Page 22 - Pharmacy College Admission Test Review Book

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5 5 5 6. (b) 3x - 10 = 5x -5 + 2 therefore,
5
( ) (2x) + ( ) (2x) 5−1 ( ) + ( ) (2x) 5−2 2

0 1 2
5 3x - 5x = 10 - 5 + 2

+ ( ) (2x) 5−3 3
3 - 2x = 7
5 7
+ ( ) (2x) 5−4 4 x = −

4 2
5

+ ( ) (2x) 5−5 5 7. (d) Subtract 4 from all parts of the inequality,
5
and then divide everything by 3 to find out the
5
2
5
4
3
= 1(32 ) + (16 ) + 5∗ 4 (8 ) + value of x.
1 1∗2
5∗4∗3 5∗4∗3∗2
3
5
4
2
(4 ) + (2 ) + 1
1∗2∗3 1∗2∗3∗4 -10  3x + 4 < 20
-10 - 4  3x + 4 -4 < 20 - 4
3 2
5
2 3
4
= 32x + 80x y + 80x y + 40x y + -14  3x < 16
5
4
10x y + y -14  x < 16
3 3
4. (d) 9.
Therefore, the answer in interval form should be (-

3. ( ) = 7. ( ), therefore: 14/3, 16/3).
3 2

! ! 8. (c) [1,  or (-, -9/5]
3 = 7
3! ( − 3)! 2! ( − 2)!
5x + 2  7 or 5x + 2  -7
3 2! ( − 3)!
= 5x  5 or 5x  -9
7 3! ( − 2)! x  5 or x  -9
5 5
1 1
= x  1 or x  -9
7 ( − 2) 5
9. (b) In order to solve this problem, first we need
( − 2) = 7 to find out the value of (f-g)(x).

( ) = 9 (f-g)(x) = f(x) -g(x) = x - 5x -6 -x -3
2
= x -6x -9
2
5. (c) (f-g)(-1) = (-1) -6(-1) - 9
2

n = 1 + 6 - 9 = -2
nPr = 24 ( )
r
10. (b) To solve this problem, we have to substitute
n! n! x for g(x) in f(x).
= 24
(n − r)! r! (n − r)!
2
f(g (x)) = f (x - 1) = (x-1) -2(x - 1) + 12
r! = 24
2
= x -2x + 1 -2x + 2 + 12
2
r! = 1 x 2 x 3 x 4 = 4! = x -4x + 15

r = 4 11. (d) To find out the inverse function, we should
follow the steps below:

1. Rewrite the function using y instead of f(x).
2. Switch the x and y variables.
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