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Derivatives and Their Applications · Paper 1 — Free Maths AA SL Practice Questions

1FoundationSAQ-SDefinition of a derivative (rate of change)5 marksPaper 1~8 min
The graph of y=f(x)y = f(x) for −2≤x≤4-2 \leq x \leq 4 is shown below. The tangent to the curve at point PP, where x=1x = 1, is also drawn.
(a)
State the geometric meaning of f′(1)f'(1). [1 mark]
(b)
Find the value of f′(1)f'(1). [2 marks]
(c)
The normal to the curve at PP intersects the xx-axis at point QQ. Find the coordinates of QQ. [2 marks]
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Solutions

2FoundationSAQ-SRules of differentiation (power, product, quotient, chain rule)s5 marksPaper 1~8 min
A company models the profit P(x)P(x), in thousands of dollars, from selling xx hundred units of a product as P(x)=(2x+3)(x−1)2,x≥1.P(x) = (2x + 3)(x - 1)^2, \quad x \geq 1. -
(a)
Find P′(x)P'(x). Write your answer in the form P′(x)=2(x−1)(ax+b)P'(x) = 2(x-1)(ax + b), where aa and bb are integers. - [3 marks]
(b)
Find the value of xx for which P′(x)=0P'(x) = 0 and x>1x > 1, or state that no such value exists. Justify your answer. - [1 mark]
(c)
The company considers increasing production beyond the level found in (b). Using your results from (a) and (b), determine whether profit is increasing or decreasing for x>1x > 1, and interpret what this means for the company. [1 mark]
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Solutions

3MasterySAQ-SDefinition of a derivative (rate of change)8 marksPaper 1~12 min
A water tank has a small hole at its base. The volume of water, VV litres, remaining in the tank tt minutes after the hole is unplugged is given by V(t)=100(1−t20)2,0≤t≤20.V(t) = 100\left(1 - \frac{t}{20}\right)^2, \quad 0 \leq t \leq 20.
(a)
Show that the rate of change of volume is V′(t)=−10(1−t20)V'(t) = -10\left(1 - \dfrac{t}{20}\right). [3 marks]
(b)
Find the time at which the water is draining at a rate of 55 litres per minute. [2 marks]
(c)
The average rate of change of volume over the entire interval 0≤t≤200 \leq t \leq 20 is compared with the instaneous rate found in part (b). Determine the average rate of change of VV over [0,20][0, 20] and explain why it is consistent with the answer to part (b). [3 marks]
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Solutions

4ChallengeSAQ-LDefinition of a derivative (rate of change)8 marksPaper 1~12 min
A particle moves along a straight line such that its displacement, ss metres, from a fixed point OO at time tt seconds is given by s(t)=t3−6t2+9t+2,t≥0.s(t) = t^3 - 6t^2 + 9t + 2, \quad t \geq 0.
(a)
Using the definition of the derivative as a limit, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, show that the velocity of the particle is v(t)=3t2−12t+9v(t) = 3t^2 - 12t + 9. [4 marks]
(b)
Determine the values of tt at which the particle is instaneously at rest. [2 marks]
(c)
Determine the intervals within 0≤t≤40 \leq t \leq 4 during which the particle moves in the positive direction and the intervals during which it moves in the negative direction. [1 mark]
(d)
Hence determine the total distance travelled by the particle in the first 44 seconds. [1 mark]
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Solutions

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