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5.1 Limits and the derivativeIB Maths: Applications and Interpretation HL: Subtopic test

10 questions, 27 marks

IB Maths: Applications and Interpretation HL

5.1 Limits and the derivative

Total 27 marks

Name

Class

Date

  1. 1
    The function ff is defined by f(x)=x+9−3xf(x)=\frac{\sqrt{x+9}-3}{x} for x≠0x\neq0. A table of values gives f(−0.1)=0.16713f(-0.1)=0.16713, f(−0.01)=0.16671f(-0.01)=0.16671, f(0.01)=0.16662f(0.01)=0.16662 and f(0.1)=0.16621f(0.1)=0.16621 (5 s.f.). Use your GDC where needed.
    (a)
    Use the table to estimate lim⁡x→0f(x)\lim_{x\to0}f(x).
    [1 mark]
    • A00
    • B33
    • CThe limit does not exist
    • D0.1670.167
    (b)
    Which statement about ff is correct?
    [1 mark]
    • AThe limit is f(0)f(0)
    • Bf(0)=0.167f(0)=0.167
    • CThe limit exists although f(0)f(0) is not defined
    • DThe values approach different numbers from the left and right
    (c)
    Use your GDC to evaluate f(0.0001)f(0.0001) and f(−0.0001)f(-0.0001), and hence write down the limit as x→0x\to0 to 33 significant figures.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A ball is thrown upwards. Its height above the ground is h(t)=20t−5t2h(t)=20t-5t^2 metres, tt seconds after it is thrown. Use your GDC or calculator.
    (a)
    Find the average rate of change of hh between t=1t=1 and t=1.1t=1.1.
    [1 mark]
    • A0.950.95 m s−1^{-1}
    • B9.59.5 m s−1^{-1}
    • C15.9515.95 m s−1^{-1}
    • D0.8640.864 m s−1^{-1}
    (b)
    The time interval starting at t=1t=1 is made smaller and smaller. The average rate of change over the interval approaches which quantity?
    [1 mark]
    • AThe gradient of the tangent at t=1t=1, written h′(1)h'(1)
    • BThe height h(1)h(1)
    • CThe average height over the interval
    • DThe area under the curve
    (c)
    Find the average rate of change of hh between t=1t=1 and t=1.001t=1.001 and hence estimate h′(1)h'(1), stating its units.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The volume of water in a tank is VV litres, tt minutes after a tap is opened. The model gives V(5)=60V(5)=60, V(5.1)=61.18V(5.1)=61.18 and dVdt=12\frac{dV}{dt}=12 when t=5t=5.
    (a)
    Interpret dVdt=12\frac{dV}{dt}=12 when t=5t=5, including units, and use it to estimate the volume of water when t=5.5t=5.5.
    [3 marks]
    (b)
    (i) A student writes 'dVdt=12\frac{dV}{dt}=12 means there are 1212 litres in the tank when t=5t=5'. Explain the error.
    (ii) Find the average rate of change of
    VV between t=5t=5 and t=5.1t=5.1, and explain why it is not exactly 1212.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    The distance, ss metres, travelled by a car tt seconds after it starts to accelerate is modelled by s(t)=0.2t3s(t)=0.2t^3 for 0≤t≤80\leq t\leq8. Use your GDC.
    (a)
    (i) Find the average velocity of the car between t=6t=6 and t=8t=8.
    (ii) Find the average velocity between
    t=6t=6 and t=6.1t=6.1, and between t=6t=6 and t=6.01t=6.01.
    (iii) Hence estimate the velocity of the car at
    t=6t=6, and explain how your values from (ii) lead to this estimate.
    [6 marks]
    (b)
    It is given that dsdt=21.6\frac{ds}{dt}=21.6 when t=6t=6 and dsdt=38.4\frac{ds}{dt}=38.4 when t=8t=8.
    (i) State what
    dsdt\frac{ds}{dt} represents, including its units.
    (ii) Comment on what the two given values show about the motion of the car.

    (iii) Use
    dsdt\frac{ds}{dt} at t=6t=6 to estimate s(6.5)s(6.5). State whether this is an underestimate or an overestimate, with a reason.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).