GRE_Math_Bible_eBook

Functions 481

16.At time t = 0, a projectile was fired upward from an initial height of 10 feet. Its height after t seconds is given by the function h(t) = p −10(q − t)2 , where p and q are positive constants. If the projectile

reached a maximum height of 100 feet when t = 3, then what was the height, in feet, of the projectile when t = 4 ?

(A)62

(B)70

(C)85

(D)89

(E)90

y

the graph at points A and B and the area of the square is 16, what is the value of a ?

(A)2

(B)4

(C)6

(D)8

(E)10

Very Hard

18.If the function f(x) is defined for all real numbers x as the maximum value of 2x + 4 and 12 + 3x, then for which one of the following values of x will f(x) actually equal 2x + 4 ?

(A)–4

(B)–5

(C)–6

(D)–7

(E)–9

482GRE Math Bible

Answers and Solutions to Problem Set BB

Medium

1. We are given the function rules f(x, y) = 2x + y and g(x, y) = x + 2y. Swapping arguments in g yields g(y, x) = y + 2x = f(x, y).

Hence, f(3, 4) = g(4, 3). The answer is (E).

2. Choice A: The point A represents x = 1. Now, f(1) = (1 – 1.5)(1 – 2.5)(1 – 3.5)(1 – 4.5) = (–0.5)(–1.5)(–2.5)(–3.5) = product of four (an even number of) negative numbers. The result is positive. Reject.

Choice B: The point B represents x = 2. Now, f(2) = (2 – 1.5)(2 – 2.5)(2 – 3.5)(2 – 4.5) = (0.5)(–0.5)(–1.5)(–2.5) = product of a positive number and three (an odd number of) negative numbers. The result is negative. Hence, correct.

Choice C: The point C represents x = 3. Now, f(3) = (3 – 1.5)(3 – 2.5)(3 – 3.5)(3 – 4.5) = (1.5)(0.5)(–0.5)(–1.5) = Product of two positive numbers and two (an even number of) negative numbers. The result is positive. Reject.

Choice D: The point D represents x = 4.5. Now, f(4.5) = (4.5 – 1.5)(4.5 – 2.5)(4.5 – 3.5)(4.5 – 4.5) = 3 × 2 × 1 × 0 = 0, not a negative number. Reject.

Choice E: The point E represents x = 5.5. Now, f(5.5) = (5.5 – 1.5)(5.5 – 2.5)(5.5 – 3.5)(5.5 – 4.5) = 4 × 3 × 2 × 1 = product of positive numbers. The result is positive. Reject.

The answer is (B).

3. By the given rule, f(x) = (x – 1)(x – 2)(x – 3)(x – 4). Hence, Column A equals

f(2.5) = (2.5 – 1)(2.5 – 2)(2.5 – 3)(2.5 – 4) = (1.5)(0.5)(–0.5)(–1.5)

The product of two positive numbers (1.5 and 0.5) and two negative numbers (–0.5 and –1.5) is positive.

Column B equals

f(3.5) = (3.5 – 1)(3.5 – 2)(3.5 – 3)(3.5 – 4) = (2.5)(1.5)(0.5)(–0.5)

The product of three positive numbers (2.5, 1.5, and 0.5) and a negative number (–0.5) is negative.

Since a positive number is greater than a negative number, Column A > Column B. The answer is (A).

4. f(3, 4) = the average of 3 and 4 = 3+24 = 72 = 3.5.

g(3, 4) = the greater number of 3 and 4, which is 4.

Hence, f(3, 4) + g(3, 4) = 3.5 + 4 = 7.5.

The answer is (D).

5. We are given the function rules f(x, y) = 2x + y andg(x, y) = x + 2y. Swapping the arguments of g yields g(y, x) = y + 2x = f(x, y). Hence, we have that f(x, y) = g(y, x). Hence, f(3, 4) = g(4, 3), and f(4, 3) = g(3, 4).

Subtracting the two equations yields f(3, 4) – f(4, 3) = g(4, 3) – g(3, 4).

Adding f(4, 3) and g(3, 4) to both sides yields

f(3, 4) + g(3, 4) = f(4, 3) + g(4, 3)

Hence, Column A equals Column B, and the answer is (C).

Functions 483

Column A = f (1) × f (2) = 12 × 23 = 13 .

Column B = f (2) × f (3) = 23 × 43 = 42 = 12 .

Since 1/2 is greater than 1/3, Column B is greater than Column A. The answer is (B).

7.Let’s change the fractional notation to radical notation: g(x) = 42x − 3 +1 . Since we have an even root, the expression under the radical must be greater than or equal to zero. Hence, 2x – 3 ≥ 0. Adding 3 to both sides of this inequality yields 2x ≥ 3. Dividing both sides by 2 yields x ≥ 3/2. The answer is (C).

8.We need to plug the x table values into each given function to find the one that returns the function values in the bottom row of the table. Let’s start with x = 0 since zero is the easiest number to calculate

with. According to the table f(0) = 3. This eliminates Choice (A) since f (0) = −2(0)2 = −2(0) = 0; and it eliminates Choice (D) since f (0) = −2(0 )2 − 3 = −2 0 − 3 = 0 − 3 = −3 . Now, choose x = 1. The next

easiest number to calculate with. According to the table f(1) = 1. This eliminates Choice (B) since

f (1) = 12 + 3 = 1+ 3 = 4; and it eliminates Choice (C) since f (1) = −(1)2 + 3 = −1 + 3 = 2 . Hence, by process of elimination, the answer is (E).

9. The graph has a height of 2 for every value of x between 2 and 3; it also has a height of 2 at about x = –2. The only number offered in this interval is 2.5. This is illustrated by the dot and the thick line in the following graph:

The answer is (D).

15. Function f is defined to be the geometric mean of its arguments, and the geometric mean of a and b is ab . Hence, Column A equals f (a,b) = ab . Function g with arguments a and b is defined to be the

LCM of a and b, and we are given that a and b are different prime numbers. Hence, the least common multiple is ab. For example, the LCM of 5 and 7 is 35 (= 5 7). Thus, Column B equals g(a, b) = ab.

The square root of a number (here ab) greater than 1 (since a and b are primes, their product is greater than 1) is always less than the number itself. Hence, ab < ab and therefore Column A < Column B. The answer is (B).

16. Method I:

Recall that when a quadratic function is written in the form y = a( x − h)2 + k , its vertex (in this case, the

maximum height of the projectile) occurs at the point (h, k). So let’s rewrite the function

h(t) = p −10(q − t)2 in the form h(t) = a(t − h)2 + k . Notice that we changed y to h(t) and x to t.

h(t) = p −10(q − t)2

=−10(q − t)2 + p

=−10(−[−q + t])2 + p

=−10(−1)2 (t − q)2 + p

=−10(+1)(t − q)2 + p

=−10(t − q)2 + p

In this form, we can see that the vertex (maximum) occurs at the point (q, p). We are given that the maximum height of 100 occurs when t is 3. Hence, q = 3 and p = 100. Plugging this into our function yields

h(t) = −10(t − q)2 + p = −10(t − 3)2 + 100

We are asked to find the height of the projectile when t = 4. Evaluating our function at 4 yields h(4) = −10(4 − 3)2 + 100

=−10(1)2 +100

=−10 1+ 100

=−10 +100

=90

The answer is (E).

486 GRE Math Bible

Method II:

In this method, we are going to solve a system of two equations in two unknowns in order to determine the values of p and q in the function h(t) = p −10(q − t)2 . At time t = 0, the projectile had a height of 10 feet. In other words, h(0) = 10. At time t = 3, the projectile had a height of 100 feet. In other words, h(3) = 100. Plugging this information into the function h(t) = p −10(q − t)2 yields

Now, we solve this system of equations by subtracting the bottom equation from the top equation:

10 = p − 10q2

(−) 100 = p −10(q − 3)2 −90 = −10q2 + 10(q − 3)2

Solving this equation for q yields

−90 = −10q2 + 10(q − 3)2 −90 = −10q2 + 10(q 2 − 6q + 9)

−90 = −10q2 + 10q2 − 60q + 90

–90 = –60q + 90 –180 = –60q 3 = q

Plugging q = 3 into the equation 10 = p − 10q2 yields

10 = p − 10 32

10 = p − 10 9

10 = p – 90

100 = p

Hence, the function h(t) = p −10(q − t)2 becomes h(t) = 100 − 10(3 − t)2 . We are asked to find the height of the projectile when t = 4. Evaluating this function at 4 yields

h(4) = 100 − 10(3− 4)2

=100 − 10(−1)2

=100 − 10 1

=100 − 10

=90

The answer is (E).

490 GRE Math Bible

8.John was born on February 28, 1999. It was a Sunday. February of that year had only 28 days, and the year had exactly 365 days. His brother Jack was born on the same day of the year 2000. February of the year 2000 had 29 days. On which day was Jack born?

(A)Monday

(B)Tuesday

(C)Friday

(D)Saturday

(E)Sunday