PSAT Math : PSAT Mathematics
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Example Questions
Example Question #921 : Psat Mathematics
Factor the following:
Since you have an addition for the last element, you know that you will have either two subtractions or two additions in your factored groups. Since the second term is positive, your factored groups will contain two additions:
Now, all you have to do are go through the factors of 30 and find the pair that adds up to 11:
Therefore, your answer will be:
Example Question #1 : How To Find F(X)
If f(x) = x2 – 5 for all values x and f(a) = 4, what is one possible value of a?
3
14
1
8
11
3
Using the defined function, f(a) will produce the same result when substituted for x:
f(a) = a2 – 5
Setting this equal to 4, you can solve for a:
a2 – 5 = 4
a2 = 9
a = –3 or 3
Example Question #2 : How To Find F(X)
If the function g is defined by g(x) = 4x + 5, then 2g(x) – 3 =
8x + 7
8x + 2
6x + 7
4x + 2
6x + 2
8x + 7
The function g(x) is equal to 4x + 5, and the notation 2g(x) asks us to multiply the entire function by 2. 2(4x + 5) = 8x + 10. We then subtract 3, the second part of the new equation, to get 8x + 7.
Example Question #2 : How To Find F(X)
If f(x) = x2 + 5x and g(x) = 2, what is f(g(4))?
4
14
36
39
2
14
First you must find what g(4) is. The definition of g(x) tells you that the function is always equal to 2, regardless of what “x” is. Plugging 2 into f(x), we get 22 + 5(2) = 14.
Example Question #3 : How To Find F(X)
f(a) = 1/3(a3 + 5a – 15)
Find a = 3.
9
1
27
19
3
9
Substitute 3 for all a.
(1/3) * (33 + 5(3) – 15)
(1/3) * (27 + 15 – 15)
(1/3) * (27) = 9
Example Question #42 : Algebraic Functions
Evaluate f(g(6)) given that f(x) = x2 – 6 and g(x) = –(1/2)x – 5
–8
30
58
50
–25
58
Begin by solving g(6) first.
g(6) = –(1/2)(6) – 5
g(6) = –3 – 5
g(6) = –8
We substitute f(–8)
f(–8) = (–8)2 – 6
f(–8) = 64 – 6
f(–8) = 58
Example Question #43 : Algebraic Functions
If f(x) = |(x2 – 175)|, what is the value of f(–10) ?
275
75
–275
–75
15
75
If x = –10, then (x2 – 175) = 100 – 175 = –75. But the sign |x| means the absolute value of x. Absolute values are always positive.
|–75| = 75
Example Question #3 : How To Find F(X)
If f(x)= 2x² + 5x – 3, then what is f(–2)?
–21
–1
–5
7
–5
By plugging in –2 for x and evaluating, the answer becomes 8 – 10 – 3 = -5.
Example Question #5 : How To Find F(X)
If f(x) = x² – 2 and g(x) = 3x + 5, what is f(g(x))?
3x² – 1
9x² + 23
9x² + 30x + 25
9x² + 30x + 23
9x² + 30x + 23
To find f(g(x) plug the equation for g(x) into equation f(x) in place of “x” so that you have: f(g(x)) = (3x + 5)² – 2.
Simplify: (3x + 5)(3x + 5) – 2
Use FOIL: 9x² + 30x + 25 – 2 = 9x² + 30x + 23
Example Question #2 : How To Find F(X)
f(x) = 2x2 + x – 3 and g(y) = 2y – 7. What is f(g(4))?
57
33
0
42
-33
0
To evaluate f(g(4)), one must first determine the value of g(4), then plug that into f(x).
g(4) = 2 x 4 – 7 = 1.
f(1) = 2 x 12 + 2 x 1 – 3 = 0.
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