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CalculusIB Maths: Analysis and Approaches SL: Topic test

20 questions, 54 marks

IB Maths: Analysis and Approaches SL

Calculus topic test

Total 54 marks

Name

Class

Date

  1. 1
    Let f(x)=x3−6x2+9x+2f(x) = x^{3} - 6x^{2} + 9x + 2, for x∈Rx \in \mathbb{R}.
    (a)
    Find f′(x)f'(x).
    [1 mark]
    • A3x2−12x+93x^{2}-12x+9
    • B3x2−6x+93x^{2}-6x+9
    • Cx2−12x+9x^{2}-12x+9
    • D3x2−12x3x^{2}-12x
    (b)
    Find the interval on which ff is decreasing.
    [1 mark]
    • Ax<1x<1 or x>3x>3
    • B1<x<31<x<3
    • Cx<1x<1
    • Dx>3x>3
    (c)
    Find the equation of the tangent to the curve y=f(x)y = f(x) at the point where x=0x = 0.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A function hh is defined by h(x)=x2+3xh(x) = x^{2} + 3x, for x∈Rx \in \mathbb{R}. The table shows values of h(2+d)−h(2)d\dfrac{h(2+d)-h(2)}{d} for small positive dd: d=0.1d=0.1 gives 7.17.1; d=0.01d=0.01 gives 7.017.01; d=0.001d=0.001 gives 7.0017.001.
    (a)
    Using the table, estimate h′(2)h'(2).
    [1 mark]
    • A7.17.1
    • B1414
    • C77
    • D00
    (b)
    Find the equation of the normal to the curve y=h(x)y=h(x) at the point where x=2x=2.
    [1 mark]
    • Ay−10=7(x−2)y-10=7(x-2)
    • By−10=17(x−2)y-10=\frac{1}{7}(x-2)
    • Cy−10=−17(x+2)y-10=-\frac{1}{7}(x+2)
    • Dy−10=−17(x−2)y-10=-\frac{1}{7}(x-2)
    (c)
    Find h′′(x)h''(x), and state whether hh is concave-up or concave-down for all x∈Rx \in \mathbb{R}.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    Let g(x)=x2exg(x) = x^{2}e^{x}, for x∈Rx \in \mathbb{R}.
    (a)
    Find g′(x)g'(x).
    [3 marks]
    (b)
    Hence find the coordinates of the stationary points of the curve y=g(x)y = g(x), and use the second derivative to determine their nature.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A ball is thrown vertically upwards from the top of a building. Its height above the ground, hh metres, tt seconds after it is thrown, is modelled by h(t)=−5t2+20t+25h(t) = -5t^{2} + 20t + 25, for 0≤t≤T0 \le t \le T, where TT is the time at which the ball hits the ground.
    (a)
    Find the maximum height reached by the ball, and the time at which it occurs. Justify that this is a maximum.
    [6 marks]
    (b)
    Find the time TT at which the ball hits the ground, and find the speed of the ball at this instant.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    The rate of growth of a plant's height is modelled by dhdt=3t2−2t\dfrac{dh}{dt} = 3t^{2} - 2t, where hh is the height in cm and tt is the time in weeks after planting, for t≥0t \ge 0. When t=1t=1, the height is h=6h=6 cm.
    (a)
    Find h(t)h(t).
    [1 mark]
    • At3−t2+6t^{3}-t^{2}+6
    • B3t3−t2+63t^{3}-t^{2}+6
    • Ct3+t2+6t^{3}+t^{2}+6
    • Dt3−t2t^{3}-t^{2}
    (b)
    Find the height of the plant after 3 weeks.
    [1 mark]
    • A1818 cm
    • B2424 cm
    • C4242 cm
    • D7878 cm
    (c)
    Find ∫13dhdt dt\displaystyle\int_{1}^{3} \dfrac{dh}{dt}\,dt, and interpret this value in context.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    Let f′(x)=4x(x2+3)3f'(x) = 4x(x^{2}+3)^{3}, for x∈Rx \in \mathbb{R}.
    (a)
    Find f(x)f(x), given that f(0)=5f(0) = 5.
    [1 mark]
    • A(x2+3)4−76(x^{2}+3)^{4}-76
    • B(x2+3)42+5\dfrac{(x^{2}+3)^{4}}{2}+5
    • C(x2+3)4−712\dfrac{(x^{2}+3)^{4}-71}{2}
    • D2(x2+3)4−1572(x^{2}+3)^{4}-157
    (b)
    Find ∫sin⁡(3x) dx\displaystyle\int \sin(3x) \, dx.
    [1 mark]
    • A13cos⁡(3x)+C\frac{1}{3}\cos(3x)+C
    • B−3cos⁡(3x)+C-3\cos(3x)+C
    • C−cos⁡(3x)+C-\cos(3x)+C
    • D−13cos⁡(3x)+C-\frac{1}{3}\cos(3x)+C
    (c)
    Find ∫6x22x3+1 dx\displaystyle\int 6x^{2}\sqrt{2x^{3}+1}\,dx.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    The curve CC has equation y=4−x2y = 4 - x^{2}, for −3≤x≤3-3 \le x \le 3.
    (a)
    Find the area of the region enclosed by CC and the xx-axis between x=−2x=-2 and x=2x=2.
    [3 marks]
    (b)
    Find the total area of the regions enclosed by CC, the xx-axis, and the lines x=−3x=-3 and x=3x=3.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    A cylindrical can with a closed top and base is designed to hold 500π500\pi cm3^{3} of liquid. The radius of the base is rr cm and the height is hh cm.
    (a)
    Show that the surface area of the can is given by A(r)=2πr2+1000πrA(r) = 2\pi r^{2} + \dfrac{1000\pi}{r}, and find the value of rr that minimises AA.
    [6 marks]
    (b)
    Use the second derivative to show that this value of rr gives a minimum surface area, and find the minimum surface area correct to 3 significant figures.
    [6 marks]

    Total for question 8: 12 marks

End of questions