Math2c Problems

[parikhs]

<p>Please explain in detail how you got the problems you answered. Cramming for Math2c :(.</p>

<p>If f: (x,y) -> (x + 2y, y) for every pair (x,y) in the plane, for what points
(x,y) is it true that (x,y) -> (x,y)?</p>

<p>(A) set of points (x,y) such that x=0
(B) set of points (x,y) such that y=0
(C) set of points (x,y) such that y=1
(D) (0,0) only
(E) (-1,1) only</p>

<p>If f(x) = ax^2 + bx + c for all real numbers x and if f(0) = 1 and f(1) = 2, then a+b = </p>

<p>(A) -2
(B) -1
(C) 0
(D) 1
(E) 2</p>

<p>If f(2x+1) = 2x-1 for all real numbers x, then f(x) = </p>

<p>(A) -x+1
(B) x-1
(C) x-2
(D) 2x-1
(E) (1/2)x - 1</p>

<p>If 3x-4y+7 = 0 and 2y-x^2=0 for x >=0, then x=?</p>

<p>(A) 1.27
(B) 2.07
(C) 2.77
(D) 4.15
(E) 5.53</p>

<p>If f(x) = log2X for x > 0, then f^-1(x) = </p>

<p>(A) 2^x
(B) x^2
(C) x/2
(D) 2/x
(E) logx2</p>

[parikhs]

<p>bump-------------</p>

[jnh]

<p>Can I know where you find these problems?
Are you sure the first one has no error?</p>

<p>No. 2 - D
f(0) = a(0)^2 + b(0) + c = c = 1
f(1) = a(1)^2 + b(1) + c = a+ b + c = (a + b) + c = (a + b) + 1 = 2</p>

<p>No. 3 - C
Method 1: Find the correct answer among answers given
f(x) = x-2 is the answer since
f(2x+1) = (2x+1)-2 = 2x-1
Method 2:
f(2x+1) = 2x-1 = a(2x+1)+b
a = 1 and b = -2
so f(x) = 1x-2 = x-2</p>

<p>No. 4 - C
2y = x^2
4y = 2x^2
3x - 2x^2 + 7 = 0
2x^2 -3x - 7 = 0
Use a calculator or formula: x = {-(-3) + [(-3)^2 - 4(2)(-7)]^1/2} / 2 = 2.77 (since x >= 0)</p>

<p>No 5 - A
x = 2^y
y = 2^x
f^-1(x) = 2^x</p>

[parikhs]

<p>Thanks a lot man! For the last problem...whenever I see inverse f(x) do I just flip flop whatever I have?</p>