Free AB Calculus with
Solutions and Explanations - Sample 1

Detailed solutions and explanations to the AB calculus questions solutions in sample 1. .

  1. AB test sample 1, problem 1 =

    Solution

    As h ----> 0

    lim (e4 eh - e4) / h = (e4 e0 - e4) / 0

    = 0 / 0 , indeterminate form

    Another approach is needed. Let f(x) = ex. The given limit may be written as follows: as x ----> h

    lim (e4 eh - e4) / h = lim (e4 + h - e4) / h = limit [ f(4+h) - f(4) ] / h

    which is the definition of the first derivative of f(x) = ex at x = 4. Hence as x ---> h

    lim (e4 eh - e4) / h = e 4


  2. The graph of function g defined by

    AB test sample 1, problem 2

    will have vertical asymptotes at

    Solution

    Let us first simplify, if possible, the given rational function

    g(x) = (x3 + 2x2 - 3x) / (x2 + 2x - 3)

    = x (x2 + 2x - 3) / (x2 + 2x - 3) = x

    Function g has no vertical asymptotes



  3. Given that

    AB test sample 1, problem 3-1

    find

    AB test sample 1, problem 3-2

    Solution

    Using the theorem that states that the limit of a sum is equal to the sum of the limits. Hence as x ----> 0,

    lim (x + 4x2 + sin x) / 3x = lim (x / 3x) + lim (4x2 / 3x) + (1/3) sin x / x

    Simplify

    = lim (1/3) + lim (4x) + (1/3) sin x / x

    = 1/3 + 0 + (1/3)*1 = 2/3


  4. Function f is defined by

    AB test sample 1, problem 4.

    Find df(x) / dx.

    Solution

    Using the theorem that states that the derivative of a sum of functions is the sum of the derivatives, we can write

    d/dx [ 2x3sin(x) + (1/x)tan(x) + x sec(x) + 2 ]

    = d/dx [ 2x3sin(x) ] + d/dx [ (1/x)tan(x) ] + d/dx [ x sec(x) ] + d/dx [ 2 ]

    we now calculate the derivative of each term above

    d/dx [ 2x3sin(x) ] = 2 [ 3x2 sin(x) + x3 cos(x) ] = 6x2 sin(x) + 2x3 cos(x)

    d/dx [ (1/x)tan(x) ] = - (1/x2) tan(x) + (1/x) sec2(x))

    d/dx [ x sec(x) ] = sec(x) + x sin(x) sec2(x)

    d/dx [ 2 ] = 0

    Hence

    df/dx = 6x2 sin(x) + 2x3 cos(x) - (1/x2)tan(x) + (1/x sec2(x)) + sec(x) + x sin(x) sec2(x)



  5. Curve C is described by the equation 0.25 x2 + y2 = 9. Determine the y coordinates of the points on curve C whose tangent lines have slope equal to 1.

    Solution

    Let us calculate the first derivative. Differentiate both sides of the given equation

    0.25 (2x) + 2 y y ' = 0

    y ' = - 0.5 x / (2 y)

    We now solve the given equation 0.25 x2 + y2 = 9 for x

    x = + or - sqrt [ (9 - y2) / 0.25 ]

    Substitute x in y ' = - 0.5 x / (2 y) by + or - sqrt [ (9 - y2) / 0.25 ]

    y ' = - 0.5 (+ or - sqrt [ (9 - y2) / 0.25 ] / (2 y)

    The slope of the tangent is equal to 1. Hence

    - 0.5 (+ or - sqrt [ (9 - y2) / 0.25 ] )/ (2 y) = 1

    Solve the above for y. Two solutions

    y = 3 sqrt(5) / 5 , y = - 3 sqrt(5) / 5

    The graph of 0.25 x2 + y2 = 9 is shown below with the two tangent lines.

    AB test sample 1, problem 5, graph of tangents


  6. Find the solution to the differential equation dy/dx = cos(x) / y2 , where y(π/2) = 0.

    Solution

    The variable in the given differential equation may be separated as follows

    y2 dy = cos(x) dx

    Integrate both sides

    ∫ y2 dy = ∫ cos(x) dx

    (1/3) y3 = sin(x) + C , constant of integration

    We now use the condition y(π/2) = 0 to find the constant C

    (1/3) y3(π/2) = sin(π/2) + C

    0 = 1 + C

    C = - 1

    Substitute C by -1 in (1/3) y3 = sin(x) + C and solve for y

    (1/3) y3 = sin(x) - 1

    y3 = 3(sin(x) - 1)

    y = (3 sin(x) - 3)1/3


  7. AB test sample 1, problem 7

    Solution

    Let u = cos x and therefore du/dx = - sin x. We now substitute cos x by u and sin x by -du/dx in the given integral. Hence

    ∫cos4(x)sin(x) dx = ∫ u4 (-du/dx) dx

    = - ∫ u4du

    = (-1/5)u5 + C , C constant of integration

    = (-1/5)cos5(x) + C


  8. AB test sample 1, problem 8

    Solution

    Let u = 2x and therefore du/dx = 2 or dx = du / 2. Hence the given integral becomes

    d/dx ∫ 32xsin(t2 + 1) dt = 2 d/du ∫ 3usin(t2 + 1) dt

    using the fundamental theorem of calculus, we obtain

    = 2 sin(u2 + 1)

    Substitute u by 2x

    = 2 sin(4x2 + 1)



  9. AB test sample 1, problem 9

    Solution

    We first to analyze the signs of the expressions 4 - x and 2 - 2x between the limits of integration 0 and 10. 4 - x changes sign at x = 4 and 2 - 2x changes sign at x = 1.

    for x between 0 and 4: 4 - x is positive and hence |4 - x| = 4 - x

    for x between 4 and 10: 4 - x is negative and hence |4 - x| = -(4 - x)

    for x between 0 and 1: 2 - 2x is positive and hence |2 - 2x| = 2 - 2x

    for x between 1 and 10: 2 - 2x is negative and hence |2 - 2x| = -(2 - 2x)

    We now rewrite the given integral as a sum of two integrals as follws.

    010 (|4 - x|+|2 - 2x|) dx =

    010 (|4 - x|) dx + ∫ 010 (|2 - 2x|) dx

    We now calculate each of the individual integrals above as follows.

    010 (|4 - x|) dx = ∫ 04 (4 - x) dx + ∫ 410 -(4 - x) dx = 8 + 18 = 26

    and

    010 (|2 - 2x|) dx = ∫ 01 (2 - 2x) dx + ∫ 110 -(2 - 2x) dx = 1 + 81 = 82

    We now have

    010 (|4 - x|) dx + ∫ 010 (|2 - 2x|) dx = 26 + 82 = 108


  10. Evaluate the integral

    AB test sample 1, problem 10

    Solution

    Let u = 5 + x3/4 and therefore du/dx = (3/4) 1/x1/4 and substitute in the given integral

    ∫ (5 + x3/4)9 / (x1/4) dx = ∫ [ ( u9 ) / (x1/4) ] (4/3) x1/4 du

    = (4/3) ∫ u9 du

    = (4/3) (1/10) u10

    = (2/15) (5 + x3/4)10


  11. Given that function h is defined by

    AB test sample 1, problem 11

    find h'(x).

    Solution

    Let u = arctan(x3 + 1) + 2x. Hence function h can be written as

    h(x) = u4 h '(x) = 4 u3 u'

    We now let v = arctan(x3 + 1) and calculate u '

    u ' = (v ')( 1 / (1 + v2) )

    = (3x2) / (1 + (x3 + 1)2)

    = (3x2) / (x6 + 2x3 + 2)

    Hence

    h '(x) = 4 (arctan(x3 + 1) + 2x)3 (3x2) / (x6 + 2x3 + 2)


  12. The graph of function h is shown below. How many zeros does the first derivative h' of h have?

    AB test sample 1, problem 12

    Solution

    Whenever the graph of h has a local maximum or local minimum h '(x) is equal to 0. The given graph has 3 local minima and 2 local maxima and therefore h ' has 5 zeros.


  13. The graph of a polynomial f is shown below. If f' is the first derivative of f, then the remainder of the division of f'(x) by x - b is more likely to be equal to

    AB test sample 1, problem 13

    Solution

    The graph of f has a local maximum at b and therefore f'(b) = 0. Since f is a polynomial then f ' is also a polynomial function such that f '(b) = 0 and according to the remainder theorem the division of f '(x) by x - b is equal to o.


  14. The set of all points (ln(t - 2) , 3t), where t is a real number greater than 2, is the graph of

    Solution

    The given parametric equations may written as

    x(t) = ln(t - 2) and y(t) = 3t

    Solve y(t) = 3t for t

    t = y / 3

    Substitute t by y / 3 in x(t) = ln(t - 2)

    x = ln(y / 3 - 2)

    Solve for y

    y/3 - 2 = ex y = 3 ( ex + 2 )


  15. Let P(x) = 2 x3 + K x + 1. Find K if the remainder of the division of P(x) by x - 2 is equal to 10.

    Solution

    The Remainder theorem states that the division of P(x) by x - 2 is equal to P(2). Hence

    P(2) = 2 (2)3 + K (2) + 1 = 10

    Solve for K

    K = - 7/2


  16. Function f is defined by

    AB test sample 1, problem 16
    .

    where C is a constant. What must the value of C be equal to for function f to be continuous at x = 0?

    Solution

    For function f to be continuous at x = 0, lim f(x) as x approaches must be equal to f(0). We first find the limit of f(x) as x approaches 0. As x ---> 0,

    lim [ sqrt(4x + 4) - sqrt(2x + 4) ] / 2x = 0 / 0 , indeterminate

    Another approach is needed. Multiply numeartor and denominator by sqrt(4x + 4) - sqrt(2x + 4), simplify and find the limit. As x ---> 0,

    lim [ sqrt(4x + 4) - sqrt(2x + 4) ] / 2x

    = lim [ sqrt(4x + 4) - sqrt(2x + 4) ] [ sqrt(4x + 4) + sqrt(2x + 4) ]/ [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]

    = lim [4x + 4 - 2x - 4] / [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]

    = lim 2x / [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]

    = lim 1 / [ sqrt(4x + 4) + sqrt(2x + 4) ]

    = 1 / [ 2 + 2 ] = 1/4

    In order for f to be continuous, we need to have

    C = 1/4


  17. f and g are functions such that f '(x) = g(x) and g '(x) = f(x). The second derivative of (f. g)(x) is equal to

    A) f "(x) g "(x)

    B) g '(x) g(x) + f(x) f '(x)

    C) 4 g(x) f(x)

    D) 2 g(x) f(x)

    E) g(x) f(x)

    Solution

    First derivative of (f. g)(x)

    (f. g) ' = f ' g + f g '

    Second derivative of (f. g)(x)

    (f. g) " = (f ' g + f g ') '

    = f "g + f 'g' + f 'g' + f g"      (I)

    Note that since f '(x) = g(x) and g '(x) = f(x), we have

    f " = g' and g " = f'

    Substitute f " and g " in (I) above to obtain

    (f. g) " = g 'g + f 'g' + f 'g' + f f'

    We now substitute g ' by f and f ' by g to obtain

    (f. g) " = fg + gf+ fg + f g = 4 fg


  18. The average rate of change of the function f defined by f(x) = sin(x) + x on the closed interval [0 , pi] is equal to

    Solution

    The average rate of change of a function from a to b is defined by

    (f(b) - f(a)) / (b - a)

    Apply the above definition to the question above

    (f(pi) - f(0)) / (pi - 0) = [ (sin(pi) + pi) - (sin(0)+ 0) ] / (pi - 0) = 1


  19. The figure shows the graphs of y = sin(x) over half a period and the line y = 1/2. Find area of the shaded region.

    AB test sample 1, problem 19
    .

    Solution

    The x coordinates of the points of intersection the line y = 1/2 and the graph of f are found by solving

    sin(x) = 1/2 , for 0 ≤ x ≤ pi

    solutions: x = pi / 6 and x = 5pi / 6

    We first split the area to be calculated into 3 parts as shown below.

    AB test sample 1, problem 19
    .

    Calculate the area A adding the three parts (NOTING that the area in the middle is that of a rectangle):

    A = ∫0pi/3 sin(x) dx+ 1/2(5Pi/6 - Pi/6) + ∫5pi/6pi sin(x) dx

    = [ - cos(x) ]0pi/3 + pi / 3 + [- cos(x)]5pi/6pi

    = 2 + pi / 3 - sqrt(3)


  20. Functions f, g and h are defined as follow: g(x) = f(x2), f(x) = h(x3 + 1) and h'(x) = 2x + 1. g'(x) =

    Solution

    Let us express g(x) in terms of h

    sin(x) = 1/2 , for 0 ≤ x ≤ pi

    g(x) = f(x2) = h( (x2)3 + 1 )

    = h(x6 + 1)

    Let u = x6 + 1. Hence

    g(x) = h(u) , with u = x6 + 1

    Use chain rule to write

    g '(x) = (du/dx) (dh/du)

    If h'(x) = 2x + 1, then

    dh/du = h '(u) = 2u + 1

    Hence

    g '(x) = 6x5 [ 2u + 1 ]

    Substitute u by x6 + 1 in g '(x)

    g '(x) = 6x5 [ 2(x6 + 1) + 1 ] =

    = 12 x11 + 18 x5

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