34 #3

f) An MCQ of the following type is routinely observed: Find the absolute value of the difference of the roots of ax² + bx + c = 0, which is simply |x₁ – x₂| where x₁ and x₂ are the 2 roots obtained from the Quadratic Formula! A variation is: Find the absolute value of the difference of the Real [or Imaginary] parts of roots of ax² + bx + c = 0 where a, b and / or c is an imaginary number as in ix² + (2 + i)x + 1 = 0, say.

5. Usually, 1 MCQ based on Direct and Inverse Variation is in the following form: If p varies jointly as q and the square–root of r, and inversely as the square of t, how shall p change if q is doubled, r is increased 4 times and t halved? Or how would q change if r is reduced by one–half, t is doubled and p remains unchanged? For practice try: SET 1 #2, SET 33 #10, SET 34 #2, SET 35 #7 and SET 41 #2.

Problem. The maximum safe load for a horizontal beam varies jointly 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?

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

6. At least 1 Q on Linear Programming shall be encountered either in Free Response style – like SET 2 #1–3, SET 4 #3 & 4, SET 16 #10 & 11 and SET 39 #2 – or in the MCQ section – examine the 1^st few Qs in the MCQ section of the Q Bank [#1–15, 28–31] especially ones requiring the identification of the inequalities constituting the shaded region! At times, one is only required to set up the set of constraints from a Word problem as in the Qs #371–375 of the MCQ part.

7. About 2–3 Qs are based on Inverse & Composite Functions, for which it is vital to know how to find the Inverse of a function; Graphing inverse functions; Finding the Domain and Range of an Inverse Function; Calculating Composite Functions and their Domains. Study the Notes for this topic! Typical Qs are like those outlined in MCQs # 83, 90–103, 175–179, 210, 217, 228 and 229 and Free Response Qs in SET 1 #3 & 4 and SET 32 #17. Another style of Qs are those in SET 16#7 and SET 30 #2–7.

a) If f(x) and g(x) are inverse functions, then the y = x line is the perpendicular bisector of the segment joining any point (a, b) on f(x) and the corresponding point (b, a) on g(x).

b) If f(x) and g(x) are inverse functions that intersect [they may not!], then the point of intersection lies on the y = x line so that if (p, q) is where f(x) and g(x) intersect, (obviously!) p = q!

Solution. To find the inverse of y = mx + b, switch x and y and solve for y.

Since a function and its inverse – if they intersect – do so on the y = x line, the point of intersection should be: (b/(1 – m), b/(1 – m)).

Clever applications of Composite Functions [which may, otherwise, be resolved using common–sense!] have been recently observed.

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

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

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

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.

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

8. One Free Response Q is assured in the topic of Mathematical Induction like those in SET 3 #2–#10, #14–16, especially those involving fractions / rational expressions on the left side [#15 & #16]. A variation is in SET 6 #19 & 20. [Ones like #11 – #14 are less likely to be seen!].

At times, a simple MCQ requiring you to "predict" a subsequent step of an Induction situation is presented, for which it is vital to grasp the 3 Steps underlying all Induction Qs.

9. In SET 12 #2–5 are Number Theory style Qs that appear almost invariably as Free Response Qs. In lieu of this, a Q based on the Euclidean Algorithm is all but assured.

Problem: Use Euclid's Algorithm to find the Greatest Common Divisor (GCD) / Factor (GCF), d , of 2 #, a and b. Specifically, use Euclid's Algorithm to find the Greatest Common Divisor (GCD) / Factor (GCF), d, of 2 #, 758 and 242.
Solution. Here, a = 758 and b = 242
Then, 758 = 242(3) + 32

[Dividing the larger by the smaller and noting the Quotient & Remainder]

And 242 = 32(7) + 18

[Dividing the Divisor of the previous step by the Remainder of the previous step, and noting the new Quotient & Remainder]

And 32 = 18(1) + 14

[Dividing the Divisor of the previous step by the Remainder of the previous step, and noting the new Quotient & Remainder]

And 18 = 14(1) + 4

[Dividing the Divisor of the previous step by the Remainder of the previous step, and noting the new Quotient & Remainder]

And 14 = 4(3) + 2

[Dividing the Divisor of the previous step by the Remainder of the previous step, and noting the new Quotient & Remainder]

And 4 = 2(2) + 0

Therefore 2 is the required GCD.

Problem: Use Euclid's Algorithm to express the GCD, d, of 2 numbers a and b in the form: d = ma + nb. Specifically, find the GCD, d, for 726 and 275, and express it in the form d = m. 726 + n.275
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.

Now, from (5):

11 = 77 – 22(3)

= 77 – [99 – 77(1)](3)
[substituting for 22 using (4): 22 = 99 – 77(1)]

= 77(4) – 99(3)
[Using Distributive Law for the 3 and simplifying by factoring out 77]

= [176 – 99(1)](4) – 99(3)
[substituting for 77 using (3): 77 = 176 – 99(1)]

= 176(4) – 99(7)
[Using Distributive Law for the 4 and simplifying by factoring out 99]

= 176(4) – [275 – 176(1)](7)
[substituting for 99 using (2): 99 = 275 – 176(1)]

= 176(11) – 275(7)
[Using Distributive Law for the 7 and simplifying by factoring out 176]

= [726 – 275(2)](11) – 275(7)
[substituting for 176 using (1): 176 = 726 – 275(2)]

= 726(11) + 275(–29)

Therefore, 11 = 726(11) + 275(–29)

Problem:
1. Show that the GCD of 237 and 81 is 3 and 3 = 237(13) + 81(–38)
2. Show that the GCD of 616 and 427is 7 and 7 = 616(–9) + 427(13)
3. Show that the GCD of 936 and 666 is 18 and 18 = 936(5) + 666(–7)

Otherwise, I strongly recommend procuring the book Schaum's Outlines ABSTRACT ALGEBRA (2^nd Edition) by Frank Ayres & Lloyd Jaisingh and studying pages pages 58 – 61, especially: Section 5.3 on Greatest Common Divisor (GCD) and The Division Algorithm. Example 6 and Theorem II on P 60 and Example 7 on P 61 outline Euclid's Algorithm which is used to (i) find the GCD of 2 numbers, a and b [denoted as d = (a, b)] and (ii) express the GCD, d, of 2 numbers, a and b as a linear combination of 2 other numbers, M and N ie. d = Ma + Nb.

Other key Number Theory skills are finding the Greatest Common Factor (GCF ~ GCD) and the Least Common Multiple (LCM) of a set of 4–5 numbers. You are sure to see at least one Q from this topic. This site provides good examples of GCF using the principle of Prime Factorization, and this for LCM. This pdf document has good examples, too. Finally, this website [purplemath] is an old reliable!

It's infernally useful to know that the product of 2 numbers is equal to the products of their LCM and GCF.

By the way, 2 numbers are Relative Primes [or Coprimes] if their GCF is 1, as in 3 and 5. This term has been observed in recent MCQs.

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

10. With regard to Arithmetic & Geometric Series, usually only 1 Q is asked. This takes the form represented in SET 8, #6 & 7, SET 19 #3–11 and SET 20 #1–7. Additionally, SET 1 # 5–7, SET 21 #1, 5, 6 and 11, SET 38 #4 are word–problems discussing applications.

11. Binomial Theorem Qs are usually relatively straightforward like those in SET 10 #1–7. I strongly suggest getting familiar with Pascal's Triangle and observing the pattern therein! Remember: Row n represents the expansion of (a + b)^n–1since the top–most row is 1, which is the coefficient of (a + b)⁰! For Pascal's Triangle, attempt these Qs:
a) find the 2 terms for which the coefficient is 10 in the 6th row of Pascal's Triangle for the expansion: (3x – 4/y)
b) find the terms for which the coefficient is 20 in Pascal's Triangle for the expansion: (2 x² – 3/x)

12. Complex Numbers and their manipulation (especially, multiplication, division, reciprocals) along with their graphical representation [Real component – x axis; Imaginary Component – y axis] are key concepts! SET 2 #4–9 and SET 4 #6–8 and MCQs 112–119 offer practice for this topic.

13. With regard to Matrices at least a couple of Qs – even 3 or 4 at times! – are asked. Translating a word problem into a system of equations, and then representing them in matrix form, AX = B, is a fundamental skill; solving the system of equations by finding the Inverse first, or directly by performing elementary row transformations on the augmented matrix, [A:B] is to be mastered. SET 5 #1 &2, SET 13 #5, SET 14 #13–16, SET 20 #15, SET 25 #15, SET 36 #10, SET 39 #4–7 and MCQs #4, 5, 15–27, 339–343, assess basic skills!

At times, a system in the form AkB = C is given, with the object being to determine matrix, k, of a suitable order. Once the order is determined [using the Rule of Matrix Multiplication], proceed to denote the elements [ie. the actual numbers!] of matrix, k, suitably as a, b, c, d, etc. Then perform the Multiplication operation: AkB and set the result to match C.

Alternately, depending on whether 1 or both of A and B is a Square matrix whose Inverse may be found: pre–multiply both sides by A^–1 first to get kB = A^–1 C, and then post–multiply both sides by B^–1: k = A^–1 CB^–1. Again, this relies on A and / or B being Square Matrices so that A^–1 and / or B^–1 may be computed!

Two "proofs" of extraordinary importance are SET 26 #2 and SET 33 #12. Master them, please! Finally, it is critical to know when a system of equations has no solution, exactly one or infinitely many solutions: while performing elementary row transformations, if any row consists entirely of zeros in an intermediate / final step (that for the coefficient matrix as well as the constant matrix [0 0...0 : 0], then the system has infinite solutions; if, however, for any row the coefficient matrix consists of zeros but the constant term is non–zero [0 0...0 : 2, say] then the system has no solution.

14. A Reflection Q of the type in SET 5 #4–6 has been observed in the past. If a point, P(a, b) is reflected about the line y = x, then the reflected point is Q(b, a).One approach is: let the reflected point be R(m, n). We shall prove that m = b and n = a. Since y = x is the line of symmetry, it is the perpendicular bisector of segment PR. If S is the point of intersection of PR and line y = x on the line, then S may be denoted as S(c, c) [since it lies on the y = x line!]. Since S is the mid–point of PR , using the mid–point formula: c = ½ (a + m) and c = ½ (b + n). Setting them equal and simplifying: a + m = b + n (*).

Also, since PR and y = x are perpendicular, slope of PR = (n – b)/(m – a) = –1 [since slope of y = x is 1!]. Simplifying this: n – b = a – m (**)

The Q above employ this very useful nugget: in general, that if a point P (x, y) lies on a function, f(x) [this could be a line, a parabola, etc.] then it can be represented as P(x, f(x)). This concept is also applied in Qs like SET 13 #3 & 4, SET 18 #1, 3, 6, 7 and 9. For example, P(x, √(x + 1)) lies on the y = √(x + 1) graph so that only one variable need be manipulated!

15. In the topic of Exponential and Logarithmic Functions, one must recognize that they are Inverse Functions of each other; and possess the capacity to convert a function expressed in one form into the other [Exponential –> Logarithmic and Logarithmic –> Exponential]. Knowledge of Properties of both "operators" is essential: the Exponent and Log "rules".

Problem. 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.

Observe that by replacing y with x and vice versa, this is equivalent to finding the inverse of f(x) = A/(B + C•e ^–Kx): f^–1(x) = y = –1/K·ln [(A – Bx)/Cx].

While MCQs are usually of the type suggested in SET 31 #8–12, word problems like SET 26 #10–12, SET 29 #9 and SET 30 #16 also assess similar skills. On several occasions, a Free Response Q has called for graphing ability as in SET 25 #11–14, SET 30 #8–15 and SET 31 #1–7. You must be familiar with the domain, range and general shape of exponential and log graphs to make rough but reasonably accurate sketches. In the MCQ section have a dekko at #61–76 and #203–223.

Problem. How long shall it take for an Amount, invested at 4%, to double, if compounded continuously?

Problem: How long shall it take for an Amount, invested at 2%, to triple, if compounded quarterly? For this, one needs to be familiar with the Change–of–Base formula: log_ba = log₁₀ a / log₁₀ b.
Solution. A_t
=3A₀ so that

16. One of the most vital topic areas is that of Polynomial Functions. One ought to be able to graph them [using x– and y–intercepts; End Behaviour; and Intervals for which y = f(x) is > 0 and < 0 via a Sign–Table] as in SET 6 #1–18, SET 37 #3–6.

Finding the roots of Polynomials like SET 7 # 1–8 and SET 37 #7–9 is a fundamental requirement. SET 21 #2, 7–10 offers general practice. A couple of MCQs emanate from this concept and very often 1 Free Response! For this, it is critical to know the Remainder, Factor and Rational Roots Theorems [it's a very good idea to know their proofs!!] and Synthetic Division. MCQs #77–82, 232–237, 240–255, 261–268, 277–296, 299–308 test skills based on this significant topic!

Applications of the Complex Conjugates Theorem are invariably tested: if a + b i is a root of a polynomial, f(x), then, so is a – bi. [In other words, complex roots always occur as conjugates: if 2 – 7i and 4i are roots of f(x), then so are 2 +7i and –4i!] . A variation is the Irrational Conjugates Theorem: if a + √b is a root of a polynomial, f(x), then, so is a – √b . [In other words, irrational roots always occur as conjugates: if 2 – √7 and √3 are roots of f(x), then so are 2 + √7 and –√3! ]. These 2 Theorems enable one to construct the polynomial "backwards".

Problem: Find the 4th degree polynomial f(x) if 2 +7 i and √3 are 2 of its roots and f( –1) = –8.

Solution. This requires one to employ the Difference of Squares factoring [ (a + b)(a – b) = a² – b²] very cleverly:

f(x) = p [x – (2 +7i)][x – ( 2 –7i)] [x – ( √3)][x – (– √3)] where p is the unknown leading constant.

Problem: Find the 6th degree polynomial f(x) if 3 – 4i, 1 – √3 and 2i are 3 of its roots with the y–intercept being 14.

Problem: Graph the parabola y = ax² + bx + c if (4 – 3i) is one x–intercept and the y–intercept is –50.

Remember that the following terms convey similar meanings:
Solution ≈ Root ≈ Zero ≈ x–intercept!

As a general concept note, remember that to find the Point of Intersection of 2 functions, y ₁ = f(x) and y₂ = g(x) simply solve them simultaneously: y₁ = y₂=>f(x) = g(x) = > f(x) – g(x) = 0 which is a polynomial function whose roots may be calculated using the Rational Roots Theorem! For example, find the point of intersection of y = 10x⁴ – 2x³ – 15x² + 3x + 6 and y = x³+ 14x² – 2x – 6 is equivalent to finding the Rational Roots of the function, y = 10x⁴ – 3x³ – 29x² + 5x + 12.

Profound observation: when one is finding the roots of a polynomial, f(x), one is implicitly finding the points of intersection of f(x) and y = 0 [er, so that to find the roots, we set f(x) = 0!].

A recent MCQ on the CSET Subtest I has the following flavour.

Problem: what is the highest number of the points of intersection between g(x) = ax⁵ + bx⁴ + cx³ + dx² + ex + f and h(x) = px³ + qx²+ rx + s.
Solution. Five. [Because the highest degree of both polynomials is 5!]

To clarify the issue, a polynomial of degree n can intersect a polynomial of degree m (n > m) in at most n points.

For instance, a parabola [degree 2] may intersect a line in at most 2 points. Similarly, 2 parabolas may intersect at no more than 2 points!

Likewise, a cubic polynomial [degree 3] may intersect a line [degree 1] / parabola [degree 2]/ another cubic [degree 3] in at most 3 points.

Examining the situation from another perspective: a polynomial of lower degree intersects another (of degree, n, say) one more time than the maximum # of times the latter changes direction; um, since "the maximum # of times the a polynomial of degree n changes direction" would be (n – 1), one more would be the degree of the latter, which is n! ["Playing around" with various graphs should illuminate the matter!]

Even more intuitively – and algebraically – if a 4th degree polynomial intersects with a 7th degree polynomial, then the points of intersection would be obtained by setting the 2 equations equal and solving for the roots of the resulting equation [after transposing all the terms to 1 side].

But the resulting equation would be of 7th degree! Therefore, there can be at most 7 points of intersection...

Problem. If f(x) = x⁷– 5x⁶ + 3x⁵ + 4x⁴ – 2x³ + 8x² + x – 6 and g(x) = x⁴+ 2x³– 7x² +5x – 11, then, 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 which clearly has 7 roots [implying 7 points of intersection].

At the other extreme, one might have two 7–th degree polynomials all of whose terms except the constant terms cancel out when they're set equal! In this situation, the 2 functions shall not intersect at all.

Problem. If p(x) = x⁷– 5x⁶ + 3x⁵ + 4x⁴ – 2x³ + 8x² + x – 6
and
q(x) = x⁷– 5x⁶ + 3x⁵ + 4x⁴ – 2x³ + 8x² + x + 1

then, to find the points of intersection of p(x) and q(x), setting p(x) = q(x): –6 = 1 implying no points of intersection!

And for those that've grasped the concept of transformations, isn't q(x) essentially p(x) shifted up 7 units [examine the constants!]. Um, then why would they intersect?! One would lie right on top of the other!

Interesting in way, but then again, not, if you paused to contemplate the underlying algebra / graph correspondence!

Likewise, to determine Intervals for which the graph of a function y₁ = f(x) lies above that of y₂ = g(x), simply solve the inequality y₁ > y₂=>f(x) > g(x) => f(x) – g(x) > 0, which is an inequality that can be solved by factoring the Left Hand Side, finding its roots and then using Test Values to determine the requisite intervals for which f(x) – g(x) > 0.

For instance, find the intervals for which y = 4x³ – 4x² + 7x + 2 lies below y = 3x³ + 2x² – 4x + 8 may be resolved by dealing with x³ – 6x² + 11x – 6 < 0. Likewise, find the intervals for which y = 1/(x + 2) lies above y = 3/(x + 1) can be solved by working equivalently on (–2x – 5)/(x + 2)( x + 1) > 0. The ability, in general, to solve Inequalities is a vital skill! See 18 below for Qs on this topic.

17. Of comparable importance to dealing with Polynomials is the ability to graph Rational Functions: SET 7 #9–17, 238, 239 explore this! SET 29 #10–15 test skills of graphing Irrational Functions [it helps to determine the domain first, and know the "standard" or "base" graph of these functions!].

One part of a Free Response question has dwelt on graphing functions of the form y = √(x – a)(x – b) or y = 1/(x – a)(x – b)(x – c) .

While the latter is a straightforward rational function [find the x– and y–intercepts, Vertical and Horizontal Asymptotes and graph!], for the former, determine the domain, first; then make a table and cleverly choose values for x. Keep in mind that a) the range for y = √(x – a)(x – b) is y> 0, and b) the graph shall have a right "component" [> a / b depending on their sign and value] and a left "component" [< a / b depending on their sign and value] with a "stretch" in the middle empty. To confirm this, graph: y = √(x – 2)(x – 5) , y = √(x + 3)(x – 6) & y = √(x + 1)(x – 4) by hand as well as on a graphing calculator!

Problem. 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.

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.

18. One MCQ is often devoted to finding the Domain of [Inverse, Composite, Rational, Exponential / Logarithmic or Irrational] functions, involving the resolution of Inequalities. SET 11 #1–14, SET 15 #19–22, SET 33 #4–6, SET 34 #4–11 offer practice for this. In the MCQ section, Qs #83, 84, 97, 98,104–109, 228, 229, 269–276, 352, 353 focus on this.

19. A MCQ on manipulation of Straight Lines or use of Distance Formula is almost de rigueur! MCQs like #153–160, 180–182, 185–198 and SET 23 #4–6 are about this.

Problem. At what point do the perpendicular bisectors of the line segments joining A(0, 0) and B(–8, –10) and C(3, –7) and D(–3, 1) intersect?

Solution. The equation of the perpendicular bisector of segment AB is: y = –4x/5 – 41/5.

Solving the above 2 equations simultaneously to determine their point of intersection: (–104/31, –171/ 31).

20. With regards to Conic Sections, of late, the emphasis has diminished dramatically as it has been incorporated into Subtest II [topic: Locus and Coordinate Geometry]. For Subtest I, let the emphasis be on Parabolas! Usually, at least 1 MCQ and 1 FR Q is devoted to this. SET 24, SET 25 # 8–10, SET 27 #12–15, SET 28, SET 32 #10–13 examine this topic. If time permits, you ought to be able to derive parabolas, ellipses and hyperbolas according to their definitions as in SET 28 #1–4, SET 32 #10–13 and SET 33 #1–3: Qs like these have appeared in the past in the Free Response section. MCQs like #161–168, 315–338 (especially #334!) test basic skills. Do not ignore the concept of asymptotes of hyperbolas: Qs on this appear as MCQs. Again, work on Ellipses and Hyperbolas only if you can spare the time!

21. About 3–4 Qs MCQs are routinely devoted to Vectors. You must know the Triangle Law and Parallelogram Law of Vectors [to find the Resultant Vector]; and must be able to find the Horizontal and Vertical components of a vector if the Magnitude and Direction is given; and find the Magnitude and Direction if the Horizontal and Vertical components of the vector are given; determine the Dot–Product and Cross–Product of 2 vectors and apply them; compute the Angle between 2 vectors; and using this, determine if 2 vectors are parallel (angle is 0 or 180) or orthogonal (ie. perpendicular); calculate the 'Unit' vector, and use it to generate a vector of a specified magnitude as another in the same direction.

MCQs #37–60 assesses these concepts. And the last few Qs of almost every SET is devoted to this extremely important subject area. If you do not have study material for Cross–Product of vectors, consult a book on Calculus: concepts are discussed quite lucidly and extensively over there!

For vectors, it is vital that elementary right–triangle trigonometry be mastered. You must know:

* how to convert radian measure to degrees and vice versa: Angles are often represented in radians rather than degrees in Algebra II and beyond, and the conversion is π rad= 180°. So, simply use proportions to convert radians into degrees and vice versa.

* how to evaluate basic trigonometric functions for different degrees/radians using the notion of reference angle (which is the smallest angle the ray makes with the x–axis).

the sin, cos and tan of 0°, 30°, 45°, 60° and 90°. Here's a nifty trick to construct a table of sin, cos and tan of 0°, 30°, 45°, 60° and 90° in under a minute!

1. Make a table with 0°, 30°, 45°, 60° and 90° on top in a "row".

2. Write sin, cos and tan to the left – in a "column" – one beneath the other.

3. Fill in the sin "row" – underneath 0°, 30°, 45°, 60° and 90° respectively – with the numbers 0/4, 1/4, 2/4, 3/4 and 4/4.

4. Take Square Roots (√) and simplify:
√0/4 = 0, √1/4 = ½, √(2/4) = √2/2, √(3/4) = √3/2 and √4/4 = 1 i.e. the results are: 0, 1/2, √(2)/2, √(3)/2 and 1 which are the values for the sin of 0°, 30°, 45°, 60° and 90°, respectively!

5. For cos: write the sin row backwards! [because sin and cos are complementary / cofuntions: cos(x) = sin(90–x).]

6. Since tan x = sin x/cos x, simply find the ratio (mentally!) of the sin and cos of the angle desired!
* the definitions only of the 3 trigonometric ratios – sin, cos and tan of an angle – and the basic Pythagorean Identity: sin² A + cos² A = 1
* Use Inverse Trigonometric Functions: ie. working backwards, for instance tan x = –1 in the 2nd quadrant, so x = ?, here's how to work them:
You see, if you can construct the sin–cos–tan "table of values" – as described above! – you just need to a) find the reference angle, and b) consult the table and work backwards! [You are not going to encounter messy angles like 20° or 145° or 219° degrees! The ones encountered shall all be "reducible" to 0°, 30°, 45°, 60° and 90°...]

For instance if cos x = – √(3) / 2, and you know that the vector – from the given sketch or some other given info – is in, say, the 2nd quadrant, then, consulting the Trigonometric Table of Values constructed, cos (30°) = √(3)/2 => the angle makes a 30° angle with the x–axis in the 2nd quadrant => Required Angle = 180° – 30° = 150°. VOILA!

Likewise, if tan x = √(3), and if it is known that the vector – from the given sketch or some other given info – is in, say, the 3rd quadrant, then, consulting the Trigonometric Table of Values, tan (60°) = √(3) => The angle is 60° with the x–axis in the 3rd quadrant => Required Angle = 180° + 60° = 240°. VOILA!

Problem. Three forces act upon an object simultaneously with following magnitudes and direction: 1^st: 70lbs at an angle of –30, 2^nd: 40lbs at an angle of 45, and 3^rd: 60lbs at an angle of 135. Find the magnitude and direction of the resultant force, R. [Express your answer in terms of √2 and √3.]

Solution. Resolving the 3 forces into their horizontal and vertical components:

22. Study the NOTES on Abstract Algebra! You've received it as part of the Q Bank or shall get it in a moment (?!). The 1st 3 MCQs shall be devoted to this!