Question

1. the complex number z = -3 + 4i. its modulus |z| is

a. 1
b. 5
c. √7
d. √25

1. let p(x) = 2x³ - 5x² + 4x - 3. the remainder when p(x) is divided by x - 2 is

a. -1
b. 0
c. 1
d. 3

1. let f(x) = x³ - 4x² + ax + 6. if x = 2 is a zero of f(x), then a =

a. -1
b. 0
c. 1
d. 2

1. for the rational function f(x) = (x + 1)/(x - 3), the vertical asymptote is

a. x = -1
b. x = 0
c. x = 1
d. x = 3

1. for f(x) = (x² - 1)/(x² - 4), the domain is

a. all real numbers
b. all real numbers except x = 1 and x = -1
c. all real numbers except x = 2 and x = -2
d. all real numbers except x = -2, -1, 1, 2

Explanation:

Question 8

Step1: Recall modulus formula for complex numbers

For a complex number \( z = a + bi \), the modulus \( |z| \) is given by \( \sqrt{a^2 + b^2} \). Here, \( a=-3 \) and \( b = 4 \).

Step2: Substitute values into the formula

\( |z|=\sqrt{(-3)^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5 \)

Step1: Recall Remainder Theorem

The Remainder Theorem states that if a polynomial \( P(x) \) is divided by \( x - c \), the remainder is \( P(c) \). Here, \( c = 2 \) and \( P(x)=2x^3-5x^2 + 4x-3 \).

Step2: Evaluate \( P(2) \)

\( P(2)=2(2)^3-5(2)^2+4(2)-3=2(8)-5(4)+8 - 3=16-20 + 8-3=1 \)

Step1: Recall the definition of a zero of a function

If \( x = c \) is a zero of \( f(x) \), then \( f(c)=0 \). Here, \( c = 2 \) and \( f(x)=x^3-4x^2+ax + 6 \).

Step2: Substitute \( x = 2 \) into \( f(x) \) and solve for \( a \)

\( f(2)=(2)^3-4(2)^2+a(2)+6=8-16 + 2a+6=2a - 2 \). Set \( f(2)=0 \), so \( 2a-2=0\Rightarrow 2a=2\Rightarrow a = 1 \). Wait, no, wait: \( 8-16+2a + 6= (8 + 6-16)+2a=(-2)+2a \). So \( -2 + 2a=0\Rightarrow 2a=2\Rightarrow a = 1 \)? Wait, no, let's recalculate: \( 2^3=8 \), \( -4\times2^2=-4\times4=-16 \), \( a\times2 = 2a \), \( +6 \). So \( 8-16+2a + 6= (8 + 6-16)+2a=(-2)+2a \). Set to zero: \( -2+2a=0\Rightarrow 2a=2\Rightarrow a = 1 \). But wait, the options have A. -1, B. 0, C. 1, D. 2. So C? Wait, no, maybe I made a mistake. Wait, \( f(2)=0 \): \( 8-16 + 2a+6=0\Rightarrow(8 + 6-16)+2a=0\Rightarrow(-2)+2a=0\Rightarrow 2a=2\Rightarrow a = 1 \). So the answer is C? Wait, but let's check again. \( x = 2 \) is a zero, so \( f(2)=0 \). \( 2^3=8 \), \( -4*(2)^2=-16 \), \( a*2=2a \), \( +6 \). So \( 8-16+2a + 6= (8+6) - 16+2a=14-16 + 2a=-2 + 2a \). So \( -2 + 2a=0\Rightarrow a = 1 \). So answer is C.
Wait, but the original problem's option A is -1, maybe I miscalculated. Wait, \( f(x)=x^3-4x^2+ax + 6 \). If \( x = 2 \) is a zero, then \( 2^3-4(2)^2+a2 + 6=0 \). \( 8-16 + 2a+6=0 \). \( (8 + 6-16)+2a=0 \). \( (-2)+2a=0 \). \( 2a=2 \). \( a = 1 \). So answer is C.

Answer:

B. 5

Question 9