Solutions to Practice Midterm 2

Solution:

(a) Probability that both marbles are red:

• First draw: P(red) = 5/10 = 1/2
• Second draw: P(red) = 4/9 (since one red marble is already removed)
• P(both red) = (1/2) (4/9) = 4/18 = 2/9

(b) Probability that one marble is red and one is blue:

• We need to consider two cases: red first then blue, or blue first then red
• Case 1: P(red first) = 5/10 = 1/2, then P(blue) = 3/9 = 1/3
• Case 2: P(blue first) = 3/10, then P(red) = 5/9
• P(one red and one blue) = (1/2) (1/3) + (3/10) (5/9) = 1/6 + 15/90 = 1/6 + 1/6 = 2/6 = 1/3

(c) Probability that at least one marble is green:

• It's easier to calculate P(no green marbles) and subtract from 1
• P(no green marbles) = P(both are red or blue)
• Total non-green marbles = 5 + 3 = 8
• P(first not green) = 8/10 = 4/5
• P(second not green) = 7/9
• P(no green marbles) = (4/5) (7/9) = 28/45
• P(at least one green) = 1 - 28/45 = 17/45

Solution:

(a) Finding f(2):

• Substitute x = 2 into the function: f(2) = 3(2) - 2(2) + 1
• Calculate: f(2) = 3(4) - 4 + 1 = 12 - 4 + 1 = 9

(b) Finding x when f(x) = 7:

• Set f(x) = 7: 3x - 2x + 1 = 7
• Rearrange: 3x - 2x - 6 = 0
• Using the quadratic formula: x = [2 (4 - 4(3)(-6))]/6
• Calculate: x = [2 76]/6 = [2 219]/6 = [1 19]/3
• Therefore, x = (1 + 19)/3 or x = (1 - 19)/3

(c) Finding the vertex of the parabola:

• The x-coordinate of the vertex is given by x = -b/2a
• For f(x) = 3x - 2x + 1, a = 3 and b = -2
• x = -(-2)/(23) = 2/6 = 1/3
• Find f(1/3) = 3(1/3) - 2(1/3) + 1 = 3(1/9) - 2/3 + 1 = 1/3 - 2/3 + 1 = 2/3
• The vertex is at (1/3, 2/3)

(d) Finding the domain and range:

• For a polynomial function, the domain is all real numbers: (-, )
• Since the coefficient of x is positive (3), the parabola opens upward
• The vertex represents the minimum point of the function
• The y-coordinate of the vertex (2/3) is the minimum value of y
• Therefore, the range is [2/3, )

Solution:

• Let x be the length of the side parallel to the wall
• Let y be the length of each side perpendicular to the wall
• Total fencing: x + 2y = 100 (equation 1)
• Area: A = xy (equation 2)
• From equation 1: x = 100 - 2y
• Substitute into equation 2: A = (100 - 2y)y = 100y - 2y
• To maximize A, find dA/dy and set it to zero: dA/dy = 100 - 4y = 0
• Solve: 4y = 100, so y = 25
• Find x: x = 100 - 2(25) = 50
• Maximum area: A = 50 25 = 1250 square feet

The dimensions that maximize the area are 50 feet parallel to the wall and 25 feet perpendicular to the wall.

Verification:

• To confirm this is a maximum, check the second derivative: dA/dy = -4
• Since the second derivative is negative everywhere, y = 25 gives a maximum

Solution:

Starting from the left side:

• tan(x) + 1 = (sin(x)/cos(x)) + 1
• = (sin(x)/cos(x)) + (cos(x)/cos(x))
• = (sin(x) + cos(x))/cos(x)
• = 1/cos(x) (using the Pythagorean identity sin(x) + cos(x) = 1)
• = sec(x) (since sec(x) = 1/cos(x))

Therefore, tan(x) + 1 = sec(x) (Q.E.D.)

Alternative Approach:

• Starting from the identity sin(x) + cos(x) = 1
• Divide both sides by cos(x): sin(x)/cos(x) + cos(x)/cos(x) = 1/cos(x)
• Simplify: tan(x) + 1 = sec(x) (Q.E.D.)

Solution:

• Given function: y = x - 2x + 5
• First, verify that (1, 4) is on the curve: y = (1) - 2(1) + 5 = 1 - 2 + 5 = 4
• Find the derivative: dy/dx = 3x - 2
• Calculate the slope at x = 1: m = 3(1) - 2 = 3 - 2 = 1
• Using point-slope form: y - y = m(x - x)
• Substitute m = 1 and (x, y) = (1, 4): y - 4 = 1(x - 1)
• Simplify: y - 4 = x - 1
• Rearrange to get the final equation: y = x + 3

The equation of the tangent line to the curve at the point (1, 4) is y = x + 3.

Solution:

• First, find the antiderivative: (3x + 2x) dx = x + x + C
• Apply the Fundamental Theorem of Calculus from 0 to 2:
• [x + x] = (2 + 2) - (0 + 0)
• = (8 + 4) - (0 + 0)
• = 12

Therefore, (3x + 2x) dx = 12

Geometric Interpretation:

• This integral represents the area under the curve y = 3x + 2x between x = 0 and x = 2
• We can verify this result by applying the Power Rule for integration to each term
• 3x dx = 3x/3 = x
• 2x dx = 2x/2 = x
• Combining: (3x + 2x) dx = x + x + C

Solution:

• Convert the logarithmic equation to exponential form: 2 = x - 5x + 6
• Simplify: 8 = x - 5x + 6
• Rearrange: x - 5x - 2 = 0
• Using the quadratic formula: x = [5 (25 - 41(-2))]/2
• Calculate: x = [5 33]/2
• Therefore, x = (5 + 33)/2 or x = (5 - 33)/2

Checking the solutions:

We need to verify that these values lie in the domain of the original logarithmic function, which requires that x - 5x + 6 > 0.

• For x = (5 + 33)/2 5.37, x - 5x + 6 = 28.8 - 26.8 + 6 = 8 > 0
• For x = (5 - 33)/2 -0.37, x - 5x + 6 = 0.14 + 1.85 + 6 = 7.99 > 0

Both solutions are valid.

Solution:

First, factor the denominator: x - 4 = (x - 2)(x + 2)

Express the integrand as partial fractions:

• 1/((x - 2)(x + 2)) = A/(x - 2) + B/(x + 2)
• 1 = A(x + 2) + B(x - 2)
• 1 = (A + B)x + 2A - 2B

Equate coefficients:

• For x terms: A + B = 0
• For constant terms: 2A - 2B = 1

Solve the system of equations:

• From A + B = 0, we get B = -A
• Substitute into 2A - 2B = 1: 2A - 2(-A) = 1
• 2A + 2A = 1 4A = 1 A = 1/4
• Since B = -A, B = -1/4

Now we can integrate:

• [1/4(x - 2) - 1/4(x + 2)] dx = 1/4 [1/(x - 2) - 1/(x + 2)] dx
• = 1/4[ln|x - 2| - ln|x + 2|] + C
• = 1/4 ln|(x - 2)/(x + 2)| + C

Therefore, 1/(x-4) dx = 1/4 ln|(x - 2)/(x + 2)| + C