SAT Math Level 2. Question. HELP!!

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Hi, EVERYONE!!

GREATEST INTEGER FUNCTION [x] Questions.
Please! someone guide me how to solve these type of questions? The solution on the book requires a graphic calculator.
how can you solve these question without a graphic calculator? PLEASE HELP!!!

Q1) If f(x)=j, where j is an integer such that j ≤x< j+1, and g(x)= f(x) - | f(x) |, what is the maximum value of g(x)?
Answer choices are { -1,0,1,2,j } and correct answer is 0. How?

Q2) If |x| is defined to represent the greatest integer less than or equal to x, and f(x) = | x - [x] - (1/2) |, what is period of f(x)?
Answer choices are { 1/2,1,2,4,f is not a period function } and correct answer is 1. How?

TRIGONOMETRY Questions

Q1) What is the smallest positive x-intercept of the graph of y = 3sin2( x+(2π/3) )?
Correct answer is 1.05. What I don’t understand is that the books say to put ( x+(2π/3) ) = π. But the function is sin2x, so ( x+(2π/3) ) = π/2. NO? can someone please tell me what I am doing wrong here?

Q2) On the interval [-π/4 , π/4 ], the function f(x) = ( √1+sin^2 (x) ) has maximum value of?
Correct answer is 1.2. How do you find the maximum value of function sin square(x)? Secondly what if sin^2 (x) was replaced by sin^2 (2x) and question ask you about the period of the function. Please help me understand this.

INEQUALITIES

Q1) Which of the following is the solution set for x(x-3)(x+2) > 0?
Correct answer is -2 <x< 0="" or="" x="">3. How? </x<>

Q2) How many positive integers are there in the solution set x/(x-2) > 5?
Correct Answer is 0. Why? I simplified the equation into x< 5/2. And answered 2 integers because 1,2 are two positive integers less than 2.5. Right? Please tell what I am doing wrong here.

Any tip related to these topic or SAT MATH would be greatly appreciated. :smiley:

Thankyou.

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@BaderC

[x] Q1: This question seems really awkwardly worded. Is j a constant? Is j ≤ x < j+1 for all x?

Assuming f(x) = ⌊x⌋ (floor value of x), then f(x) is necessarily an integer and f(x) - |f(x)| is either negative or zero (depending on whether ⌊x⌋ is negative or not) and so its maximum value is 0.

Q2: You can solve this without a graphing calculator by noting that x - ⌊x⌋ represents the fractional part of x, and when graphed on the xy-plane, represents a “sawtooth” wave. Subtracting 1/2 simply translates the sawtooth wave down by 1/2, and taking its absolute value converts f(x) into a different periodic function whose period is 1.

Trig Q1: Assuming you meant y = 3 sin 2(x + (2π/3)), note that you can find an x-intercept by setting x + 2π/3 = π/2. But the question is asking for the smallest positive x-intercept. Solving x + 2π/3 = π/2 gives x = -π/6.

Q2: sin^2(x) has a maximum value of 1 over reals.

However sin^2(x) never reaches 1 on the interval [-π/4, π/4]. So you will have to find the maximum a different way.
Hint sin x is odd, so sin^2(x) is even (i.e. sin^2 (x) = sin^2(-x) so you only need to consider [0, π/4]. On what interval is f(x) increasing/decreasing? This can be answered without a calculator.

Inequalities Q1: Note that x(x-3)(x+2) = 0 at x = -2, 0, 3. Now determine which intervals x(x-3)(x+2) > 0 or < 0.
You can do this by picking a point in each of the intervals (-∞, -2), (-2, 0), (0, 3), (3, ∞) and testing whether the expression is positive or negative. Because the expression can’t evaluate to 0, and because it is a continuous function, you can determine the interval.

Q2: Your error came when you multiplied both sides of the inequality by x-2, which could be positive or negative.
You can simply check that 1,2 don’t work by plugging them back in.

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THANKS @MITer94 for the reply mate.

GREATEST INTEGER FUNCTION.
Q2) But how did you find the period of the function? could you please explain this? please!

TRIG.
Q1) I got that we get x= -π/6 which is -ve obvious. But isn’t the question asking for smallest positive x-intercept, not the value of x. Right? As the f(x) =3sin2A [let (x + (2π/3)) =A], having period of “π”. Shouldn’t we put 2A = π [which yields A=π/2 ], to get first positive x-intercept? because graph sin(x) graph sin(x) cuts graph at x=π

INEQUALITIES
I am very bad at Inequalities. could you please elaborate the method a little? How did you choose the points and how do you test these points? what does -ve or +ve expression tells ?

BTW thanks again. very informative. :smiley:

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@BaderC
Q2: First graph the function using the method I described earlier (by graphing y = x - ⌊x⌋, then shifting down 1/2 unit, then taking its absolute value). The period of the function is the smallest number t for which f(x) = f(x+t) for all x, or for which the function “repeats”. You simply need to look at the graph of the function to find its period.

An x-intercept is simply a value of x such that f(x) = 0.

The reason why we can’t set x + 2π/3 = π/2 is because we obtain x = -π/6, which is obviously negative. x = -π/6 is indeed an x-intercept, but it’s not the one we’re looking for. So setting x + 2π/3 to the next smallest value, π, gives us the correct answer x = π/3 (which is about 1.05).

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Many algebra textbooks will have an entire section dedicated to this.

Basically what you’re doing is, you want to find the entire set of numbers x for which P(x) > 0 (or < 0, depending on the problem) where P(x) is some polynomial (in this case, P(x) = x(x-3)(x+2)).

The first step is to find the real zeroes of P(x). In this problem it’s easy because the polynomial is factored, so the zeroes are -2, 0, and 3. Equivalently, P(x) ≠ 0 for all other real x.

To determine the actual solution set, the observation is that, between any two consecutive zeroes (by ordering, e.g. -2,0 or 0,3), P(x) must be either strictly positive or strictly negative. This has to do with the fact that P(x) is a continuous function, and if P(x) weren’t strictly positive/negative then there would be an additional zeroes between the two zeroes. It might help visually to plot the real zeroes on a number line.

For example, knowing whether P(-1) is positive or negative tells us about the sign of P(x) between -2 and 0. Since P(-1) = -1(-4)(1) = 4 > 0, then -2 < x < 0 is part of the solution set.

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THANKS @MITer94 that was really helpful.:smiley: GREAT EXPLANATIONS

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Hi, @MITer94 please help me to understand this problem.

If $5,000 is invested at a rate of 5.8% compounded
daily, how much will the investment be worth in
3 years?
(A) 5,117
(B) 5,597
© 5,921
(D) 5,943
(E) 5,950

Answer is E. How? aren’t the answer choice incorrect because 5.8 % of 5000 =290. The amount added to original increase every day. so it must be in 10 thousands. right?
Explanation on the book provides a formula. Can someone provide a logic based answer? please help me understand this types of question :smiley:

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@BaderC are you sure it’s not a yearly interest rate?

Compounded daily, the total amount after 3 years is (assuming no other transactions)

5000(1 + (0.058/365))^(3653) = 5950.19…