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CalculusIB Maths: Analysis and Approaches HL: Topic test

20 questions, 54 marks

IB Maths: Analysis and Approaches HL

Calculus topic test

Total 54 marks

Name

Class

Date

  1. 1
    Let f(x)=5x4−8x3+2xf(x) = 5x^4 - 8x^3 + 2x, for x∈Rx \in \mathbb{R}.
    (a)
    Find f′(x)f'(x).
    [1 mark]
    • A20x3−24x2+220x^3-24x^2+2
    • B20x3−24x220x^3-24x^2
    • C5x3−8x2+25x^3-8x^2+2
    • D20x3−24x2+2x20x^3-24x^2+2x
    (b)
    Find f′′(x)f''(x).
    [1 mark]
    • A60x2−48x+260x^2-48x+2
    • B60x2−48x60x^2-48x
    • C60x3−48x260x^3-48x^2
    • D20x2−24x20x^2-24x
    (c)
    Find the equation of the tangent to the curve y=f(x)y=f(x) at the point where x=1x=1.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    Let g(x)=x2e4xg(x) = x^2e^{4x}, for x∈Rx \in \mathbb{R}.
    (a)
    Find g′(x)g'(x).
    [1 mark]
    • A2xe4x+x2e4x2xe^{4x}+x^2e^{4x}
    • B8xe4x8xe^{4x}
    • C2xe4x+4x2e4x2xe^{4x}+4x^2e^{4x}
    • D2xe4x2xe^{4x}
    (b)
    Find the gradient of the curve y=g(x)y=g(x) at the point where x=0x=0.
    [1 mark]
    • A11
    • B44
    • C22
    • D00
    (c)
    Show that gg is an increasing function for all x>0x>0.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A particle moves in a straight line so that its velocity, in m s−1^{-1}, at time tt seconds is v(t)=3t2−12t+9v(t)=3t^2-12t+9, for t≥0t\ge0. When t=0t=0 the particle's displacement from a fixed point OO is s=4s=4 m.
    (a)
    Find the times at which the particle is instantaneously at rest.
    [3 marks]
    (b)
    Find the displacement of the particle from OO when t=3t=3.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    An open-topped rectangular tank is to be made from thin sheet metal. The base is a square of side xx metres and the height is hh metres. The tank must hold 3232 m3^3 of water, and the total area of metal used, in m2^2, is A(x)=x2+128xA(x)=x^2+\dfrac{128}{x}, for x>0x>0.
    (a)
    Show that A′(x)=2x−128x2A'(x)=2x-\dfrac{128}{x^2}, and hence find the value of xx that minimises AA, justifying that it gives a minimum.
    [6 marks]
    (b)
    (i) Using x=4x=4, find the minimum area of metal required and the height hh of the tank. (ii) The tank is then filled with water at a constant rate of 22 m3^3 per minute. Let yy metres be the depth of water at time tt minutes, so that the volume of water is V=16yV=16y. Find the rate at which the depth of the water is increasing.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    Let f′(x)=6x2−2x3f'(x)=6x^2-\dfrac{2}{x^3}, for x≠0x\neq0, and f(1)=5f(1)=5.
    (a)
    Find the indefinite integral of f′(x)f'(x).
    [1 mark]
    • A2x3−1x2+C2x^3-\dfrac{1}{x^2}+C
    • B2x3+1x2+C2x^3+\dfrac{1}{x^2}+C
    • C2x3+23x2+C2x^3+\dfrac{2}{3x^2}+C
    • D12x+6x4+C12x+\dfrac{6}{x^4}+C
    (b)
    Given that f(1)=5f(1)=5, find f(x)f(x).
    [1 mark]
    • A2x3+1x2+52x^3+\dfrac{1}{x^2}+5
    • B2x3+1x2−22x^3+\dfrac{1}{x^2}-2
    • C2x3+1x2+22x^3+\dfrac{1}{x^2}+2
    • D2x3+1x2+32x^3+\dfrac{1}{x^2}+3
    (c)
    Find ∫12f′(x) dx\displaystyle\int_1^2 f'(x)\,dx, and interpret this value in terms of ff.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    Let f(x)=x2−5xf(x)=x^2-5x, for x∈Rx\in\mathbb{R}.
    (a)
    Using the definition of the derivative from first principles, find the value of lim⁡h→0f(3+h)−f(3)h\displaystyle\lim_{h\to0}\dfrac{f(3+h)-f(3)}{h}.
    [1 mark]
    • A66
    • B−1-1
    • C−2-2
    • D11
    (b)
    Find lim⁡x→5f(x)x−5\displaystyle\lim_{x\to5}\dfrac{f(x)}{x-5}.
    [1 mark]
    • A55
    • B00
    • CThe limit does not exist
    • D∞\infty
    (c)
    State, with a reason, whether ff is differentiable at x=5x=5.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    Let h(x)=arctan⁡(3x)h(x)=\arctan(3x), for x∈Rx\in\mathbb{R}.
    (a)
    Find h′(x)h'(x).
    [3 marks]
    (b)
    Hence, using integration by parts, find ∫arctan⁡(3x) dx\displaystyle\int \arctan(3x)\,dx.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    A biologist models the population NN (in thousands) of a bacterial colony by the differential equation dNdt=0.02N(100−N)\dfrac{dN}{dt}=0.02N(100-N), for t≥0t\ge0 hours, where N=10N=10 when t=0t=0.
    (a)
    Solve the differential equation to show that N=100e2t9+e2tN=\dfrac{100e^{2t}}{9+e^{2t}}.
    [6 marks]
    (b)
    Using the differential equation directly (without using the result of part (a)), find the Maclaurin series for N(t)N(t) up to and including the term in t2t^2.
    [6 marks]

    Total for question 8: 12 marks

End of questions